case study questions ray optics

Case Study Questions Class 12 Physics Ray Optics and Optical Instruments

Case study questions class 12 physics chapter 9 ray optics and optical instruments.

CBSE Class 12 Case Study Questions Physics Ray Optics and Optical Instruments. Term 2 Important Case Study Questions for Class 12 Board Exam Students. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Ray Optics and Optical Instruments.

CBSE Case Study Questions Class 12 Physics Ray Optics and Optical Instruments

As we know that, when light ray travels from one medium to another changes its direction of path due to the change in optical density of the medium. When light ray travels from optically denser medium to the optically rarer medium then some of the light get reflected back in the same medium and remaining light get refracted in the second medium and such a phenomenon of light is called as internal reflector light. We know that, when a ray of light travels from denser medium to rarer medium then it get bended away from the normal. If we increase the angle of incidence slowly then angle of refraction also get increased and at one stage the angle of refraction is 90° for some angle of incidence. And further if we increase the angle of incidence then there will be no refraction of light and the ray will be totally internally get reflected. Such phenomenon of reflection of light is called as total internal reflection of light. And the angle of incidence in denser medium for which the angle of refraction in rarer medium is 90°, that angle of incidence is called as critical angle. So we define total internal reflection as, if the angle of incidence exceeds the critical angle then the total light get internally reflected. Mirage is the best real life example of total internal reflection of light. The highly brilliance of diamonds is only due to the total internal reflection of light through them which is get total internally reflected many times and causes brilliance of diamonds. Prisms are also made by using total internal reflection of light. We all know that, optics is the vast branch of physics which has vast application in our daily life such as optical fibre. Optical fibre are used for long distance transmission of audio and video signal also. And they are mainly constructed on the basis of total internal reflection of light to reduce the decrease in amplitude of the sending signal.

Q 1.) For critical angle the value of corresponding angle of refraction is____

Q 3.) In case of total internal reflection, for angle of incidence greater than critical angle the refraction of light is not possible because____

Answer key:

Q 1.) d) 90°

Lower is the refractive index of the medium then lower it’s optical density also. Such medium with lower refractive index is called as optically rarer medium. While speed of light is higher in rarer medium than the speed of light in denser medium.

We all know that, the different colours of light are due to the different wavelength or frequency of light. The visible spectrum shows different colours from which red light is having longer wavelength and violet light is having shortest wavelength. As the wavelength of the light increases the corresponding refractive index of the medium decreases. Because that glass is dispersive medium while vacuum is not. As the speed of light depends on the refractive index and there by on the wavelength also. So it is responsible for the desperation of light in that medium. We see in rainy season, after raining there is a rainbow in the sky which is only due to the dispersion of sunlight through water droplets present in the atmosphere. The observer can see the rainbow when he is seeing opposite to the direction where sun is present. The blue colour of the sky is due to the scattering of light by the atmospheric dust particles. Light having shorter wavelength scatters more than the light having longer wavelength. Hence, violet light having shorter wavelength get scattered more than blue light but the light reaches to our eyes is blue mostly. Because our eyes cannot be sensitive to violet light. The sun looks reddish at sunset and sunrise only because of the scattering of light. The red light having longer wavelength get scattered less and directly reach to our eyes while other light having greater wavelength get scattered more and doesn’t reach to our eyes that’s why we see the sun as reddish in colour at sunset and sunrise near the horizon.

a) frequency

b) wavelength

Q 2.) b) wavelength

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12th Class Physics Ray Optics Question Bank

Done case based (mcqs) - ray optics and optical instruments total questions - 25.

Directions (Q. No. 1 to 5):

Read the following text and answer the following questions on the basis of the same:

Sparking Brilliance of Diamond:

The total internal reflection of the light is used in polishing diamonds to create a sparking brilliance.

By polishing the diamond with specific cuts, it is adjusted the most of the light rays approaching the surface are incident with an angle of incidence more than critical angle. Hence, they suffer multiple reflections and ultimately come out of diamond from the top. This gives the diamond a sparking brilliance.

A) its critical angle with reference to air is too large. done clear

B) its critical angle with reference to air is too small. done clear

C) the diamond is transparent. done clear

D) rays always enter at angle greater than critical angle. done clear

question_answer 2) The critical angle for a diamond is \[{{24.4}^{o}}\]. Then its refractive index is:

A) 2.42 done clear

B) 0.413 done clear

C) 1 done clear

D) 1.413 done clear

question_answer 3) The basic reason for the extraordinary sparkle of suitably cut diamond is that:

A) it has low refractive index. done clear

B) it has high transparency. done clear

C) it has high refractive index. done clear

D) it is very hard. done clear

question_answer 4) A diamond is immersed in a liquid with a refractive index greater than water. Then the critical angle for total internal reflection will:

A) depend on the nature of the liquid. done clear

B) decrease. done clear

C) remains the same. done clear

D) increase. done clear

The following diagram shows same diamond cut in two different shapes.

The brilliance of diamond in the second diamond will be:

A) less than the first. done clear

B) greater than first. done clear

C) same as first. done clear

D) will depend on the intensity of light. done clear

Directions (Q. No. 6 to 10):

Read the following text and answer the following questions on the basis of the same:

Photometry:

The measurement of light as perceived by human eye is called photometry. Photometry is measurement of a physiological phenomenon, being the stimulus of light as received by the human eye, transmitted by the optic nerves and analysed by the brain. The main physical quantities in photometry are (i) the luminous intensity of the source, (ii) the luminous flux or flow of light from the source and (iii) illuminance of the surface. The SI unit of luminous intensity (I) is candela (cd).

The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency \[540\text{ }\times \text{ }{{10}^{12}}\] Hz and that has a radiant intensity in that direction of 1/683 watt per steradian, If a light source emits one candela of luminous intensity into a solid angle of one steradian, the total luminous flux emitted into that solid angle is one lumen (1m). A standard 100 watt incandescent light bulb emits approximately 1700 lumens.

A) Measurement of light as perceived by human eye done clear

B) Measurement of number of photons emerging from a light source done clear

C) Measurement of electrons emitted by photosensitive surface done clear

D) Measurement of photosensitivity done clear

question_answer 7) Light received by human eye is analysed by:

A) retina done clear

B) brain done clear

C) optic nerve done clear

D) nervous system done clear

question_answer 8) The SI unit of luminous intensity is:

A) Dioptre done clear

B) Steradian done clear

C) Candela done clear

D) Lumen done clear

question_answer 9) Unit of luminous flux is:

A) Candela done clear

C) Nit done clear

question_answer 10) A standard 100 watt incandescent light bulb emits approximately:

A) 1700 Lumen done clear

B) 700 Lumen done clear

C) 1200 Lumen done clear

D) 1000 Lumen done clear

Directions (Q. No. 11 to 15):

Read the following text and answer the following questions on the basis of the same:

Optical Fibre:

Optical fibre works on the principle of total internal reflection. Light rays can be used to transmit a huge amount of data, but there is a problem here - the light rays travel in straight lines. So unless we have a long straight wire without any bends at all, harnessing this advantage will be very tedious.

Instead, the optical cables are designed such that they bend all the light rays' inwards (using TIR).

Light rays travel continuously, bouncing off the optical fibre walls and transmitting end to end data. It is usually made of plastic or glass.

Modes of transmission: Single-mode fibre is used for long-distance transmission, while multi- mode fiber is used for shorter distances. The outer cladding of these fibres needs better protection than metal wires. Although light signals do degrade over progressing distances due to absorption and scattering. Then, optical Regenerator system is necessary to boost the signal.

Types of Optical Fibres: The types of optical fibers depend on the refractive index, materials used, and mode of propagation of light. The classification based on the refractive index is as follows:

Step Index Fibres: It consists of a core surrounded by the cladding, which has a single uniform index of refraction.

Graded Index Fibres: The refractive index of the optical fibre decreases as the radial distance from the fibre axis increases.

A) scattering of light. done clear

B) diffraction of light. done clear

C) total internal reflection of light. done clear

D) dispersion of light. done clear

question_answer 12) For long-distance transmission:

A) single mode fibre is used. done clear

B) multi-mode fibre is used. done clear

C) both single mode and multi-mode are used. done clear

D) any one of single mode or multi-mode may be used. done clear

question_answer 13) Optical fibre is made of:

A) copper done clear

B) semiconductor done clear

C) plastic or glass done clear

D) superconductors done clear

question_answer 14) In graded index optical fibre:

A) the refractive index of the optical fibre increases as the radial distance from the fibre axis increases. done clear

B) the refractive index of the optical fibre decreases as the radial distance from the fibre axis increases. done clear

C) the refractive index of the optical fibre remains same throughout. done clear

D) inner side of cladding is mirrored to ensure reflection. done clear

question_answer 15) Light signal through optical fibre may degrade due to:

A) refraction. done clear

B) refraction and reflection. done clear

C) diffraction and scattering. done clear

D) scattering and absorption. done clear

Directions (Q. No. 16 to 20):

Read the following text and answer the following questions on the basis of the same:

Negative Refractive Index:

One of the most fundamental phenomena in optics is refraction. When a beam of light crosses the interface between two different materials, its path is altered depending on the difference in the refractive indices of the materials. The greater the difference, the greater the refraction of the beam.

For all known naturally occurring materials the refractive index assumes only positive values. But does this have to be the case?

In 1967, Soviet physicist Victor Veselago hypothesized that a material with a negative refractive index could exist without violating any of the laws of physics.

Veselago predicted that this remarkable material would exhibit a wide variety of new optical phenomena. However, until recently no one had found such a material and Veselago's ideas had remained untested. Recently, meta-material samples are being tested for negative refractive index. But the experiments show significant losses and this could be an intrinsic property of negative- index materials.

Snell's law is satisfied for the materials having a negative refractive index, but the direction of the refracted light ray is 'mirror-imaged' about the normal to the surface.

There will be an interesting difference in image formation if a vessel is filled with "negative water" having refractive index - 1.33 instead of regular having refractive index 1.33.

Say, there is a fish in a vessel filled with negative water. The position of the fish is such that the observer cannot see it due to normal refraction since the refracted ray does not reach to his eye.

But due to negative refraction, he will be able to see it since the refracted ray now reaches his eye.

A) Joseph Von Fraunhofer done clear

B) Augustin-Jean Fresnel done clear

C) Thomas Moore done clear

D) Victor Veselago done clear

question_answer 17) Is Snell's law applicable for negative refraction?

A) Yes done clear

B) No done clear

C) Unpredictable done clear

D) Yes, only for normal incidence done clear

question_answer 18) A ray in incident on normal glass and "negative glass" at an angle \[{{60}^{o}}\]. If the magnitude of angle of refraction in normal glass is \[{{45}^{o}}\] then, what will be the magnitude of angle of refraction in the "negative glass"?

A) Less than \[{{45}^{o}}\] done clear

B) More than \[{{45}^{o}}\] done clear

C) \[{{45}^{o}}\] done clear

D) Unpredictable done clear

question_answer 19) When the angle of incidence will be equal to angle of refraction for material having negative refraction index?

A) When angle of incidence = \[{{90}^{o}}\] done clear

B) When angle of incidence = \[{{0}^{o}}\] done clear

C) It will vary from material to material done clear

D) It is never possible done clear

question_answer 20) Which of the following is the intrinsic property of negative-index materials?

A) Significant gain of light energy due to refraction done clear

B) No loss of light energy due to refraction done clear

C) Significant loss of light energy due to refraction done clear

D) Loss of energy due to refraction in intermittent done clear

Directions (Q. No. 21 to 25):

Read the following text and answer the following questions on the basis of the same:

First Surface Mirror:

Normally we use back surface mirrors. These are considered low precision mirrors because they actually have two reflecting surfaces. The first reflecting surface is the initial surfaces on the pane of glass where a small percentage of light is reflected off the surface. The second reflecting surface is the aluminium coating where a high percentage of light is reflected off the surface.

This dual reflection effect of a low precision mirror causes a loss of contrast and image distortion that is undesirable in high precision applications like rear projection systems, scanners and reflecting telescopes. In these cases good image quality is highly preferred, and this is where a front surface mirror is desired for clarity and single image reflection. First surface mirrors are quite common in professional optics. However, compared with back surface mirrors, they have the important disadvantage of being substantially more sensitive.

The front surface may be touched, and a metal coating on the front surface is substantially more sensitive than a bare glass surface. For example, fingerprints can easily cause oxidation of the metal.. Also, moisture or may cause oxidation of the mirror coating.

A) high. done clear

B) low. done clear

C) depends on intensity of light. done clear

D) similar to first surface mirror. done clear

question_answer 22) Light incident on back surface mirror suffers:

A) two reflections. done clear

B) one reflection. done clear

C) two reflections and two refractions. done clear

D) one refraction and one reflection. done clear

question_answer 23) In professional optics:

A) first surface mirrors are used. done clear

B) back surface mirrors are used: done clear

C) both type of mirrors are used. done clear

D) mirrors are not used. done clear

question_answer 24) Image formed of front coated mirror:

A) suffers astigmatism done clear

B) is brilliant done clear

C) has low contrast done clear

D) is a dual image done clear

question_answer 25) The front surface coating:

A) is susceptible to moisture. done clear

B) is not so sensitive. done clear

C) is worse than back surface coating. done clear

D) cannot prevent dual reflection. done clear

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Ray Optics and Optical Instruments Important Questions | Class 12 | Physics | Chapter 9 2024

Last updated on July 14th, 2024 at 04:53 pm

Ray Optics Questions and Answers

Below are some of the very important NCERT Class 12 Physics Chapter 9 Ray Optics Questions and Answers. These Class 12 Ray Optics Questions and Answers have been prepared by expert teachers and subject experts based on the latest syllabus and pattern of term 2 questions with answers to help students understand the concept.

These Questions for Class 12 Physics Ray Optics Questions and Answers are very important for the latest CBSE term 2 pattern. These Class 12 Q and A are very important for students who want to score high in CBSE Board.

We have put together these NCERT Questions of Class 12 Physics Chapter 8 Electromagnetic Waves Questions and Answers for practice on a regular basis to score high in exams. Refer to these Questions with Answers here along with a detailed explanation.

Short Answer (SA) Type Questions

1. When monochromatic light travels from a rarer to denser medium, explain the following, giving reasons.

(i) Is the frequency of reflected and refracted ray same as the frequency of incident light? (ii) Does the decrease in speed imply a reduction in the energy carried by light wave?

2. Mention any two situation in which Snell’s law of refraction fails.

3. A ray of light is incident at an angle of 45 o on one face of rectangular glass slab of thickness 10 cm and refractive index 1.5. Calculate the lateral shift produced.

4. Why does the sun rising in the sky appear oval in shape?

5. Show analytically from the lens equation that when the object is at the principal focus, the image is formed at infinity.

6. A student measures the focal length of a convex lens by putting an object pin at a distance ‘u’ from the lens and measuring the distance ‘v’ of the image pin. What will be the graph drawn between ‘u’ and ‘v’.

7. A magician during a show makes a glass lens n = 1.47 disappear in a trough of liquid. What is the refractive index of the liquid? Could the liquid be water?

8. A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 upto the same height, by what distance will the microscope have to be moved to focus on the needle again?

9. What should be the position of the object relative to the biconcave lens, so that this lens behaves like a magnifying glass?

10. How does the magnification of a magnifying glass differ from its magnifying power?

11. Calculate the radius of curvature of an equi-concave lens of refractive index 1.5, when it is kept in a medium of refractive index 1.4, to have a power of -5D.

12. An equi-concave lens of focal length ‘f’ is cut into two equal halves in thickness. What is the focal length of each half?

13. What is the focal length of convex lens of focal length 30 cm in contact with a concave lens of focal length 20cm? Is this system a converging or diverging lens? Ignore thickness of the lenses.

14. The figure shows a ray of light falling normally on the face AB of an equilateral glass prism having refractive index 3/2, placed in water of refractive index 4/3. Will this ray suffer total internal reflection on striking the face AC. Justify your answer.

Ray Optics and Optical Instruments Important Questions | Class 12 | Physics | Chapter 9 2024

15. A ray PQ incident normally on the refractive face BA is refracted in the prism BAC made of material of refractive index 1.5. Complete the path of rays through the prism.

From which face will the ray emerge? Justify your answer.

16. An equilateral glass prism has a refractive index 1.6 in air. Calculate the angle of minimum deviation of the prism, when kept in a medium of refractive index (4√2)/5.

17. Is it possible to increase the range of telescope by increasing the diameter of the objective lens?

18. Explain two advantages of a reflecting telescope over a refracting telescope.

19. Small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

20. The objective of an astronomical telescope has a diameter of 150 mm and a focal length of 4 m. The eyepiece has a focal length of 25 mm. Calculate the magnifying power of telescope (λ = 6000 A o for yellow colour).

21. Define power of a lens. Write its units. Deduce the relation for two thin lenses kept in contact coaxially.

22. A symmetric bi-concave lens of radius of curvature R and made of glass of refractive index 1.5, is placed on a layer of liquid placed on top of a plane mirror as shown in the figure below.

An optical needle with its top tip on the principal axis of the lens is moved along the axis until it’s real and inverted image coincides with the needle itself. The distance of the needle from the lens is measured to be ‘x’.

On removing the liquid layer and repeating the experiment the distance is found to be ‘y’. Obtain the expression for the refractive index of the liquid in terms of ‘x’ and ‘y’.

23. State the condition of total internal reflection. Refractive indices of the given prism material for red, blue and green colours are 1.39, 1.48 and 1.42 respectively. Trace the path of rays through the prism.

24. A ray of light incident on the face AB of an isosceles triangle prism makes an angle of incidence ‘i’ and deviates by angle as shown in the figure. Show that in the position of minimum deviation ∠β=∠α. Also find out the condition, when the refracted ray QR suffers total internal reflection.

25. (i) A ray of light incident of face AB of an equilateral glass prism, shows minimum deviation of 30 o . Calculate the speed of light through the prism.

(ii) Find the angle of incidence at face AB, so that the emergent ray grazes along the face AC.

26. An optical instrument uses and optic objective lens of power 100 D and an eyepiece of power 40 D. The final image is formed at Infinity when the tube length of instrument is kept at 20 cm. Identify the optical instrument. Calculate the angular magnification produced by the instrument.

27. Draw a labelled ray diagram of an astronomical telescope in the near point adjustment position. A giant refracting telescope at an observatory has an objective lens of focal length 15 m and an eyepiece of focal length 1 cm. If this telescope is used to view the moon, find the diameter of the image of the moon formed by the objective lens. The diameter of the moon is 3.48 x 10 6 m and the radius of lunar orbit is 3.8 x 10 8 m.

Short Answer Type Question Answer

1. (i) The frequency of reflected and refracted light remains same as that of the incident light because frequency only depends on the source of light.

(ii) Since, the frequency remain same, has there is no reduction in energy.

2. Snell’s law of refraction fails in 2 situations

(i) when total internal reflection takes place at an angle greater than the critical angle (ii) when light is incident normally on the surface

3. Given i = 45 o , t = 10 cm = 0.1 m and μ = 1.5

By Snell’s law, μ = (sin i / sin r) sin r = sin i / μ = sin45 o / 1.5 = 0.4713 r -1 = sin -1 (0.4713) = 28.12 o

Lateral shift = (tsin(i – r)) / (cos r) = [0.1 x sin (45 o – 28.12 o )] / cos 28.12 o

Lateral shift = 0.033 m

4. It is due to the refraction of sunlight as it travels through different layers of the earth’s atmosphere. Refraction of light by these layers can make the sun appear flattened or distorted. The rays of light from the upper part and lower part of the periphery of the sun bend unequally on the traveling through Earth’s atmosphere, making the sun appear oval in shape.

5. Given u = -f

From lens maker’s formula

7. If μ 1 = μ 2 , then f = ∞

Hence the lens in the liquid arts like a plane sheet, when refractive index of the lens and the surrounding medium is the same. Therefore, μ 1 = μ 2 = 1.47. Hence the liquid medium is not water. Refractive index for water is 1.33.

8. Case 1 : When tank is filled with water.

Given, the apparent depth = 9.4 cm Height of water, t = 12.5 cm Real depth = 12.5 cm

Refractive index of water, μ w = (Real depth / Apparent depth) = 12.5 / 9.4 = 1.33

Case 2 : When tank is filled with the liquid.

Refractive index of the liquid, μ l = (Real depth / Apparent depth) Apparent depth = 12.5 / 1.63 = 7.67 cm

Therefore, The microscope is shifted by 9.4 – 7.67 = 1.73 cm.

9. Whenever object is placed within the focus of the bi-concave lens, we will obtain enlarged image, hence the by concave lens behaves like a magnifying lens.

10. The magnification of a magnifying glass depends upon, where it is placed between the user’s eye and the object being viewed and the total distance between them, while the magnifying power is equivalent to angular magnification.

13. Given, focal length of convex lens, f 1 = 30 cm Focal length of concave lens, f 2 = -20 cm

Using the formula of combination of lens

1/f = 1/f 1 + 1/f 2 = 1/30 -1/20 = -1/60 f = – 60 cm

Since, the focal length of combination is negative in nature, so, the combination acts like a diverging lens, i.e. as a concave lens.

14. Given refractive of water, μ w = 4/3 Refractive index of glass prism, μ g = 3/2

For total internal reflection occurrence the incident angle must be greater than critical angle.

As the critical angle 61.6 o is greater than the angle of incidence 60 o , TIR will not occur.

15. Given refractive index of the material of the prism, μ = 1.5

From the ray diagram it is clear that angle of incidence i = 30 o < C.

Therefore, the ray incident at the face AC will not suffer TIR and merges out of this face.

17. By increasing the diameter of the objective lens, we can increase the range of the telescope because as the diameter of the lens increases, the area covered by the lens also increases. Lens is able to focus on a large area thereby helping us to view the object better.

18. Advantage of reflecting telescope over refracting telescope are:-

(i) reflecting telescope image formed is free from chromatic aberration defect. So, it is sharper than image formed by a refracting telescope. (ii) mirror is easy to produce with a large diameter so that it can intercept rays crossing a large area and direct them to the eyepiece.

19. Given, focal length of the objective lens, f o = 144 cm Focal length of eyepiece, f e = 6 cm

Magnifying power of the telescope in normal adjustment, m = -f o /f e = -144/6 = -24

Separation between lenses, L = f o + f e = 144 + 6 = 150 cm

21. The power of a lens is equal to the reciprocal of its focal length, when it is measured in metre. Power of a lens, P = 1/f and its SI unit is dioptre (D).

Consider two lenses A and B of focal lengths f 1 and f 2 placed in contact with each other. An object is placed at a point ‘O’ on the focus of the first lens A.

The first lens produces an image at I, which serves as a virtual object for the second lens B producing the final image at I.

Since, the lenses are thin, we assume the optical centres P of the lenses to be coincident. For the image formed by the first lens A, we obtain

22. First measurement gives the focal length (f eq = x) of combination of convex lens and the plano-convex liquid lens. Second measurement gives the focal length (f 1 = y) of the convex lens.

Focal length (f 2 ) of plano-convex lens is given by

23. There are 2 conditions for total internal reflection as follows:

(i) Light must travel from denser to rarer medium. (ii) Angle i > i c

Therefore, angle of incidence at face AC is 45 o which is more than the critical angle for blue and green colours. Therefore blue and green colours will undergo total internal reflection but red colour will reflect to other medium.

(i) Since the focal length of eyepiece is more than the focal length of objective, so the optical instrument is compound microscope.

(ii) Since the final image is formed at infinity, so the angular magnification is given by

27. The ray diagram of an astronomical telescope in the near point adjustment position is shown below.

Long Answer (LA) Type Questions

1. (i) A point object ‘O’ is kept in a medium of refractive index n 1 in front of convex spherical surface of radius of curvature ‘R’ with separates the second medium of refractive index n 2 from the first as shown in the figure.

Draw the ray diagram showing the image formation and reduce the relationship between the object distance and image distance in terms of n 1 , n 2 and R.

(ii) When the image formed above acts as a virtual object for concave spherical surface separating the medium n 2 from n 1 (n 2 >n 1 ), draw this ray diagram and write the similar (similar to (i)) relation. Hence, obtain the expression for the lens makers formula.

2. (i) Define the term focal length of a mirror. With the help of a ray diagram, obtain the relation between its focal length and radius of curvature.

(ii) Calculate the angle of emergence (e) of the ray of light incident normally on the face AC of a glass prism ABC of refractive index √3. How will the angle of emergence change qualitatively, if the ray of light images from the prism into a liquid of refractive index 1.3 instead of air?

3. (i) Under what condition is the phenomenon of total internal reflection of light observed? Obtain the relation between the critical angle of incidence and the refractive index of the medium.

(ii) Three lenses of focal length +10 cm, -10 cm and +30 cm are arranged coaxially as in the figure given below. Find the position of the final image formed by the combination.

4. (i) Two thin lenses are placed coaxially and contact. Obtain the expression for the focal length of this combination in terms of the focal length of the two lenses.

(ii) A converging lens of refractive index 1.5 has a power of 10 D when it is completely immersed in a liquid it behaves as a diverging lens of focal length 15 cm find the refractive index of the liquid.

5. A ray PQ of light is incident on the face AB of a glass prism ABC and emerges out of the face AC. Trace the path of the ray. Show that i + e = A + δ

where δ and e denote the angle of deviation and angle of emergence respectively. Plot a graph showing the variation of angle of deviation as a function of angle of incidence. State the condition under which ∠δ is minimum.

6. Define magnifying power of a telescope. Write its expression. A small telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If this telescope is used to view a 100 m high tower 3 km away, find the height of the final image, when it is formed 25 cm away from the eyepiece.

7. Draw a ray diagram to show the working of a compound microscope. Deduce an expression for the total magnification, when the final image is formed at the near point.

In a compound microscope, an object is placed at a distance of 1.5 cm from the objective of focal length 1.25 cm. If the eyepiece has a focal length of 5 cm and the final image is formed at the near point. Estimate the magnifying power of the microscope.

8. How is the working of a telescope different from that of a microscope? The focal length of objective and eyepiece of a microscope at 1.25 cm and 5 cm, respectively. Find the position of the object relative to the objective in order to obtain an angular magnification of 30 in normal adjustment.

9. (i) Draw a labelled ray diagram showing the image formation of a distant object by refracting telescope.

Deduce the expression for its magnifying power when the final image is formed at infinity.

(ii) The sum of focal lengths of the two lenses of refracting telescope is 105 cm. The focal length of one lens is 20 times that of the other. Determine the total magnification of telescope with the final image is formed at infinity.

Long Answer Type Question Answer

1. (i) Let a spherical surface separate a rarer medium of refractive index n 1 from the second medium of refractive index n 2 . Let C be the centre of curvature and R=MC be the radius of the surface.

Consider a point object ‘O’ or lying on the principal axis of the surface. Let a ray starting from ‘O’ incident normally on the surface along OM and pass straight. Let another ray of light incident on NM along ON and refract along NI. From M, draw MN perpendicular to OI.

The above figure shows the geometry of the formation of image ‘I’ of an object ‘O’ and the principal axis of a spherical surface with centre of curvature ‘C’ and the radius of curvature ‘R’. Here, we have to make following assumptions :-

(a) the aperture of the surface is small as compared to other distance involved. (b) NM will be taken as nearly equal to the length of the perpendicular from the point N on the principal axis.

(ii) Now, the image I’ acts as a virtual object for the second surface that will form a real at I. As refraction takes place from denser to rarer medium,

2. (i) The distance of the principal focus from the pole of the mirror is called the focal length of the mirror.

Relation between focal length and radius of curvature of mirror :-

Considered a ray parallel to the principal axis striking the mirror at point M, then CM will be perpendicular to the mirror at point M.

Let Ө be the angle of incidence and MD be perpendicular to the principal axis.

3. (i) Following are the criteria for total internal reflection.

(a) Light must pass from a optically denser to a optically rarer medium. (b) Angle of incidence in denser medium is must be greater than critical angle for two media.

5. (i) Let PQ and RS are incident and emergent rays and incident ray get deviated by δ by the prism. i.e. ∠TMS = δ

Let δ 1 and δ 2 are deviation produced at refractions taking place at AB and AC, respectively.

6. Magnifying power of a telescope is the ratio of the angle β subtended at the eye by the image to the angle α subtended at the eye by the object.

7. A compound microscope consists of two convex lenses co-axially separated by some distance. The lens nearer to the object is called the objective. The lens through which the final image is viewed is called the eyepiece. The focal length of objective lens is smaller than eyepiece.

The objective lens forms real and inverted magnified image A’B’ of object AB fall some, where between pole and focus of eye lens.

So, A’B’ acts as an object for eyepiece and its virtual magnified image A”B” formed by the lens.

The magnifying power of a compound microscope is defined as the ratio of the visual angle subjected by final image of eye and the visual angle subtended by object at naked eye, when both are at the least distance of distinct vision from the eye.

CHARACTERISTIC	TELESCOPE	MICROSCOPE
Position of object	At infinity	Near objective at a distance lying between f and 2f
Position of image	Focal plane of objective	Beyond 2f , where f is the focal length of objective

9. It consists of an objective lens of a large focal length (f o ) and large aperture, also an eyepiece of small aperture and focal length.

(i) Magnification when final image is formed at infinity, magnification, m = -f o /f e and length of telescope, L = |f o | + |f e |

Case Study Based Questions

1. Refraction involves change in the path of light due to change in the medium.

When a beam of light encounters and other transparent medium, a part of light gets reflected back into the first medium, while the rest enters the other. The direction of propagation of an oblique incident ray of light, that enters the other medium, changes at the interface of two media. This phenomenon is called refraction of light.

(i) For the same value of angle of incidence, the angle of refraction in three media A, B and C are 15 o , 25 o and 35 o respectively. In which medium, would the velocity of light be minimum?

(ii) Why does a crack in a glass window pane appear silvery?

(iii) The refractive index of diamond is much higher than that of glass. How does a diamond cutter make use of this fact?

(iv) What is the apparent position of an object below a rectangular block of glass 6 cm thick, if a layer of water 4 cm thick is on the top of the glass?

1. (i) From Snell’s law, μ = sin i / sin r = c / v v ∝ sin r

Smaller the angle of refraction, smaller the velocity of light in medium.

Velocity of light is minimum in medium A, as the angle of refraction is minimum, i.e. 15 o .

(ii) Whenever rays of light travels through glass, they strike the glass-air interface at an angle greater than critical angle of glass. They are totally reflected, hence cracks appear silvery.

(iii) The refractive index of diamond is much higher than that of glass. Due to high refractive index, the critical angle for diamond-air interface is low. The diamond is cut suitably, so that the light entering the diamond from any face suffers multiple total internal reflections at the various surfaces. This gives sparkling effect to the diamonds.

(iv) Here, μ = thickness of object / apparent depth

Now due to refraction at 2 different boundaries, the apparent depth of object = (thickness of glass/μ glass ) + (thickness of water/μ water ) = (6/1.5) + (4/1.3) = 4 + 3 = 7

Final Words

From the above article, you have practiced Class 12 Physics Chapter 9 Ray Optics Questions and Answers. We hope that the above-mentioned notes and Q & A for term 2 will surely help you in your exam.

If you have any doubts or queries regarding Class 12 Physics Chapter 9 Ray Optics Questions and Answers, feel free to reach us and we will get back to you as early as possible.

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CBSE Class 12 Physics Case Study Questions PDF Download

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CBSE (Central Board of Secondary Education) is a renowned educational board in India that designs the curriculum for Class 12 Physics with the goal of promoting a scientific temperament and nurturing critical thinking among students. As part of their Physics examination, CBSE includes case study questions to assess students’ ability to apply theoretical knowledge to real-world scenarios effectively.

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Case Study Ray Optics & Optical Instruments Class 12 Physics PDF Fully Solved

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TWo Questions from case study each of 4 Marks will come in Board exam.so total Weightage of case study in board exam is 8.

So Now you can understand importance of case study type Questions.

In 2023-24 one Question of case study was from Ray optics & second question was from Current electricity.

So we are providing few Questions of case study from Ray optics with solutions.

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Teachers can guide you they can do 40% work. They can give u a direction, rest work is yours.

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CBSE Important Questions Class 12 Physics Chapter 9

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Important Questions for CBSE Class 12 Physics Chapter 9- Ray Optics and Optical Instruments

With important questions for Class 12 physics chapter 9, students will get elaborated and authentic solutions to their doubts regarding ray optics and optical instruments.

After preparing chapter 9 class 12 Physics important questions, students will be able to solve CBSE Sample papers, CBSE Past year question papers, and NCERT EXEMPLAR Questions.

In addition, Physics Class 12 chapter 9 important questions also contain important formulas, short derivations and CBSE extra questions that can help students to test their understanding.

CBSE CLASS 12 PHYSICS CHAPTER-9 IMPORTANT QUESTIONS

Important questions class 12 physics chapter 9.

In these important question notes, we will talk about the phenomena of reflection, refraction and dispersion of light, using the ray diagram of light. Using the basic laws of reflection and refraction, we shall understand image formation by plane and spherical reflecting and refracting surfaces.

1-MARK QUESTIONS

A short pulse of white light is incident from air to a glass slab at normal incidence.

What would the first colour after emergence from a glass slab be?

Ans. Since, V ∝ 𝛌, the light of red colour is of the highest wavelength and therefore, the highest speed.

Hence, red would emerge first.

CONCEPT- When light rays go from one medium to another medium, the frequency of light remains unchanged.

The speed of the yellow light in a certain liquid is 2.4×10⁸m/s. Find the refractive index of the liquid.

Ans. Given, cꞌ=2.4×10⁸m/s

We know, c = speed of light =3×10⁸m/s

We have to find, the refractive index “𝜇” of the liquid

Since,𝜇=c/cꞌ

𝜇=3×10⁸/2.4×10⁸

A substance has a critical angle of 30° for a yellow light. Find the refractive index.

Ans. We know that refractive index (𝜇)= 1 sin C

Substituting the values, we have

𝜇= 1 sin30 = 1 1/2

At a given time, it is recorded that in NASA’s Hubble telescope, the focal length of the objective and the eyepieces are 80cm and 10cm respectively. What is its magnification power?

Ans. Given ƒ₀= 80cm

ƒₑ=10cm,

We know that magnification(M)= – ƒ₀ ƒₑ = – 80 10 = -8

Hence, M=-8

CONCEPT – To find tube length (L), Use formula L=ƒ₀+ƒₑ

Define power of a lens. What is the S.I. unit of power?

Ans. The power P of a lens is defined as the tangent of the angle by which it converges or diverges a beam of light parallel to the principal axis falling at a unit distance from the optical centre.

P= 1 f

The S.I. unit of power is 1/m or dioptre.

CONCEPT- the power of a lens is positive for a converging lens and negative for a diverging lens.

Two thin lenses of power +4D and -2D are in contact. What is the focal length of the combination?

Ans. The combined power of the two thin lenses is given by the formula

P= P₁ + P₂ = 4 +(-2) = +2D

Since focal length f= 1 P

Hence, f = ½ = 0.5 m = 50 cm

State the conditions for the phenomenon of total internal reflection to occur.

Ans. Two essential conditions for total internal reflection to occur are-

Light should travel from an optically denser medium to an optically rarer medium.
The angle of incidence in the denser medium must be greater than the critical angle for the two media.
What is the difference between refraction and reflection?

Ans. Refraction can b e defin ed as the process of the shift of light when it passes through a medium leading to the bending of light. Whereas reflection can be simply defined as the bouncing back of light when it strikes a medium on a plane.

2-MARK QUESTIONS

An object is placed 10cm in front of a concave mirror. If the radius of curvature is 15cm

Find the position, nature, and magnification of the image.

Ans. Given the radius of curvature(R)=15cm

Hence focal length(ƒ)= – R 2 = – 15 2 = -7.5cm

Given, object distance (u) = -10cm

Applying mirror formula –

1 v + 1 u = 1 ƒ

Putting the values,

1 v + 1 -10 = 1 -7.5

After solving,ign we get, v= -30cm

Negative sign of v shows that object and image lie on the same side (left side) of the mirror.

Also, magnification (m) = – v u = – -30 -10 = -3

The negative sign of magnification shows that the image is magnified, real and inverted.

Define the term optical fibre. State one its important uses.

Ans. An optical fibre is a thin, long and transparent rod usually made of glass or plastic through which light can propagate.

Optical fibre work on the principle of total internal reflection and thus, they avoid loss in the transfer of information.

Use of optical fibre –

Optical fibres are often used in medical investigations i.e. one can examine the inside view of the stomach and intestine by a method called endoscopy.

A simple microscope has a magnifying power of 3.0 when the image is formed at the near point (25cm) of a normal eye.
Find its focal length.
What will its magnifying power be if the image is formed at infinity?

Ans. Given D = 25cm

m= 3

For simple microscope, m= 1+ D ƒ

Hence f = 12.5 cm

(b)

Ans. When the image is formed at infinity m= D ƒ ( the adjustment is normal)

m= 25 12.5

m= 2.0

So if the magnifying power is 2 if the image is formed at infinity.

Explain the mechanism of mirage formation.

Ans. On hot summer days, the air near the ground becomes hotter and less dense

than the air at higher levels.

The refractive index of air increases with its density. If the air currents are small, the optical density at different layers of air increases with height.

As a result, light from a tall object such as a tree passes through a medium whose refractive index decreases towards the ground. Thus, a ray of light from such an object successively

bends away from the normal and undergoes total internal reflection, if the angle of incidence for the air near the ground exceeds the critical angle.

To a distant observer, light appears to be coming from somewhere below the ground. The observer assumes that light is being reflected from the ground by a pool of water near the tall object.

Such inverted images cause an optical illusion to the observer. This phenomenon is called mirage.

Why does bluish colour predominate in a clear sky?

Ans. BLUE COLOR OF THE SKY – The scattering of light by the atmosphere is a colour-dependent phenomenon. The scattering of the sky is primarily based on Rayleigh’s Law which shows the intensity of scattered light I ∝ 1 𝛌⁴ , where 𝛌 is the wavelength of the light.

Since blue light is scattered much more strongly than red light, the colour of the sky becomes blue. The blue component of light is proportionately more in the light coming from different parts of the sky. This gives the impression of a blue sky.

3-MARK QUESTIONS

If the focal length of a plane convex lens is 0.3m and the refractive index of the material of the lens is 1.5, find the radius of curvature of the convex surface of a plane convex lens.

Ans. Given that, 𝜇 = 1.5

ƒ= 0.3m

For plane convex lens,

R₂ = – ∞ and let R₁ = R

Substituting these values in the formula for focal length,

1 ƒ = (𝜇 – 1) ( 1 R₁ – 1 R₂ )

⇒ 1 0.3 = 1.5-1 ( 1 R + 1 ∞ )

⇒ ( 1 R ) 0.5 = 1 0.3

⇒ R = 0.15m

Thus, the radius of curvature is R = 0.15m.

At what angle should a ray of light be incident on the face of a prism of refracting angle 60°, so that, it just suffers total internal reflection at the other face? (Given the refractive index of the material of the prism is 1.524)

Given, angle of the prism, ∠A = 60°

Refractive index of the prism, 𝜇 = 1.524

i₁= Incident angle

r₁= Refracted angle

r₂= Angle of incidence at the face AC

e= Emergent angle = 90°

According to Snell’s law, for the face AC, we can have:

sine sin r₂ = 𝜇

According to Snell’s law, for face AC, we can have:

Sin r₂ = 1 𝜇 × sin 90°

= 1 1.524 ×1 = 0.6562

Therefore, r₂ = sin⁻¹ 0.6562 ≈ 41°

It is clear from the figure that angle A = r₁ + r₂

Therefore, r₁ = A – r₂ = 60 – 41 = 19°

According to Snell’s law, we have the relation:

𝜇 = sini₁ sin r₁

Sin i₁ = 𝜇 sin r₁

= 1.524 × sin 19° = 0.496

Therefore, i₁ = 29.75°

Hence the angle of incidence is 29.75°.

A myopic person has been using spectacles of power -1.0 dioptre for distant vision.

During old age, he also needs to use a separate reading glass of power +2.0 dioptres.

Explain what may have happened.

The power of the spectacles used by the myopic person, P = -1.0 D

Focal length of the spectacles, ƒ = 1 p = 1 -1✕10⁻² = -100 cm

Hence, the far point of the person is 100 cm. He might have a normal near point of 25 cm.

When he uses the spectacles, the objects placed at infinity produce virtual images at 100 cm.

He uses the ability of accommodation of the eye lens to see the objects placed between 100 cm and 25 cm.

During old age, the person uses reading glasses of power, Pꞌ = +2D

The ability to accommodation is lost in old age. This defect is called presbyopia. As a result, he is unable to see clearly the objects placed at 25 cm.

There is a prism made up of two different materials such as crown glass and flint glass with a wide variety of angles. What can be the combinations of prisms which will:
Deviate a pencil of white light without many dispersions.
Disperse and displace a pencil of white light without much deviation.

Ans. a. Place the two given prisms next to each other.

Make sure that their bases are on the opposite sides of the incident white light, with their faces touching each other.

When the white light is incident on the first prism, it will get dispersed. This acts as the incident light for the second prism and the dispersed light this time will recombine to give white light as a result of the combination of the two prisms.

Ans. b. Take the system of the two prisms as suggested in answer (a).

Adjust (increase) the angle of the flint glass prism so that the deviations due to the combination of the prisms become equal.

This combination will now disperse the pencil of white light without much deviation.

What are the factors that affect the refractive index of a medium?

Ans. Following are the factors that affect the refractive index of a medium –

The refractive index of the medium will depend on the nature and temperature of the medium. It also depends on the color of the light ray.
Refractive index is an optical property therefore any impurity added to the medium will alter the refractive index of the medium.
The absolute refractive index of the medium is the ratio of the velocity of light in air or vacuum to that in the given medium. The velocity of light is maximum in vacuum. The velocity in any other medium is less than the value in air. Thus the absolute refractive index of the medium is always greater than the unity.

5-MARK QUESTIONS

Briefly explain the formation of a rainbow.

Ans. The rainbow is an example of the dispersion of sunlight by the water drops in the atmosphere. This a phenomenon due to the combined effect of dispersion, refraction, and reflection.

The conditions for observing a rainbow are that the sun should be shining in one part of the sky, while it should be raining in the opposite part.

MECHANISM –

Consider figure A, in which sunlight is first refracted, which causes the different wavelengths of white light to separate. Longer wavelengths of light (RED) are bent the least while the shorter wavelength (VIOLET) is bent the most.

If the refracted angle is greater than the critical angle, then the light will get internally reflected.

It is found that violet light emerges at an angle of 40° while red light emerges at an angle of 42°.

Consider figure B, in which the formation of the primary rainbow takes place. The observer sees a rainbow with red color on the top and violet on the bottom which is a result of three-step processes i.e., refraction, reflection, and refraction.

When light rays undergo two internal reflections inside a raindrop, instead of one as in the primary rainbow, a secondary rainbow is formed, as shown in figure C .

It is due to a four-step process.

The intensity of light is reduced. Hence, the secondary rainbow is fainter.

A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

Actual depth of the needle in water, h₁ = 12.5 cm

Apparent depth of the needle in water, h₂ = 9.4 cm

Refractive index of water = 𝜇

The value of 𝜇 can be obtained as follows:

= 12.5 / 9.4 = 1.33

Hence, 1.33 is the refractive index of water.

Now water is replaced by a liquid of refractive index 1.63

The actual depth of the needle remains the same, but its apparent depth changes.

Let x be the new apparent depth of the needle.

Therefore, y = h₁ 𝜇ꞌ e

= 12.5 / 1.63 = 7.67 cm

The new apparent depth of the needle is 7.67 cm. It is observed that the value is less than h₂,

therefore, the needle needs to be moved up to the focus again.

Distance to be moved to focus = 9.4 – 7.67 = 1.73 cm

The figure depicted below shows a biconvex lens ( of refractive index 1.50 ) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed, and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?

The focal length of the convex lens, f₁ = 30 cm

The liquid acts as a mirror. The focal length of the liquid = f₂

Focal length of the system ( convex lens + liquid ) f = 45cm

The equivalent focal length of the pair of optical systems that are in contact are given as:

1 ƒ = 1 ƒ₁ + 1 ƒ₂

1 f₂ = 1 ƒ – 1 ƒ₁

1 45 – 1 30

– 1 90

Therefore, f₂ = -90 cm

Let 𝜇₁ be the refractive index, and the radius of curvature of one surface be R.

Hence, -R will be the radius of the curvature of the other surface.

R can be obtained by using the relation:

1 ƒ₁ = ( 𝜇₁ -1 ) ( 1 R + 1 -R )

Draw a ray diagram showing the image formation by a compound microscope. Then obtain the expression for total magnification when the image is formed at infinity.

Ans. Ray diagram:

(a) Ray diagram of a compound microscope: A schematic diagram of a compound microscope is shown in the figure. The lens nearest the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens, the eyepiece, which functions essentially like a simple microscope or magnifier, and produces the final image, which is enlarged and virtual. The first inverted image is thus near (at or within) the focal plane of the eyepiece, at a distance appropriate for final image formation at infinity, or a little closer for image formation at the near point. Clearly, the final image is inverted with respect to the original object.

Magnification due to a compound microscope.

The ray diagram shows that the (linear) magnification due to the objective, namely h’/h, equals

Here h’ is the size of the first image, the object size being h and f 0 being the focal length of the object. The first image is formed near the focal point of the eyepiece. The distance L, i.e., the distance between the second focal point of the objective and the first focal point of the eyepiece (focal length f e ), is called the tube length of the compound microscope.

As the first inverted image is near the focal point of the eyepiece, we use for a simple microscope to obtain the (angular) magnification me due to it when the final image is formed at the near point, is

When the final image is formed at infinity, the angular magnification due to the eyepiece, me = (D//e)

Thus, the total magnification from equation (i) and (iii), when the image is formed at infinity, is

RELEVANCE OF RAY OPTICS IN FUTURE

G eometrical optics or ray optics is a model that describes light propagation in terms of rays.

Ray optics is a sub-branch that is significantly important in the field of fiber optics. Optical fiber is used as a medium for telecommunication and computer networking.

It is especially advantageous for long-distance communications.

Ray optics is also used in the field of astronomy. All microscopes, telescopes, and other optical instruments work under the phenomena of reflection and refraction.

Recently, a black hole was discovered by NASA was found by the James Webb telescope.

We can say that ray optics is an important area of study for the generations to come because of its significant role in ARTIFICIAL INTELLIGENCE and quantum computing.

WHY IS IT IMPORTANT TO STUDY IMPORTANT QUESTIONS FOR RAY OPTICS CLASS 12?

Benefits of preparing NCERT Chapter 9 Ray Optics and optical instruments from Important questions for RAY OPTICS CLASS 12.

These physics Class 12 chapter 9 important question notes of NCERT give us a wide coverage of the topic keeping in mind the CBSE syllabus.
Important questions for Ray optics Class 12 cover a wide range of CBSE extra questions and also the CBSE past year question paper.
The questions have been set up keeping in mind the CBSE Sample papers, and also the CBSE revision notes have been taken into consideration.
The important question for the chapter Ray Optics covers all the important formulas, derivations, and necessary diagrams.
Class 12 Physics chapter 9 important questions will benefit students in covering this whole chapter in a lucid and understanding manner.

HOW WILL STUDYING FROM EXTRAMARKS IMPORTANT QUESTION FOR RAY OPTICS CLASS 12 CBSE HELP STUDENT?

Extramarks Class 12 physics chapter 12 important questions for Ray Optics Class 12 are very significant in preparing for CBSE Class 12 Board exam and also other competitive exams. These important questions consist of NCERT Solutions, CBSE sample paper questions, and CBSE past year questions also.

Extramarks provides important questions for the chapter Ray Optics which widely cover the CBSE syllabus; this would prove to be very beneficial for the students. Extramarks important questions are compiled using important formulas, derivations, and explanations that go in very lucid and understandable patterns.

HOW TO STUDY FOR RAY OPTICS FROM EXTRAMARKS NOTES?

Students are advised to first go through the basic text for the chapter on ray optics using the NCERT Textbook Class 12. After glancing through the chapter, they should refer to the Extramarks Important Question for Ray Optics. These important questions widely cover the CBSE Syllabus, CBSE Past year question papers, etc., which would help students to know the kind of questions asked from the topic of ray optics.

Students can first refer to the important questions and try solving them on their own and, after that, go through the detailed solutions provided by Extramarks subject experts.

INTRODUCTION TO RAY OPTICS

A lightwave can be considered to travel from one point to another along a straight line joining them. The path is called a ray of light, and a bundle of such rays constitutes a beam of light.

The branch of Physics that deals with the nature, properties, sources, and effects of light is called optics. Optics is broadly divided into two branches namely physical optics, which is the study of the wave-like nature of light and the interactions between light and matter.

Ray optics is the study of simple properties of light and optical instruments by assuming that light travels in a straight line.

Ray optics deals with the geometry of light. In ray optics, we study the image formed by mirrors, lenses, and prisms.

What is a ray?

A ray is a part of the line having one fixed point, and the other point does not have an end..

In optics, a ray is an idealized geometrical model of light obtained by choosing a curve, i.e., is perpendicular to the wavefronts of the actual light and that points in the direction of energy flow.

CONCLUSION –

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Q1- Two thin lenses P 1 and P 2 are made in contact to get the focal length of 60 cm. If the focal length of one of the lens is 20 cm, then the power of the other lens is

a–3.34 D

b–1 D

ans – -3.34 D

Q2- The refractive indices of glass and water are 1.62 and 1.32 respectively. The critical angle for a ray going from glass to water is

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Faqs (frequently asked questions), 1. what are the defects of the human eye.

The defects of the human eye are as follows –

Myopia – It is also known as short-sightedness in which nearby objects appear to be seen clearly, and far objects appear to be seen blurred. It can be corrected by a concave lens.

Hypermetropia – It is also called farsightedness in which far away appear to be seen clearly and nearby objects appear to be blurred. It can be corrected using the convex lens.

Astigmatism – This defect causes impaired vision due to which some parts of the picture may appear clear, whereas some parts may appear blurred. It can be corrected with the help of spherical or cylindrical lenses.

2. What is illuminance?

Illuminance is defined as luminous flux incident per unit area on a surface.

The illuminance E produced by a source of luminous intensity I is given by

E = I / R²

Where R is the normal distance of the surface from the source.

Its S.I. unit is lux.

3. Why are diamonds known for their spectacular brilliance?

The critical angle for diamond – air interface is very small, therefore, once light enters a diamond, it is very likely to undergo Total Internal Reflection inside it.

Hence, their brilliance is mainly due to total internal reflection.

4. State the advantages of a reflecting type telescope over a refracting type of telescope.

Following are the advantages of a reflecting type telescope over a refracting type of telescope –

Due to the large aperture of the mirror used in the reflecting telescopes, it has high resolving power.
The reflecting type of telescope is free from chromatic aberration (formation of the coloured image of a white object).
The use of paraboloidal mirrors reduces spherical aberration (formation of a nonpoint, blurred image of a point object).
The image formed by a reflecting telescope is brighter than a refracting telescope.

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Class 12th Physics - Ray Optics and Optical Instruments Case Study Questions and Answers 2022 - 2023

Class 12th Physics - Ray Optics and Optical Instruments Case Study Questions and Answers 2022 - 2023 Study Materials Sep-09 , 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 12th Physics Subject - Ray Optics and Optical Instruments, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Ray optics and optical instruments case study questions with answer key.

Final Semester - June 2015

Case Study

(ii) Two thin lenses are in contact and the focal length of the combination is 80 cm. If the focal length of one lens is 20 cm, the focal length of the other would be

(iii) A spherical air bubble is embedded in a piece of glass. For a ray of light passing through the bubble, it behaves like a

(iv) Lens used in magnifying glass is

(v) The magnification of an image by a convex lens is positive only when the object is placed

Power (P) of a lens is given by-reciprocal of focal length (f) of the lens i.e., $P=\frac{1}{f}$ where fis in metre and P is in dioptre. For a convex lens, power is positive and for a concave lens, power is negative. When a number of thin lenses of powers P 1 , P 2 , P 3 , .....are held in contact with one another, the power of the combination is given by algebraic sum of the powers of all the lenses i.e., P = P 1 + P 2 + P 3 + .... (i) A convex and a concave lens separated by distance d are then put in contact. The focal length of the combination

(ii) If two lenses of power +1.5D and +1.0D are placed in contact, then the effective power of combination will be

(iii) If the power of a lens is +5 dioptre, what is the focal length of the lens?

(iv) Two thin lenses offocallengths +10 cm and -5 em are kept in contact. The power of the combination is

(v) A convex lens of focal length 25 cm is placed coaxially in contact with a concave lens of focal length 20 cm. The system will be

Total internal reflection is the phenomenon of reflection of light into denser medium at the interface of denser medium with a rarer medium. For this phenomenon to occur necessary condition is that light must travel from denser to rarer and angle of incidence in denser medium must be greater than critical angle (C) for the pair of media in contact. Critical angle depends on nature of medium and wavelength of light. We can show that $\mu=\frac{1}{\sin C} .$ (i) Critical angle for glass air interface, where $\mu$ of glass is $\frac{3}{2}$ is

(ii) Critical angle for water air interface is 48.6°. What is the refractive index of water?

(a) 1

(b) $\frac{3}{2}$

(d) $\frac{3}{4}$

(iii) Critical angle for air water interface for violet colour is 49°. Its value for red colour would be

(iv) Which of the following is not due to total internal reflection?

(v) Critical angle of glass is $\theta$ 1 and that of water is $\theta$ 2 , The critical angle for water and glass surface would be $\left(\mu_{g}=3 / 2, \mu_{w}=4 / 3\right)$

and $\theta$

The lens maker's formula is a relation that connects focal length of a lens to radii of curvature of two surfaces of the lens and refractive index of the material of the lens. It is $\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$ where $\mu$ is refractive index oflens material w.r.t. the medium in which lens is held.As , $\mu_{v}>\mu_{r}$ therefore $f_{r}>f_{v^{\prime}}$ Mean focal length of lens for yellow colour is $f=\sqrt{f_{r} \times f_{v}}$ . (i) Focal length of a equiconvex lens of glass $\mu=\frac{3}{2}$ in air is 20 cm. The radius of curvature of each surface is

(ii) A substance is behaving as convex lens in air and concave in water, then its refractive index is

(iii) For a thin lens with radii of curvatures R 1 and R 2 , refractive index nand focal length f, the factor $\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$ is equal to

$(a) \frac{1}{f(n-1)}$

$(b) f(n-1)$

$(c) \frac{(n-1)}{f}$

$(d) \frac{n}{f(n-1)}$

(iv) A given convex lens of glass $\left(\mu=\frac{3}{2}\right)$ can behave as concave when it is held in a medium of $\mu$ equal to

(v) The radii of curvature of the surfaces of a double convex lens are 20 cm and 40 cm respectively, and its focal length is 20 cm. What is the refractive index of the material of the lens?

An astronomical telescope is an optical instrument which is used for observing distinct images of heavenly bodies libe stars, planets etc. It consists of two lenses. In normal adjustment of telescope, the final image is formed at infinity. Magnifying power of an astronomical telescope in normal adjustment is defined as the ratio of the angle subtended at the eye by the angle subtended at the eye by the final image to the angle subtended at the eye, by the object directly, when the final image and the object both lie at infinite distance from the eye. It is given by, $m=\frac{f_{0}}{f_{e}}$ To increase magnifying power of an astronomical telescope in normal adjustment, focal length of objective lens should be large and focal length of eye lens should be small. (i) An astronomical telescope of magnifying power 7 consists of the two thin lenses 40 cm apart, in normal adjustment. The focal lengths of the lenses are

(ii) An astronomical telescope has a magnifying power of 10. In normal adjustment, distance between the objective and eye piece is 22 cm. The focal length of objective lens is

(iii) In astronomical telescope compare to eye piece, objective lens has

(iv) To see stars, use

(v) For large magnifying power of astronomical telescope

\((a) f_{v}<

$(b) f_{v}=f_{\mathrm{e}}$

$(c) f_{o}>>f_{\mathrm{e}}$

Refraction of light is the change in the path oflight as it passes obliquely from one transparent medium to another medium. According to law of refraction $\frac{\sin i}{\sin r}={ }^{1} \mu_{2}$ where ${ }^{1} \mu_{2}$ is called refractive index of second medium with respect to first medium. From refraction at a convex spherical surface, we have $\frac{\mu_{2}}{v}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R}$ Similarly from refraction at a concave spherical surface when object lies in the rarer medium, we have $\frac{\mu_{2}}{v}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R}$ and when object lies in the denser medium, we have $\frac{\mu_{1}}{v}-\frac{\mu_{2}}{u}=\frac{\mu_{1}-\mu_{2}}{R}$ . (i) Refractive index of a medium depends upon

(ii) A ray of light of frequency 5 x 10 14 Hz is passed through a liquid. The wavelength of light measured inside the liquid is found to be 450 x 10 -9 m. The refractive index of the liquid is

(iii) A ray of light is incident at an angle of 60° on one face of a rectangular glass slab of refractive index 1.5. The angle of refraction is

(0.95)

(0.58)

(0.79)

(0.86)

(iv) A point object is placed at the centre of a glass sphere of radius 6 cm and refractive index 1.5. The distance of the virtual image from the surface of sphere is

(v) In refraction, light waves are bent on passing from one medium to the second medium because in the second medium

A compound microscope is an optical instrument used for observing highly magnified images of tiny objects. Magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the final image and the object are situated at the least distance of distinct vision from the eye. It can be given that: $m=m_{e} \times m_{o}$ where m e is magnification produced by eye lens and m o is magnification produced by objective lens. Consider a compound microscope that consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. (i) The object distance for eye-piece, so that final image is formed at the least distance of distinct vision, will be

(ii) How far from the objective should an object be placed in order to obtain the condition described in part(i)?

(iii) What is the magnifying power of the microscope in case ofleast distinct vision?

(iv) The intermediate image formed by the objective of a compound microscope is

(v) The magnifying power of a compound microscope increases with

The lens maker's formula relates the focal length of a lens to the refractive index of the lens material and the radii of curvature of its two surfaces. This formula is called so because it is used by manufacturers to design lenses of required focal length from a glass of given refractive index. If the object is placed at infinity, the image will be formed at focus for both double convex lens and double concave lens Therefore, lens maker's formula is $\frac{1}{f}=\left[\frac{\mu_{2}-\mu_{1}}{\mu_{1}}\right]\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$ When lens is placed in air, $\mu$ 1 = 1 and $\mu$ 2 = $\mu$ . The lens maker formula takes the form $\frac{1}{f}=(\mu-1)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$ (i) The radius of curvature of each face of biconcave lens with refractive index 1.5 is 30 cm. The focal length of the lens in air is

(ii) The radii of curvature of the faces of a double convex lens are 10 cm and 1.5 cm. If focal length is 12 cm, then refractive index of glass is

(iii) An under-water swimmer cannot see very clearly even in absolutely clear water because of

(iv) A thin lens of glass ( $\mu$ = 1.5) offocallength 10 cm is immersed in water ( $\mu$ = 1.33). The new focal length is

(v) An object is immersed in a fluid. In order that the object becomes invisible, it should

A prism is a portion of a transparent medium bounded by two plane faces inclined to each other at a suitable angle. A ray of light suffers two refractions on passing through a prism and hence deviates through a certain angle from its original path. The angle of deviation of a prism is, $\delta$ = ( $\mu$ - 1) A, through which a ray deviates on passing through a thin prism of small refracting angle A. If $\mu$ is refractive index of the material of the prism, then prism formula is, $\mu=\frac{\sin \left(A+\delta_{m}\right) / 2}{\sin A / 2}$ (i) For which colour, angle of deviation is minimum?

(ii) When white light moves through vacuum

(iii) The deviation through a prism is maximum when angle of incidence is

(iv) What is the deviation produced by a prism of angle 6°? (Refractive index of the material of the prism is 1.644).

(v) A ray of light falling at an angle of 50° is refracted through a prism and suffers minimum deviation. If the angle of prism is 60°, then the angle of minimum deviation is

(ii) A ray of light will undergo rotal internal reflection inside the optical fibre, if it

(iii) If in core, angle of incidence is equal to critical angle, then angle of refraction will be

= n

> n

< n

+ n = 2

(v) If the value of critical angle is 30 ° for total internal reflection from given optical fibre, then speed of light in that fibre is

A compound microscope consists of two convex lenses one acts as a magnifying lens and is known as an objective, and another lens is called an eyepiece. The two lenses work independently. Objective lens produces a magnified image of a tiny object O. This image is further modified by an eyepiece and final image is further magnified by an eyepiece and is seen at least distance of distinct vision. (i) What type of image is produced by an objective? (ii) Where would the first image have to be produced a by the objective relative to the eyepiece such that second magnified image is produced on the same side of the eyepiece as the first image? (First image distance = u e from the eyepiece). (iii) If we have two microscopes with similar set of combination oflenses (i.e. for each f o = 1.25 cm; f e = 6.0 cm). But magnification produced by first microscope is higher than the second one. What might be the reason?

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NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

December 14, 2018 by Bhagya

NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments are part of NCERT Solutions for Class 12 Physics . Here we have given NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments.

Topics and Subtopics in NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments :


9	Ray Optics and Optical Instruments
9.1	Introduction
9.2	Reflection of Light by Spherical Mirrors
9.3	Refraction
9.4	Total Internal Reflection
9.5	Refraction at Spherical Surfaces and by Lenses
9.6	Refraction through a Prism
9.7	Dispersion by a Prism
9.8	Some Natural Phenomena due to Sunlight
9.9	Optical Instruments

NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Q1

index of diamond is much greater than that of ordinary glass, hence critical angle C for diamond is much smaller (24°) as compared to that of glass (42°). A skilled diamond cutter thus can take the advantage of such large range of angle of incidence available for total internal reflection 24° to 90°. The diamond can be cut with so many faces, to ensure that light entering the diamond does multiple total internal reflections before coming out. This behavior produce brilliance i.e., sparkling effect in the diamond.

NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Q19

Question 25. Does short-sightedness (myopia) or long¬> sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of r accommodation? If not, what might cause these defects of vision? Solution: A person with normal ability of accommodation may be myopic or hypermetropic due to defective eye structure, j- When the eye ball from front to back gets too elongated the myopic defect occur, similarly when the eye ball from front to back gets too shortened the hypermetropia defect occur. When the eye ball has normal length ‘ but the eye lens loses partially its ability of accommodation, the defect is called “presbyopia” and is corrected in the same I manner as myopia or hypermetropia.

Question 27. A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected? Solution: This defect is called astigmatism. It arises due to non spherical cornea. The eye lens is ideally spherical and has same curvature in different planes, but in an astigmatic eye due to non spherical cornea the curvature may be insufficient in different planes. In the given situation the curvature in the vertical plane is enough, so vertical lines are visible distinctly. But the curvature is insufficient in the horizontal plane, hence horizontal lines appear blurred. The defect can be corrected by using a cylindrical lens with its axis along vertical. The parallel rays in the vertical plane will suffer no extra refraction but the parallel rays in the horizontal plane will be refracted largely and converges at the retina, according to the requirement to form the clear image of horizontal lines.

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Access NCERT Exemplar Solutions for CBSE Class 12 Science (Physics) Chapter 9 - Ray Optics and Optical Instruments

1. A beam of light incident at an angle on a prism refracting face generally emerges from the opposite face. The angle of incidence is 5° if the prism is 5° and the prism is built of a material with a refractive index of 1.5.

Ans: (a) 7.5°

$A = 5^\circ $

${i_2} = 0^\circ $

${r_2} = 0^\circ $

${r_1} + {r_2} = A$

\[{r_1} = A - {r_2}{\text{ }} = 5 - 0 = 5^\circ \]

\[\mu = sin{\text{ }}{i_1}sin{\text{ }}{r_1}\]

\[sin{\text{ }}{i_1} = sin{\text{ }}{r_1}\]

=>sini1=μsinr1

2. At normal incidence, a brief pulse of white light is incident from the air onto a glass slab. The first colour to appear after passing through the slab is

Ans: (d) red.

Because c=v$\lambda $ and v is constant throughout refraction that is why c $c \propto \lambda $ In glass, the velocity of red colour is at its highest as ${\lambda _R} < {\lambda _V}$

3. A particle approaches a convergent lens from the left at a consistent speed of 5 m/s and comes to a complete stop at the focus. The picture

(a) goes away from the lens at a constant 5 m/s pace.

(b) with a consistent accleration, advances away from the lens.

(d) accelerates in a non-uniform manner towards the lens.

Ans: (c): accelerates away from the lens in a non-uniform manner.

When an item approaches a lens with uniform speed, its image accelerates non-uniformly away from the lens to infinity.

4. An aircraft passenger will

(a) never witness a rainbow.

(b) perceives the major and secondary rainbows as concentric rings.

(d) is never going to witness a secondary rainbow.

Ans: (b) perceives the major and secondary rainbows as concentric rings

In an aeroplane, a passenger may observe the primary and secondary rainbows as concentric rings.

5. You are given four light sources, each of which emits a single colour — red, blue, green, and yellow. Assume that the angle of refraction for a yellow light beam corresponding to a certain angle of incidence at the interface of two mediums is 90° If the source of yellow light is changed with that of other lights without modifying the angle of incidence, which of the following assertions is correct?

(a) The red-light beam would be completely internally reflected.

(b) As the red light is refracted via the second medium, it bends back towards normal.

(d) As it passes through the second medium, the green light beam will bend away from the normal.

Ans: (c): The blue light beam would be completely internalised

If the angle of refraction is $90^\circ $for the length, then the crucial angle is the incidence angle in this case. As a result, light rays pass through a material that is denser and rarer.

As sin $c = \dfrac{1}{\mu }$so, $c \propto \dfrac{1}{\mu } > {\mu _v} > {\mu _g} > {\mu _Y} > {\mu _R}$

As a result, the critical angle for\[{C_v} < {C_g} < {C_Y} < {C_R}\] i.e., the critical angle of blue and green light is less than that of yellow light, while the critical angle of red light is bigger.

Because the angle of refraction for yellow light is specific to a specific incidence angle. This incidence angle is crucial for yellow, so make it \[{C_Y}\]. In the same way that \[{C_Y} < {C_R}\]. As a result, it will not receive complete internal reflection and\[{C_v} < {C_g} < {C_Y} < {C_R}\].

As a result, blue and green light get 100% internal reflection. So, the right response is (c).

6. A plano-convex lens' curved surface has a radius of curvature of 20 cm. If the lens' material has a refractive index of 1.5, it will

(a) serve as a convex lens only for objects that fall on its curved side.

(b) serve as a concave lens for items that pass through its curved side.

(d) operate as a concave lens regardless of which side the item is on.

Ans: (c) operate as a convex lens regardless of which side the item is on.

If the item is on the curved side, ${R_1}$ equals +20 cm and ${R_2}$ equals infinity ${\mu _1} = 1,{\mu _2} = 1.5$

$\dfrac{1}{f} = ({\mu _2} - {\mu _1}) = \left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$

$\dfrac{1}{f} = (1.5 - 1) = \left( {\dfrac{1}{{20}} - \dfrac{1}{\infty }} \right) = \dfrac{{0.5}}{{20}} = \dfrac{5}{{200}} = \dfrac{1}{{40}}$

If the item is on the curved side, ${R_1}$ equals infinity and ${R_2}$ equals -20 ${\mu _1} = 1,{\mu _2} = 1.5$

$\dfrac{1}{f} = (1.5 - 1) = \left( {\dfrac{1}{\infty } + \dfrac{1}{{20}}} \right) = \dfrac{{0.5}}{{20}} = \dfrac{5}{{200}} = \dfrac{1}{{40}}$

As a result, regardless of which side the objects are on, the lens will always behave as a convex lens.

7. The mechanisms involved in radio wave reflection by the ionosphere are analogous to

(a) light reflection by a planar mirror.

(b) during a mirage, entire internal reflection of light in the air.

c) light dispersion by water molecules during the production of a rainbow

(d) light scattering due to air particles.

Ans: (b) during a mirage, entire internal reflection of light in the air.

The ionosphere is a translucent optical medium that reflects radio waves. Total internal reflection occurs when a radio wave is reflected through a clear surface, resulting in internal reflection of the radio wave

8. PQ depicts the direction of a ray of light incident on a concave mirror, while four rays labelled 1, 2, 3, and 4 depict the directions in which the ray might go after reflection. Which of the four rays appropriately depicts the reflected ray direction?

Ans: (b) 2

Incidence Because beam PQ enters primary focus F, it must be parallel to the principal axis, i.e., 2 or 4.

Ray (4) is rejected since it cannot get behind the mirror because it is a concave mirror.

Beam 2 is therefore the reflected ray. It confirms the response (b)

We can verify it again by drawing normal QC and find the ∠ r = ∠ i. So ray (2) is the reflected ray.

9. Turpentine has a higher optical density than water while having a lower mass density. A layer of turpentine floats on top of water in a container, as shown in Figure. Which of the four rays impacting on turpentine in Figure follows the proper path?

Ans: (b): 2

\[{\mu _a} < {\mu _T} < {\mu _W}\]

In this case, the incident ray travels from air to turpentine to water, i.e., from rare to denser than denser to rarer, hence it bends towards normal first, then away from normal, resulting in the route illustrated (2).

10. On a straight road, an automobile is travelling at a constant speed of 60 km/h–1. Looking in the rear-view mirror, the driver notices that the automobile behind him is coming at a distance of 100 metres and at a speed of 5 kilometres per hour –1. To maintain track of the automobile in front, the driver alternates between looking in the back and side mirrors of his car every 2 seconds until the other car overtakes. Which of the following statements is/are accurate if the two automobiles maintain their speeds?

(a) The automobile in the back is travelling at 65 km/h.

(b) In the side mirror, the automobile in the rear appears to be approaching the driver of the leading car at a speed of 5 km/h.

(d) As the distance between the automobiles shrinks, the approaching car's speed appears to rise in the side mirror.

Ans: (d): As a result, as the back automobile approaches, it seems to be at rest at first since the image is generated at focus. As the automobile gets closer, the speed appears to increase, hence the solution is (d).

11. Certain materials with a negative refractive index have been created in laboratories (Figure). A beam incident from the air (medium 1) into such a medium (medium 2) must take the route prescribed by

Ans: (a)

The incoming beam from air (medium 1) refracts or bends differently or opposite and symmetrically to that of the positive refractive index medium in negative refractive index materials. So, the solution is (a).

MCQ -2

12. Consider a long item submerged in water within a flat trough. When viewed from close to the trough's edge, the item seems deformed because

(a) the apparent depth of the points close to the edge is closer to the water's surface than the ones farther away.

(b) the angle occupied by the item's picture in the eye is less than the actual angle occupied by the object in the air.

(d) In a trough, water acts as a lens, magnifying the item.

Ans: (a), (b) and (c)

We know that the shift (h) of an object immersed in liquid from the object is proportional to the actual distance of the object from the liquid's surface.

$h = t\left( {1 - \dfrac{1}{\mu }} \right)$

t=the object's true depth or distance from the liquid's refractive index surface.

When seen from one edge of the trough, the relative disparities in depth (distance) in water between the two ends of the item are greater than when viewed from the top or away from the edge. The choice is verified by the following calculation, which shows that the distortion of the closer end is lower than the distortion of the further end (a).

The angle occupied by an item under water is greater than its image, and as the image changes upward, choice (b).

Rays from the object to the observer move from a denser to a rarer medium, and the angle of incidence for rays from the object's further end B is greater than the angle of incidence for rays from the object's close end A. The incidence angles for rays coming from end 3 may be greater than the critical angle, causing total internal reflection and preventing the rays from reaching the observer.

13. The refractive index of a rectangular block of glass ABCD is 1.6. A pin is positioned in the middle of the AB figure's face. When seen from the AD side, the pin should-

(a) appear to be in close proximity to A.

(b) appear to be in the vicinity of D.

(d) go unnoticed at all.

Ans: (a) appear to be in close proximity to A.

(b) appear to be in the vicinity of D.

$\sin C = \dfrac{1}{\mu } = \dfrac{1}{{1.6}} = 38.7^\circ $

On the mid-point of face AB, there is a point P. The angle of incidence will be less than critical angle when seen via face AD at point A. \[{i_c} = 38.7^\circ \]. As a result, the image of P will appear at P', and the picture may be viewed at P'. P' is closer to both A and D than P, which validates option (a) (b).

Angle of incidence when observed at point D via face AD \[\;{i_2}{\text{ }} > {\text{ }}{i_c}\]. As a result, there is entire internal reflection and the item cannot be perceived. Option (d) is not confirmed since the item may be seen when viewed close to A.

14 . A dark band known as Alexandar's dark band exists between the major and secondary rainbows. This is due to the fact that (a) light diffused into this region interferes in a negative way.

(b) There is no dispersed light in this area.

(d) The dispersed rays make an angle at the eye with regard to the incident light of the sun that is between about 42° and 50° .

Ans: (d) The dispersed rays make an angle at the eye with regard to the incident light of the sun that is between about 42° and 50°

Between the major and secondary rainbows is Alexandar's black band. The light diffused into this area interferes destructively, causing this to develop. Because the primary and secondary rainbows are at right angles to one other (41° to 42°) and (51° to 54°) The dispersed rays with regard to the incident ray of the sun lie between roughly and at the observer's eye with respect to incident light ray 42° to 50°

15. Because the item to be examined may be brought closer to the eye than the typical near point, a magnifying glass is needed. As a result

(a) the item near the eye can subtend a bigger angle and so be perceived in more detail

b) the emergence of a simulated erect image

c) widening of the area of view

(d) At the near point, infinite magnification.

Ans (a) and (b) are the correct answers.

The object is put inside the focal length of a magnifying glass, and the image generated is enlarged and erect. Because the image (A'B') is enlarged, it has a greater angle at the eye than object (AB), allowing it to be seen more clearly.

A magnifying glass is used, as the object to be viewed can be brought closer to the eye than the normal near point.

16. An astronomical refractive telescope features a 20m focal length objective and a 2cm focal length eyepiece.

(a) The telescope tube is 20.02 metres long.

(b) A magnification of 1000 is used.

(d) A bigger aperture objective will boost the image's brightness while reducing chromatic aberration.

: The length of the telescope

\[L{\text{ }} = {\text{ }}{f_o}{\text{ }} + {\text{ }}{f_e}{\text{ }} = {\text{ }}20{\text{ }} + {\text{ }}00.02{\text{ }} = {\text{ }}20.02{\text{ }}m\]

\[{\text{m }} = {\text{ }}\dfrac{{{f_o}}}{{{f_e}}}{\text{ }} = {\text{ }}\dfrac{{20.02{\text{ }}m}}{{0.02m}} = \dfrac{{2000}}{2} = 1000\]

Inverted, virtual, and smaller than the object, the final image formed in a telescope (Refracting) is inverted, virtual, and smaller than the object.

17. Will the focal length of a lens for red light be more, same or less than that for blue light?

Ans: As we know that \[{\mu _v} > {\mu _r}\]

As a result, 1/f will be big for blue light and tiny for red light. As a result, for red-coloured light, f will be bigger.

18. The typical person near eyesight is 25cm. What should the microscope's power be to examine an item with an angular magnification of 10?

Ans: D=25 cm to see the final image with clear eyesight

u=-f and v=-25 for eyes lens

$m = \dfrac{v}{u}$

$10 = \dfrac{{ - 25}}{{ - f}}$

$f = + \dfrac{{25}}{{10}}$ =+2.25 cm =0.025 m

$p = \dfrac{1}{f} = \dfrac{1}{{0.025}} = \dfrac{{1000}}{{25}} = 40D$

19. The image of a point object is formed on the axis of an unsymmetrical double convex thin lens. When the lens is inverted, does the image position change?

Ans: Altering the u of f or changing the position of the picture in a convex lens affects the image's position. u is constant in this case. and$\dfrac{1}{f} = (\mu - 1)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ . Because and the curvature of the lens are the same in this case, f is constant. As a result, inverting the lens has no effect on the image's position or nature.

20. Three immiscible liquids of densities \[{d_1}{\text{ }} > {d_2}{\text{ }} > {d_3}\] and refractive indices \[\;{\mathbf{\mu }}{{\text{ }}_{\mathbf{1}}}{\text{ }} > {\text{ }}{{\mathbf{\mu }}_2}{\text{ }} > {\text{ }}{{\mathbf{\mu }}_{\mathbf{3}}}\] are put in a beaker. The height of each liquid column is $\dfrac{h}{3}$ . A dot is made at the bottom of the beaker. For near normal vision, find the apparent depth of the dot.

Ans: Because the liquids are incompatible, they must be placed

from bottom to top. ${d_1},{d_2},{d_3}$ with their refractive indices ${\mu _1},{\mu _2},{\mu _3}$ respectively separated by layers ${l_1},{l_2}$ Consider the layer ${l_1}$ . If the object is at P, then the distance between the image of P and the liquid of refractive index from layer is ${l_1}$

\[{x_1} = \dfrac{{\dfrac{h}{3}}}{{_2{\mu _1}}}\]

\[_2{\mu _1} = \dfrac{{{\mu _1}}}{{{\mu _2}}}\]

\[{x_1} = \dfrac{{\dfrac{h}{3}}}{{3\dfrac{{{\mu _1}}}{{{\mu _2}}}}} = \dfrac{h}{3}\dfrac{{{\mu _1}}}{{{\mu _2}}}\]

This image ${l_1}$ acts as object for liquid of \[{\mu _2}\] and form the image at ${l_2}$. The distance \[{x_2}\] below level ${l_2}$

\[ = \left( {\dfrac{h}{3} + {x_1}} \right)\]

\[{x_2} = \dfrac{{{\text{Real depth}}}}{{_3{\mu _2}}} = \dfrac{{{\text{Real depth}}}}{{\dfrac{{{\mu _1}}}{{{\mu _2}}}}}\]

\[{x_2} = \dfrac{{{\mu _3}}}{{{\mu _2}}}\left[ {\dfrac{h}{3} + {x_1}} \right] = \dfrac{{{\mu _3}}}{{{\mu _2}}}\left[ {\dfrac{h}{3} + \dfrac{h}{3}\dfrac{{{\mu _2}}}{{{\mu _1}}}} \right] = \dfrac{h}{3}\dfrac{{{\mu _3}}}{{{\mu _2}}}\left[ {1 + \dfrac{{{\mu _2}}}{{{\mu _1}}}} \right] = \dfrac{h}{3}\left[ {\dfrac{{{\mu _3}}}{{{\mu _2}}} + \dfrac{{{\mu _3}}}{{{\mu _1}}}} \right]\]

As the point P is seen from

\[{x_3} = \dfrac{{{\mu _0}}}{{{\mu _3}}}\left[ {\dfrac{h}{3} + {x_2}} \right] = \dfrac{1}{{{\mu _3}}}\left[ {\dfrac{h}{3} + \dfrac{h}{3}\left( {\dfrac{{{\mu _3}}}{{{\mu _2}}} + \dfrac{{{\mu _3}}}{{{\mu _1}}}} \right)} \right]\]

Since \[[{\mu _0} = 1]\] ,refractive index of air

\[ = \dfrac{h}{3}\left[ {\dfrac{1}{{{\mu _3}}} + \dfrac{1}{{{\mu _3}}}\left( {\dfrac{{{\mu _3}}}{{{\mu _2}}} + \dfrac{{{\mu _3}}}{{{\mu _1}}}} \right)} \right] = \dfrac{h}{3}\left[ {\dfrac{1}{{{\mu _3}}} + \dfrac{1}{{{\mu _2}}} + \dfrac{1}{{{\mu _1}}}} \right]\]

\[{X_3}\] is apparent depth.

21. The angle of minimum deviation for a glass prism ( $\mu = \sqrt 3 $ ) is equal to the prism's angle. Determine the prism's angle.

Ans: The necessary relation for the smallest angle of departure is

$\mu = \dfrac{{\sin \left( {\dfrac{{A + {\alpha _m}}}{2}} \right)}}{{\sin \dfrac{A}{2}}}$

${\alpha _m} = A$

$\mu = \dfrac{{\sin \left( A \right)}}{{\sin \dfrac{A}{2}}} = \mu = \dfrac{{2\sin \left( {\dfrac{A}{2}} \right)\cos \left( {\dfrac{A}{2}} \right)}}{{\sin \dfrac{A}{2}}} = 2\cos \left( {\dfrac{A}{2}} \right)$

$\sqrt 3 = 2\cos \left( {\dfrac{A}{2}} \right)$

$\cos \dfrac{A}{2} = \dfrac{{\sqrt 3 }}{2} = \cos 30^\circ $

$\dfrac{A}{2} = 30^\circ $

A = 60° is angle of prism

22. A short item of length L is positioned distant from focus along the major axis of a concave mirror. The distance between the item and the observer is u. What is the length of the picture if the mirror has a focal length f? You may take L <<| v – f |.

Ans: Because the object's average distance from the mirror is u,

\[{u_1} = u - \dfrac{L}{2}{\text{ }}\;and{\text{ }}\;{u_2} = (u + \dfrac{L}{2})\]

Allow the picture of the object's two ends to develop at a distance. As a result, the picture length on the principal axis\[\;L\prime ({v_1} - {v_2})\]

$\dfrac{1}{f} = \left( {\dfrac{1}{v} + \dfrac{1}{u}} \right)$

\[\;L\prime = ({v_1} - {v_2}) = \dfrac{{\left( {u - \dfrac{L}{2}} \right)f}}{{\left( {u - \dfrac{L}{2}} \right) - f}}{\text{ - }}\dfrac{{\left( {u + \dfrac{L}{2}} \right)f}}{{\left( {u + \dfrac{L}{2}} \right) - f}}{\text{ = f - }}\dfrac{{\left( {u - \dfrac{L}{2}} \right)f}}{{\left( {u - f - \dfrac{L}{2}} \right)}} - \dfrac{{\left( {u + \dfrac{L}{2}} \right)}}{{\left( {u - f + \dfrac{L}{2}} \right)}}\]

\[{\text{ = f}}\left[ {\dfrac{{\left( {u - \dfrac{L}{2}} \right)\left( {u - f + \dfrac{L}{2}} \right){\text{ }} - \left( {u + \dfrac{L}{2}} \right)\left( {u - f - \dfrac{L}{2}} \right)}}{{\left( {u - f - \dfrac{L}{2}} \right)\left( {u - f + \dfrac{L}{2}} \right)}}} \right]\]

\[\therefore L < < (u - f)\]

\[{L^2} = f\left[ {\dfrac{{\dfrac{{fL}}{2} + \dfrac{{fL}}{2}}}{{{{(u - f)}^2}}}} \right]\]

\[L = \dfrac{{ffL}}{{{{(u - f)}^2}}} = \dfrac{{{f^2}L}}{{{{(u - f)}^2}}}\]

23. Inside an opaque hemispherical bowl of radius 'a', a circular disc of radius 'R' is arranged co-axially and horizontally. When viewed from the bowl's edge, the disc's far edge is barely visible. The bowl is filled with a clear liquid with a high refractive index allowing the discs near edge to be seen. How far below the bowl's rim is the disc positioned?

The rays from the ends of disc AB reaching one end of the bowl at M are shown in figures AM and BM. MN is tangent at M, therefore \[{\text{ MN}} \bot {\text{AB i}}{\text{.e }}\angle {\text{N = 90}}^\circ \]

Using the incident beam BM and the refracted ray MD as examples

BN=CN−CB=OM−CB=a−R

\[MB = \sqrt {{d^2} + (a - {R^2})} \]

\[sini = \dfrac{{BN}}{{BM}} = \dfrac{{a - R}}{{\sqrt {{d^2} + {{(a - R)}^2}} }}\]

\[\angle r = \angle a = \angle AMN\]

\[sinr = cos(90^\circ - \alpha ) = \dfrac{{AN}}{{AM}}{\text{ }} = \dfrac{{a + R}}{{{d^2} + {{(a + R)}^{2}}}}\]

\[\dfrac{1}{\mu_{1}}=\dfrac{\sin i}{\sin r} \quad\]

\[\dfrac{1}{\mu_{1}}=\dfrac{\dfrac{a-r}{\sqrt{d^{2}+(a+R)^{2}}}}{\dfrac{a+r}{\sqrt{d^{2}+(a-R)^{2}}}}=\dfrac{(a-R) \sqrt{d^{2}+(a+R)^{2}}}{(a-R) \sqrt{d^{2}+(a-R)^{2}}}=\dfrac{\mu\left(a^{2}-d^{2}\right)}{\sqrt{(a+R)^{2}-\mu(a-R)^{2}}}\]

\[ = \dfrac{{(a - R)\sqrt {{d^2} + {{(a + R)}^2}} }}{{(a - R)\sqrt {{d^2} + {{(a - R)}^2}} }}\]

\[ = \dfrac{{\mu ({a^2} - {d^2})}}{{\sqrt {{{(a + R)}^2} - \mu {{(a - R)}^2}} }}\]

24. 0.5 cm above the primary axis, a thin convex lens with a focal length of 25 cm is split into two halves. The top section is at (0,0), whereas the item is at (–50 cm, 0). Find the image's coordinates.

Ans: The object is 0.5cm above the primary axis.

u = -50cm f = +25cm, v =?

$\dfrac{1}{f} = \left( {\dfrac{1}{v} - \dfrac{1}{u}} \right)$

$\dfrac{1}{v} - \dfrac{1}{{ - 50}} = \dfrac{1}{{25}}$

$ = \dfrac{{2 - 1}}{{50}} = \dfrac{1}{{50}}$

$m = \dfrac{v}{u} = \dfrac{{50}}{{ - 50}} = - 1$

As a result, the picture is the same size as the item, and because m is negative, the image is inverted.

As a result, the picture is located at (50cm, -1cm) and 0.5 cm below the X –X' axis.

25. In many experimental set-ups the source and screen are fixed at a distance say D and the lens is movable. Show that there are two positions for the lens for which an image is formed on the screen. Find the distance between these points and the ratio of the image sizes for these two points.

Ans:

\[u = - (D - v)\]

\[\dfrac{1}{f}{\text{ }} = \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{{D - v}}\]$$

\[ = \dfrac{{D - v + v}}{{v(D - v)}} = - \dfrac{D}{{[vD - {v^2}]}}\]

\[fD = vD - {v^2}\]

\[{v^2} - vD + fD = 0\]

\[v = \dfrac{{ + D \pm \sqrt {{D^2} + 4Df} }}{2} = \dfrac{D}{2} \pm \dfrac{{\sqrt {D{}^2 + 4Df} }}{2}.............(1)\]

\[u = - (D - v) = - [D - (\dfrac{D}{2} \pm \dfrac{{\sqrt {{D^2} - 4Df} }}{2})]\]

\[u = - [\dfrac{D}{2} \pm \dfrac{{\sqrt {D2 - 4Df} }}{2}].............(2)\]

When the location of object \[{u_2}\]is\[ = \dfrac{D}{2}{\text{ }} + \dfrac{{\sqrt {{D^2} + 4Df\;} }}{2}\]

When the location of object \[{v_2}\]is\[ = \dfrac{D}{2}{\text{ }} - \dfrac{{\sqrt {{D^2} - 4Df\;} }}{2}\]

When the location of object \[{u_1}\]is\[ = \dfrac{D}{2}{\text{ }} - \dfrac{{\sqrt {{D^2} - 4Df\;} }}{2}\]

When the location of object \[{u_1}\]is\[ = \dfrac{D}{2}{\text{ }} + \dfrac{{\sqrt {{D^2} - 4Df\;} }}{2}\]

\[\;d = {v_{1}} - v{}_{2}\]

\[d = \left[ {\dfrac{D}{2}{\text{ }} \pm \dfrac{{\sqrt {{D^2} - 4Df\;} }}{2}} \right] - \left[ {\dfrac{D}{2}{\text{ }} \pm \dfrac{{\sqrt {{D^2} - 4Df\;} }}{2}} \right]\]

Case 1 : ${L_1}$

\[{u_1} = \dfrac{D}{2} - - \dfrac{d}{2}\]

\[{v_1} = \dfrac{D}{2} - + \dfrac{d}{2}\]

\[{m_1} = \dfrac{{{v_1}}}{{{u_1}}} = \dfrac{{\dfrac{D}{2} - \dfrac{d}{2}}}{{\dfrac{D}{2} + \dfrac{d}{2}}} = \dfrac{{D - d}}{{D + d}}\]

Case 2 : ${L_2}$

\[{u_2} = \dfrac{D}{2}{\text{ }} + \dfrac{d}{2}\]

\[{v_2} = \dfrac{D}{2}{\text{ }} - \dfrac{d}{2}\]

\[{m_2} = \dfrac{{{v_2}}}{{{u_2}}} = \dfrac{{\dfrac{D}{2} - \dfrac{d}{2}}}{{\dfrac{D}{2} + \dfrac{d}{2}}} = \dfrac{{D - d}}{{D + d}}\]

\[\dfrac{{{m_2}}}{{{m_1}}} = \left( {\dfrac{{D - d}}{{D + d}}} \right){^2}\]

26. A clear liquid with a refractive index $(\mu )$ is placed in a jar with a height of h. On the bottom surface of the jar, there is a dot in the centre. Find the smallest diameter of a disc that will make a dot invisible when put on the top surface symmetrically around the centre.

Ans: If the light ray emanating from O does not come out, or if it grazes the surface of a liquid, or if it receives entire internal reflection, the point O will be invisible. Ray OA is incident at A with critical angle ${i_c}$ and a ninety-degree angle of refraction.

If the incidence angle is more than ${i_c}$, the beam will get entire internal reflection from the surface and will not emerge through the liquid.

As a result, a disc with a diameter of d is required to restrict the rays from 'O' escaping the liquid.

$\sin i = \dfrac{1}{\mu }$

$\dfrac{{d/2}}{h} = \tan i$

.$\dfrac{d}{2} = h\tan i = h{[\sqrt {{\mu ^2} - 1} ]^{ - 1}}$

$d = \dfrac{{2h}}{{\sqrt {{\mu ^2} - 1} }}$

27 . The far point of a myopic adult is 0.1 m. His accommodation power is 4 dioptres.

(i) How high of a magnification is necessary to observe distant objects?

(ii) Without his spectacles, what is his near point?

(iii) With his glasses on, what is his near point? (Assume a 2 cm image gap between the lens of the eye and the retina.)

i)The lens power necessary to see an item at infinity clearly.

\[U = - \infty \]

\[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\]

\[\dfrac{1}{f} = \dfrac{1}{{ - 10}} - \dfrac{1}{\infty }\]

\[P = \dfrac{1}{f}m\]

\[\dfrac{1}{f} = \dfrac{1}{{10}}\]

\[P = \dfrac{1}{{ - 0.1}}m\]

(ii ) if no corrective lenses are worn: Let the eye's capabilities be a \[{p_f}\]and \[{p_n}\] when the item is far away and a p f and pn when the object is close

\[{P_{n}} = {P_f} + {P_{a}}\] When an object is far away, a clear picture is created at the retina 2cm from the eye lens (u=1cm=0.1m, V=2cm=0.02m).

If f is the focal length of an eye lens focused at a distance, then

\[\dfrac{1}{f}{\text{ }} = \dfrac{1}{v} - \dfrac{1}{u}{\text{ }} = \dfrac{1}{{0.02}} - \dfrac{1}{{( - 0.1)}}\]

\[\dfrac{1}{f} = 50 + 10 = 60\]

\[{P_f} = 60D\]

\[\;{P_{n}} = {P_f} + {P_a} = 60 + 4 = 64D\]

\[u = {x_n}\;(v = 2cm = 0.02m)\]

\[\dfrac{1}{f}{\text{ }} = \dfrac{1}{v} - \dfrac{1}{u}\]

\[\dfrac{1}{{0.02}} + \dfrac{1}{{{x_n}}} = Power({P_n})\]

\[50 + \dfrac{1}{{{x_n}}} = 64\]

\[\dfrac{1}{{{x_n}}} = 64 - 50 = 14D\]

\[{x_n} = \dfrac{1}{{14}}m = \dfrac{{100}}{{14}}cm = 7cm\]

(iii) When corrective lenses are used: When corrective lenses are worn, the eye is able to view the object at infinity. P = $\infty $ and v=2cm=0.02m are the eye lens powers in this circumstance.

\[{P_{\infty }} = \dfrac{1}{{0.02}}{\text{ }} - \dfrac{1}{\infty } = 50\]

\[{P_n} = {P_{\infty }} + {P_a} = 50 + 4 = 54D\]

\[u = - {x_{n}}m{\text{ , }}v = + 2cm = 0.02m\]

\[\dfrac{1}{f}{\text{ }} = 54\]

\[\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}{\text{ }}\]

\[\dfrac{1}{{0.02}} - \dfrac{1}{{{x_n}}} = 54\]

${x_n} = 0.25m$

28. Demonstrate that light incident at any angle on a material with refractive index \[\;\mu \geqslant \sqrt 2 \] must be steered down a length perpendicular to the incident face.

Ans: Consider a refractive index \[\;\mu \geqslant \sqrt 2 \] slab that is rectangular in shape. At the incidence point, an incidence ray incidence at angle I on face PQ. Refracted ray BC collides with face QR, which is perpendicular to PQ and has an incidence angle of\[{i_c}\], causing refracted ray CD to pass normal to the face PQ, as required by the question. As a result, \[{i_c}\] must be a critical viewpoint.

Snell’s law at C

\[\mu = \dfrac{1}{{sin{i_c}}}\]

\[sin{i_c} \geqslant \dfrac{1}{\mu }{\text{ }}[sin{i_c} = \dfrac{1}{\mu }]\]

\[sin(90 - r) \geqslant \dfrac{1}{\mu }\]

\[[\because r + 90 + {i_c} = 180]\]

\[cos{\text{ }}r \geqslant \dfrac{1}{\mu }\]

Square both side

\[co{s^2}r \geqslant \dfrac{1}{{{\mu ^2}}}\]

\[1 - co{s^2}r \leqslant 1 - \dfrac{1}{{{\mu ^2}}}\]

\[si{n^2}r \leqslant co{s^2}r \leqslant 1 - \dfrac{1}{{{\mu ^2}}}.............(1)\]

\[\;sin{\text{ }}i = \mu sin{\text{ }}r \Rightarrow si{n^2}i = {\mu ^2}si{n^2}r\]

\[\dfrac{1}{{{\mu ^2}}}si{n^2}i = sin{}^2r..............(2)\]

\[\dfrac{1}{{\mu {}^2}}si{n^2}i \leqslant 1 - \dfrac{{1}}{{{\mu ^2}}}\]

\[si{n^2}i \leqslant {\mu ^2} - 1\;\]

\[si{n^2}90 \leqslant {\mu ^2} - 1[\because \mu \geqslant \sqrt 2 ]\]

\[1 + 1 \leqslant {\mu ^2}\]

\[2 \leqslant {\mu ^2}\]

\[\sqrt {2} \leqslant \mu \]

29. In a long vertical column (i.e., horizontal dimensions vertical dimensions), a mixing of a pure liquid and a solution generates diffusion of solute particles and hence a refractive index gradient along the vertical dimension. A beam of light that enters the column at an angle to the vertical is diverted from its intended course. Determine the deviation when travelling a horizontal distance of d h, the column's height.

Consider a long, translucent vertical column with an unlimited height (h) and thickness (dx). Consider the case of a beam AB that enters at an angle into a liquid column of height y and exits at an angle θ into liquid of height y in column of liquid and emerges at an angle (θ +d θ) at height (y+ dy).

From Snell’s law,

\[\mu \left( y \right)\sin \theta = y\left( {y + dy} \right)\sin \left( {\theta + d\theta } \right)\]

\[\mu \left( y \right)\sin \theta ,\left( {\mu \left( y \right) + \dfrac{{d\mu }}{{dy}}dy} \right)\left( {\sin \theta \cos d\theta + \cos \theta \sin d\theta } \right)\]

\[\mu \left( y \right)\sin \theta \mu \left( y \right)\cos \theta d\theta + \dfrac{{d\mu }}{{dy}}dy\sin \theta \]

\[\mu (y)cos\theta d\theta ,\,\dfrac{{d\mu }}{{dy}}dysin\theta d\theta ,\,\dfrac{{ - d\mu }}{{\mu dy}}dytan\theta \]\[\]

\[tan\theta = \dfrac{{dx}}{{dy}}\]

On solving , we have

\[\therefore \;\:\;\:d\theta = \dfrac{{ - 1d\mu }}{{\mu dy}}dx\]

Solving this variable separable form of differential equation

. \[\therefore \;\theta = \dfrac{{ - 1d\mu }}{{\mu dy}}\int_o^d {dx} = \dfrac{{ - 1d\mu }}{{\mu dy\,}}d\]

30. When light comes close to a big object, the gravitational interaction bends the beam. This may be explained by a change in the medium's effective refractive index, which is given by

\[n(r) = 1 + \dfrac{{2GM}}{{r{c^2}}}\]

where r is the distance between the point of interest and the large body's mass centre, G is the universal gravitational constant, M is the mass of the body, and c is the speed of light in vacuum. Consider a spherical object and calculate the ray's divergence from its initial path when it grazes the object.

Ans: Consider two spherical surfaces separated by r and (r + dr) from the centre of a large object with radius R and mass m

Consider two spherical surface at r and (r+dr) distance from the centre of massive object of mass m and radius R

A ray ABCD incident at B and C on two surface at r and (r+dr) then, by Snell's law

\[\mu \left( r \right)sin\theta \cong \mu \left( {r + ar} \right)sin\left( {\theta + d\theta } \right)\mu \left( r \right)\]

\[\mu \left( r \right)sin\theta \cong [\mu \left( r \right) + \dfrac{{d\mu }}{{dr}}dr](sin\theta .cosd\theta + cos\theta sin{\text{ }}d\theta )\]

\[d\theta \to 0\;{\text{ }}\;cosd\theta = 1\;and\;d\theta = 0\]

\[\mu \left( r \right)sin\theta \cong [\mu \left( r \right) + \dfrac{{d\mu }}{{dr}}dr]\left( {sin\theta + cos\theta d\theta } \right)\]

\[\mu \left( r \right)sin\theta \cong \mu \left( r \right)sin\theta + d\mu sin\theta + \mu \left( r \right)cos\theta d\theta + d\mu d\theta cos\theta \]

\[d\mu d\theta cos\theta \]is very small as\[\;d\mu \] and \[d\theta \] are very small

\[\therefore d\mu d\theta cos\theta \cong 0\]

\[\therefore \mu \left( r \right)sin\theta \cong \mu \left( r \right)sin\theta + d\mu sin\theta + \mu cos\theta .d\theta \]

\[0 \cong d\mu sin\theta + \mu \left( r \right)cos\theta .d\theta \]

\[ - d\mu sin\theta = \mu \left( r \right)cos\theta .d\theta \]

Dividing both side by dr

\[ - \dfrac{{d\mu }}{{dr}}sin\theta = \mu \left( r \right)cos\theta \dfrac{{d\theta }}{{dr}}.............\left( 1 \right)\]

\[\mu \left( r \right) = 1 + \dfrac{{2GM}}{{{c^2}}}.\dfrac{1}{r}\left( {given} \right)\]

\[\dfrac{{d\mu }}{{dr}} = 0 + \dfrac{{2GM}}{{c{}^2}}\left( { - 1} \right){r^{ - 2}}\]

\[ - \dfrac{{d\mu }}{{dr}} = [\dfrac{{2GM}}{{{c^2}{r^2}}}]...............\left( 2 \right)\]

\[\dfrac{{2GMsin\theta }}{{{c^2}{r^2}}} = [1 + \dfrac{{2GM}}{{{c^2}{r^2}}}]cos\theta \dfrac{{d\theta }}{{dr}}\]

As the G is very small and \[{c^2}\] is very large so

\[\dfrac{{GM}}{{{c^2}{r^2}}}{\text{ }} \to 0\]

\[\dfrac{{2GMcos\theta sin\theta }}{{{c^2}{r^2}}} = \dfrac{{d\theta }}{{dr}}\]

\[\dfrac{{2GMtan\theta }}{{{c^2}{r^2}}} \cong drd\theta \]

\[\dfrac{{2GM}}{{c{}^2}}\smallint \dfrac{{tan\theta }}{{{r^2}}}dr = \int\limits_0^{{\theta _0}} {d\theta } \]

\[\;{r^2} = {x^2} + {R^2}\]

Distinguish between the two sides

\[2rdr = 2x.dx\]

\[dr = \dfrac{x}{r}dx\]

∴ \[\int\limits_0^{{\theta _0}} {d\theta } = \dfrac{{2GM}}{{{c^2}}}\int\limits_\infty ^{ - \infty } {\dfrac{1}{{{r^2}}}{\text{ }}} \dfrac{R}{r}(tan\theta = \dfrac{{R}}{x}) = \dfrac{{2GM}}{{{c^2}}}\int\limits_\infty ^{ - \infty } {\dfrac{1}{{{r^2}}}{\text{ }}} \dfrac{R}{{x{r^3}}}xdx = \dfrac{{2GM}}{{{c^2}}}\int\limits_\infty ^{ - \infty } {\text{ }} \dfrac{R}{{{{({x^2} + R{}^2)}^{3/2}}}}dx\]

\[\int\limits_0^{{\theta _0}} {d\theta } = \dfrac{{2GM}}{{{c^2}}}\int\limits_\infty ^{ - \infty } R \dfrac{{dx}}{{{{({x^2} + {R^2})}^{3/2}}}}[\because x = Rtan\theta ;dx = Rse{c^2}\theta d\theta ]\]

\[{\theta _0} = \dfrac{{2GM}}{{{c^2}}}\int\limits_{ - \pi /2}^{\pi /2} {\dfrac{{R.Rse{c^2}\theta d\theta }}{{{{({R^2} + ta{n^2}\theta {R^2})}^{3/2}}}}} = \dfrac{{2GM}}{{{c^2}}}\int\limits_{ - \pi /2}^{\pi /2} \dfrac{{{R^2}se{c^2}\theta d\theta }}{{{{[{R^2}(1 + ta{n^2}\theta )]}^{3/2}}}}\]\[\int\limits_0^{{\theta _0}} {d\theta } = \dfrac{{2GM}}{{{c^2}}}\int\limits_{ - \pi /2}^{\pi /2} {\dfrac{{{R^2}se{c^2}\theta d\theta }}{{(R{}^3 + se{c^3}\theta )}}} = \dfrac{{2GM}}{{{c^2}}}\int\limits_{ - \pi /2}^{\pi /2} {\dfrac{1}{{sec\theta }}} d\theta \]

\[ = \dfrac{{2GM}}{{R{c^2}}}\int\limits_{ - \pi /2}^{\pi /2} {cos\theta d\theta } {\text{ }} = \dfrac{{2GM}}{{R{c^2}}}\left[ {sin\theta } \right]_{ - \pi /2}^{\pi /2} = \dfrac{{2GM}}{{R{c^2}}}[sin\dfrac{\pi }{2} - sin( - \dfrac{\pi }{2})]\]

\[{\theta ^0} = \dfrac{{2GM}}{{R{c^2}}}[1 + sin\dfrac{\pi }{2}] = \dfrac{{2GM}}{{R{c^2}}}\left[ {1 + 1} \right]\]

\[ \Rightarrow {\theta _0} = \dfrac{{4GM}}{{R{c^2}}}\]

31. An infinitely long cylinder of radius R is made of an unusual exotic material with refractive index –1 The cylinder is placed between two planes whose normal are along the y direction. The centre of the cylinder O lies along the y -axis. A narrow laser beam is directed along the y direction from the lower plate. The laser source is at a horizontal distance x from the diameter in the y direction. Find the range of x such that light emitted from the lower plane does not reach the upper plane.

Ans: Because the cylinder is formed of a material with a refractive index of -1 and is put in air , of μ=1 so, when beam AB collides with cylinder B, ${\theta _r}$ will be negative, implying that the refractive ray will be reflected from the normal. At incidence point C, the same thing happens, and the angle of refraction and incident at B and C will be equal. (${\theta _r}$) as OB=OC=R and of refraction at C is ${\theta _r}$.${\theta _1} = |{\theta _i}| = |{\theta _r}|$ as reflection takes place

The overall difference between the outgoing and entering rays. 4${\theta _1}$.Rays are not allowed to reach the receiving plane$\dfrac{\pi }{2} \leqslant {\theta _1} \leqslant \dfrac{{3\pi }}{2}$ angles measured from the y axis in a clockwise direction

$\sin {\theta _1} = \dfrac{{3\pi }}{2}$

$\dfrac{\pi }{8} \leqslant {\sin ^{ - 1}}\dfrac{x}{R} \leqslant \dfrac{{3\pi }}{8}$

As a result, if light produced by the source does not reach the receiving plane,$\dfrac{{R\pi }}{8} \leqslant x \leqslant \dfrac{{3R\pi }}{8}$.

32. (i) Consider a narrow lens in front of a source (S) and an observer (O) (O) Allow the lens thickness to change as needed. $W(b) = {W_0} - \dfrac{{{b^2}}}{\alpha }$ where an is the horizontal distance from the pole and b is the vertical distance from the pole. ${W_0}$ is a fixed value. Find the condition that all paraxial rays beginning from the source will converge at a point 'O' on the axis using Fermat's principle, i.e. the duration of transit for a ray between the source and observer is an extremum. Calculate the focal length.

(ii) A gravitational lens is thought to have a changing form width.

\[W(b) = {K_1}\log \left( {\dfrac{{{K_2}}}{b}} \right)({b_{\min }} < b < {b_{\max }})\]

\[ = {K_1}\log \left( {\dfrac{{{K_2}}}{{b{}_{\min }}}} \right)({b_{\min }} < b)\]

Demonstrate that an observer will view an image of a point item as a ring with an angular radius around the lens's centre.

$\beta = \sqrt {\dfrac{{(n - 1){K_1}\dfrac{u}{v}}}{{u + v}}} $

Ans: The time taken by ray from S to \[{P_1}\]

\[{t_1} = \dfrac{{\sqrt {{u^2} + {b^2}} }}{c} = \dfrac{u}{c}{\left( {1 + \dfrac{{{b^2}}}{{{u^2}}}} \right)^{\dfrac{1}{2}}}\]

\[{t_1} = \dfrac{u}{c}\left( {1 + \dfrac{{{b^2}}}{{2{u^2}}}} \right)\] (assuming b <<<u)

Similarly, time required by ray from \[{P_1}\] to 0

\[{t_2} = \dfrac{v}{c}\left( {1 + \dfrac{{{b^2}}}{{2{v^2}}}} \right)\]

The amount of time it takes for a beam to go through the lens

\[{t_3} = \dfrac{{(\mu - 1)W(b)}}{c}\]

Thus total time by ray from S to O=t=t1+t2+t3

\[t = \dfrac{u}{c}\left( {1 + \dfrac{{{b^2}}}{{2{u^2}}}} \right) + \dfrac{v}{c}\left( {1 + \dfrac{{{b^2}}}{{2{v^2}}}} \right) + \dfrac{{(\mu - 1)W(b)}}{c}\]

\[ = \dfrac{1}{c}\left[ {u + \dfrac{{{b^2}}}{{2u}} + v + \dfrac{{{b^2}}}{{2v}} + (\mu - 1)W(b)} \right]\]

\[ = \dfrac{1}{c}\left[ {u + v + \dfrac{{{b^2}}}{2}\left( {\dfrac{1}{u} + \dfrac{1}{v}} \right) + (\mu - 1)W(b)} \right]\]

If \[\dfrac{1}{D} = \dfrac{1}{v} + \dfrac{1}{u}\]

\[t = \dfrac{1}{c}\left[ {u + v + \dfrac{{{b^2}}}{{2D}} + (\mu - 1)\left( {{W_o} - \dfrac{{{b^2}}}{\alpha }} \right)} \right]\]

Fermat’s Principle

\[\dfrac{{dt}}{{db}} = \dfrac{1}{c}\left[ {0 + 0 + \dfrac{{2b}}{{2D}} + (\mu - 1)\left[ {0 - \dfrac{{2b}}{\alpha }} \right]} \right] = \dfrac{b}{{cD}} - (\mu - 1)\left[ { - \dfrac{{2b}}{\alpha }} \right]\]

But time from S to O remains constant

\[\dfrac{{2(\mu - 1)b}}{{\alpha c}} = \dfrac{b}{{cD}}\]

\[\alpha = 2(\mu - 1)D\]

Thus, the convergence lens is formed \[\alpha =2\left ( \mu -1 \right )D\].

Because it is unaffected by B, all paraxial rays from S will converge at O (for rays b <<u and b <<v).Since \[\dfrac{1}{D} = \dfrac{1}{v} + \dfrac{1}{u}\]

it is equivalent to focal length f

\[t = \dfrac{1}{c}\left[ {u + v + \dfrac{{{b^2}}}{{2D}} + (\mu - 1)W(b)} \right]\]

Here $W(b) = {K_1}{\log _e}\left( {\dfrac{{{K_2}}}{b}} \right)$

\[t = \dfrac{1}{c}\left[ {u + v + \dfrac{{{b^2}}}{{2D}} + (\mu - 1){K_1}\log \dfrac{{{K_2}}}{b}} \right]\]

\[\dfrac{{dt}}{{db}} = \dfrac{1}{c}\left[ {0 + 0 + \dfrac{{2b}}{{2D}} + (\mu - 1)\dfrac{b}{{{K_2}}}.{K_1}} \right]\dfrac{d}{{db}}\left( {\dfrac{{{K_2}}}{b}} \right)\]

\[ = \dfrac{1}{c}\left[ {\dfrac{b}{D} + (\mu - 1)\dfrac{{{K_1}}}{{{K_2}}}b.{K_2}( - 1)\dfrac{1}{{{b^2}}}} \right]\]

Because the period between S and O is constant, so, $\dfrac{{dt}}{{db}} = 0$

So \[0 = \dfrac{1}{c}\left[ {\dfrac{b}{D} - (\mu - 1)\dfrac{{{K_1}}}{b}} \right]\]

\[(\mu - 1)\dfrac{{{K_1}}}{b} = \dfrac{b}{D}\]

${b^2} = {K_1}D(\mu - 1)$

$b = \sqrt {(\mu - 1){K_1}D} $

As a result, all rays going through at a height of b will contribute to the picture. The beam path is angled.

$\beta = \dfrac{b}{v} = \dfrac{{\sqrt {(\mu - 1){K_1}D} }}{v} = \sqrt {\dfrac{{(\mu - 1){K_1}D}}{{{v^2}}}} $

$\beta = \sqrt {\dfrac{{(\mu - 1){K_1}u}}{{v(u + v)}}} \left( {D = \dfrac{{uv}}{{u + v}}} \right)$

NCERT Exemplar for Class 12 Physics - Ray Optics And Optical Instruments

NCERT Exemplar for Class 12 Physics - Ray Optics And Optical Instruments is a part of the NCERT Exemplar for Class 12 Physics. The complete study material focuses on every chapter given in the NCERT Class 12 Physics textbook. The highly important topics in these chapters are used to create the exercise section of every chapter in the NCERT Class 12 Physics textbook. And the NCERT Exemplar for Class 12 Physics provided by Vedantu provides Class 12 students with answers for those questions. These answers are expert-created and are completely relevant from CBSE's point of view.

The NCERT Exemplar for Class 12 Physics - Ray Optics And Optical Instruments is a thorough guide that students of Class 12 CBSE should study in order to score high and achieve a good rank in the Class 12 board examination.

On this very page, students can find the NCERT Exemplar for Class 12 Physics - Ray Optics And Optical Instruments in a downloadable PDF form completely free of charge. The study material is created by experts with the goal of providing the crucial knowledge Class 12 students need to perform well in the exams. The solutions given in the NCERT Exemplar for Class 12 Physics Chapter 9 - Ray Optics And Optical Instruments are created keeping in mind the NCERT guidelines and CBSE marking scheme. Any student of any Intelligence Quotient can understand and apply these solutions with ease. That is because the language used to make these solutions is simple and easy to comprehend by any Class 12 student.

Also, the topics provided in this study gear are organized in an extremely detailed manner.

The NCERT Exemplar for Class 12 Physics - Ray Optics And Optical Instruments consists of every question given in the exercise section of chapter 9 (Ray Optics And Optical Instruments), which includes multiple-choice questions, questions that require brief answers, and the questions that require long answers.

Chapter 9 - Ray Optics And Optical Instruments in the NCERT Class 12 Physics textbook is important for Class 12 students to study and understand. A lot of questions from different topics in these chapters are asked in the CBSE Class 12 physical examination.

Given Below are All the Crucial Topics Available in the Chapter 9:

9.1 Introduction

9.2 Reflection of Light by Spherical Mirrors

9.3 Refraction

9.4 Total Internal Reflection

9.5 Refraction at Spherical Surfaces and by Lenses

9.6 Refraction through a Prism

9.7 Dispersion by a Prism

9.8 Some Natural Phenomena due to Sunlight

9.9 Optical Instruments

FAQs on NCERT Exemplar for Class 12 Physics Chapter-9 (Book Solutions)

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Physics MCQs for Class 12 Chapter 9 Ray Optics and Optical Instruments

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Here we are providing MCQ Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instrument.

Q.1. Which of the following colour of white light deviated most when passes through a prism? (a) Red light (b) Violet light (c) Yellow light (d) Both (a) and (b)

Q.2. For a total internal reflection, which of the following is correct? (a) Light travels from rarer to denser medium. (b) Light travels from denser to rarer medium. (c) Light travels in air only. (d) Light travels in water only.

Q.3. Mirage is a phenomenon due to (a) refraction of light (b) reflection of light (c) total internal reflection of light (d) diffraction of light.

Q.4. Critical angle of glass is θ 2 and that of water is θ 2 . The critical angle for water and glass surface would be (μ g = 3/2, μ w = 4/3). (a) less than θ 2 (b) between θ 1 and θ 2 (c) greater than θ 2 (d) less than θ 1

Q.5. An astronomical refractive telescope has an objective of focal length 20 m and an eyepiece of focal length 2 cm. Then (a) the magnification is 1000 (b) the length of the telescope tube is 20.02 m (c) the image formed of inverted (d) all of these

Q.6. A convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. Then its focal length will (a) become zero (b) become infinite (c) become small, but non-zero (d) remain unchanged

Q.7. Which of the following forms a virtual and erect image for all positions of the object? (a) Concave lens (b) Concave mirror (d) Convex mirror (d) Both (a) and (c)

Q.8. Two lenses of focal lengths 20 cm and – 40 cm are held in contact. The image of an object at infinity will be formed by the combination at (a) 10 cm (b) 20 cm (c) 40 cm (d) infinity

Q.9. Two beams of red and violet color are made to pass separately through a prism (angle of the prism is 60°). In the position of minimum deviation, the angle of refraction will be (a) 30° for both the colors (b) greater for the violet color (c) greater for the red color (d) equal but not 30° for both the colors

Q.10. An under-water swimmer cannot see very clearly even in absolutely clear water because of (a) absorption of light in water (b) scattering of light in water (c) reduction of speed of light in water (d) change in the focal length of eye lens

Q.11. The field of view is maximum for (a) plane mirror (b) concave mirror (c) convex mirror (d) cylindrical mirror

Q.12. A person is six feet tall. How tall must a plane mirror be if he is able to see his entire length? (a) 3 ft (b) 4.5 ft (c) 7.5 ft (d) 6 ft

Q.13. The image formed by a concave mirror is (a) always real (b) always virtual (c) certainly real if the object is virtual (d) certainly virtual if the object is real

Q.14. When light is refracted into a medium, (a) its wavelength and frequency both increase (b) its wavelength increases but frequency remains unchanged (c) its wavelength decreases but frequency remains unchanged (d) its wavelength and frequency both decrease

Q.15. Which of the following phenomena is used in optical fibres ? (a) Total internal reflection (b) Scattering (c) Diffraction (d) Refraction

Q.16. Critical angle of light passing from glass to water is minimum for (a) red colour (b) green colour (c) yellow colour (d) violet colour

Q.17. If a glass prism is dipped in water, its dispersive power (a) increases (b) decreases (c) does not change (d) may increase or decrease depending on whether the angle of the prism is less than or greater than 60º

Q.18. The objective of a telescope must be of large diameter in order to (a) remove chromatic aberration (b) remove spherical aberration and high magnification (c) gather more light and for high resolution (d) increase its range of observation

Q.19. A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. The length of the image is (a) 10 cm (b) 15 cm (c) 2.5 cm (d) 5 cm

Q.20. To increase the angular magnification of a simple microscope, one should increase (a) the focal length of the lens (b) the power of the lens (c) the aperture of the lens (d) the object size

You may also check the MCQs on other chapters from the links provided below:

Hope this helps!

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Case Study Questions Class 12 Physics Ray Optics and Optical Instruments
CBSE Case Study Questions Class 12 Physics Ray Optics and Optical Instruments. Case study 1: As we know that, when light ray travels from one medium to another changes its direction of path due to the change in optical density of the medium. When light ray travels from optically denser medium to the optically rarer medium then some of the light ...
Case Study Questions for Class 12 Physics Chapter 9 Ray Optics and
Case Study Questions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Attempt any 4 sub parts from each question. Each question carries 1 mark. Case Study Question 1: A compound microscope is an optical instrument used for observing highly magnified images of tiny objects. Magnifying power of a compound microscope is defined … Continue reading Case Study Questions for Class ...
Class 12 Physics Chapter 9 Case Study Question Ray Optics and Optical
In Class 12 Boards there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Download of CBSE Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Case Study and Passage Based Questions with Answers were Prepared Based on Latest Exam Pattern.
12th Physics Ray Optics and Optical Instruments Case Study Questions
QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 12th Physics Subject - Ray Optics and Optical Instruments, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.
Case Study on Ray Optics & Optical Instruments Class 12 ...
The PDF file of the Ray Optics & Optical Instruments Case Study for Class 12 Physics with Solutions is a very important study resource that can help students better prepare for the exam and boost conceptual learning. The solutions are in the hint manner as well as contain full examples too, refer to the link to access the Case Study on Ray ...
CBSE 12th Physics Ray Optics and Optical Instruments Case Study
CBSE 12th Standard Physics Subject Ray Optics and Optical Instruments Case Study Questions 2021. 12th Standard CBSE. Reg.No. : Physics. Time : 00:30:00 Hrs. Total Marks : 20. A convex or converging lens is thicker at the centre than at the edges. It converges a parallel beam of light on refraction through it. It has a real focus.
Ray Optics
Ray Optics Case Study Questions will be discussed in this session by Educator Sunil Jangra for Class 12. He will also revise all the important questions for ... CBSE Exam, class 12.
12th Class Physics Ray Optics Question Bank
Read the following text and answer the following questions on the basis of the same: Negative Refractive Index: One of the most fundamental phenomena in optics is refraction. When a beam of light crosses the interface between two different materials, its path is altered depending on the difference in the refractive indices of the materials.
CBSE 12th Standard Physics Subject Ray Optics and Optical Instruments
QB365 Provides the updated CASE Study Questions for Class 12 , and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get more marks in Exams - Complete list of 12th Standard CBSE question papers, syllabus, exam tips, study material, previous year exam question papers, centum tips ...
[Latest] Ray Optics Questions And Answers |2024
Ray Optics Questions and Answers. Below are some of the very important NCERT Class 12 Physics Chapter 9 Ray Optics Questions and Answers. These Class 12 Ray Optics Questions and Answers have been prepared by expert teachers and subject experts based on the latest syllabus and pattern of term 2 questions with answers to help students understand the concept.
Important Questions for CBSE Class 12 Physics Chapter 9
Class 12 Physics Chapter 9 important questions deal with important questions from the chapter Ray Optics. This chapter explains various phenomenons like refraction, reflection, and dispersion with the help of diagrams of rays of light. Also, this chapter includes image formation on smooth as well as spherical surfaces along with how optical ...
CBSE Class 12 Physics Case Study Questions PDF Download
Chapter-wise Solved Case Study Questions for Class 12 Physics. Chapter 1 Electric Charges and Fields. Chapter 2 Electrostatic Potential and Capacitance. Chapter 3 Current Electricity. Chapter 4 Moving Charges and Magnetism. Chapter 5 Magnetism and Matter. Chapter 6 Electromagnetic Induction. Chapter 7 Alternating Current.
Case Study Ray Optics & Optical Instruments Class 12 Physics ...
In 2023-24 one Question of case study was from Ray optics & second question was from Current electricity. So we are providing few Questions of case study from Ray optics with solutions. We can provide Material Or we can take doubts. Rest is your hardwork. Without your hardwork any extra book, Reference books & Notes are waste.
CBSE Important Questions Class 12 Physics Chapter 9
With important questions for Class 12 physics chapter 9, students will get elaborated and authentic solutions to their doubts regarding ray optics and optical instruments. After preparing chapter 9 class 12 Physics important questions, students will be able to solve CBSE Sample papers, CBSE Past year question papers, and NCERT EXEMPLAR Questions.
Conceptual Questions Based on Class 12 Physics Ray Optics
Conceptual Questions Based on Class 12 Physics Ray Optics Here we are providing conceptual questions based on class 12 physics chapter 9 Ray Optics. These questions are prepared by experts and are very useful from exam point of view. Students can practice these questions for scoring better marks in the exam. Q.1. Why is a … Continue reading Conceptual Questions Based on Class 12 Physics Ray ...
Class 12th Physics
QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 12th Physics Subject - Ray Optics and Optical Instruments, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.
Case Study Questions of Ray Optics
Class 12 Physics 2023 : Case Study Questions of Ray Optics | Boards 2023Class 12 Boards 2023 : How to Pass in Physics in 1 Day | Sunil JangraClass 12 Physics...
Case Study on Ray Optics & Optical Instruments Class 12 ...
Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now. Case Study on Ray Optics & Optical Instruments Class 12 Physics: Here, you will get Case Study Questions on Class 12 Ray Optics & Optical Instruments PDF at free of cost. Along with you can also download Ray Optics ...
NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical
Solution: (a) The convex lens is placed in the path of convergent beam. The image 1 is formed by further converging beams at a distance of 7.5 cm from lens. (b) A concave lens is placed in the path of convergent' beam, the concave lens further diverge the light.
Assertion and Reason questions on class 12 Physics Chapter 9 Ray Optics
Directions: These questions consist of two statements, each printed as Assertion and Reason. While answering these questions, you are required to choose any one of the following four responses.(a) If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.(b) If both Assertion and Reason are correct but Reason … Continue reading Assertion and Reason ...
NCERT Exemplar for Class 12 Physics Chapter 9
The NCERT Exemplar for Class 12 Physics - Ray Optics And Optical Instruments consists of every question given in the exercise section of chapter 9 (Ray Optics And Optical Instruments), which includes multiple-choice questions, questions that require brief answers, and the questions that require long answers.
RAY OPTICS :5 Questions You Must Solve Before NSEP 2024 Exam
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Physics MCQs for Class 12 Chapter 9 Ray Optics and Optical Instruments
Q.3. Mirage is a phenomenon due to. (d) diffraction of light. Q.4. Critical angle of glass is θ2 and that of water is θ2. The critical angle for water and glass surface would be (μg = 3/2, μw = 4/3). Q.5. An astronomical refractive telescope has an objective of focal length 20 m and an eyepiece of focal length 2 cm.

Case Study Questions Class 12 Physics Ray Optics and Optical Instruments

CBSE Case Study Questions Class 12 Physics Ray Optics and Optical Instruments

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CBSE Important Questions Class 12 Physics Chapter 9

Important Questions for CBSE Class 12 Physics Chapter 9- Ray Optics and Optical Instruments

CBSE CLASS 12 PHYSICS CHAPTER-9 IMPORTANT QUESTIONS

1-MARK QUESTIONS

2-MARK QUESTIONS

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RELEVANCE OF RAY OPTICS IN FUTURE

WHY IS IT IMPORTANT TO STUDY IMPORTANT QUESTIONS FOR RAY OPTICS CLASS 12?

INTRODUCTION TO RAY OPTICS

What is a ray?

CONCLUSION –

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2. What is illuminance?

3. Why are diamonds known for their spectacular brilliance?

4. State the advantages of a reflecting type telescope over a refracting type of telescope.

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