young experiment diagram

Wave Optics

Young’s double slit experiment, learning objectives.

By the end of this section, you will be able to:

Explain the phenomena of interference.
Define constructive interference for a double slit and destructive interference for a double slit.

Although Christiaan Huygens thought that light was a wave, Isaac Newton did not. Newton felt that there were other explanations for color, and for the interference and diffraction effects that were observable at the time. Owing to Newton’s tremendous stature, his view generally prevailed. The fact that Huygens’s principle worked was not considered evidence that was direct enough to prove that light is a wave. The acceptance of the wave character of light came many years later when, in 1801, the English physicist and physician Thomas Young (1773–1829) did his now-classic double slit experiment (see Figure 1).

A beam of light strikes a wall through which a pair of vertical slits is cut. On the other side of the wall, another wall shows a pattern of equally spaced vertical lines of light that are of the same height as the slit.

Figure 1. Young’s double slit experiment. Here pure-wavelength light sent through a pair of vertical slits is diffracted into a pattern on the screen of numerous vertical lines spread out horizontally. Without diffraction and interference, the light would simply make two lines on the screen.

Why do we not ordinarily observe wave behavior for light, such as observed in Young’s double slit experiment? First, light must interact with something small, such as the closely spaced slits used by Young, to show pronounced wave effects. Furthermore, Young first passed light from a single source (the Sun) through a single slit to make the light somewhat coherent. By coherent , we mean waves are in phase or have a definite phase relationship. Incoherent means the waves have random phase relationships. Why did Young then pass the light through a double slit? The answer to this question is that two slits provide two coherent light sources that then interfere constructively or destructively. Young used sunlight, where each wavelength forms its own pattern, making the effect more difficult to see. We illustrate the double slit experiment with monochromatic (single λ ) light to clarify the effect. Figure 2 shows the pure constructive and destructive interference of two waves having the same wavelength and amplitude.

Figure a shows three sine waves with the same wavelength arranged one above the other. The peaks and troughs of each wave are aligned with those of the other waves. The top two waves are labeled wave one and wave two and the bottom wave is labeled resultant. The amplitude of waves one and two are labeled x and the amplitude of the resultant wave is labeled two x. Figure b shows a similar situation, except that the peaks of wave two now align with the troughs of wave one. The resultant wave is now a straight horizontal line on the x axis; that is, the line y equals zero.

Figure 2. The amplitudes of waves add. (a) Pure constructive interference is obtained when identical waves are in phase. (b) Pure destructive interference occurs when identical waves are exactly out of phase, or shifted by half a wavelength.

When light passes through narrow slits, it is diffracted into semicircular waves, as shown in Figure 3a. Pure constructive interference occurs where the waves are crest to crest or trough to trough. Pure destructive interference occurs where they are crest to trough. The light must fall on a screen and be scattered into our eyes for us to see the pattern. An analogous pattern for water waves is shown in Figure 3b. Note that regions of constructive and destructive interference move out from the slits at well-defined angles to the original beam. These angles depend on wavelength and the distance between the slits, as we shall see below.

The figure contains three parts. The first part is a drawing that shows parallel wavefronts approaching a wall from the left. Crests are shown as continuous lines, and troughs are shown as dotted lines. Two light rays pass through small slits in the wall and emerge in a fan-like pattern from two slits. These lines fan out to the right until they hit the right-hand wall. The points where these fan lines hit the right-hand wall are alternately labeled min and max. The min points correspond to lines that connect the overlapping crests and troughs, and the max points correspond to the lines that connect the overlapping crests. The second drawing is a view from above of a pool of water with semicircular wavefronts emanating from two points on the left side of the pool that are arranged one above the other. These semicircular waves overlap with each other and form a pattern much like the pattern formed by the arcs in the first image. The third drawing shows a vertical dotted line, with some dots appearing brighter than other dots. The brightness pattern is symmetric about the midpoint of this line. The dots near the midpoint are the brightest. As you move from the midpoint up, or down, the dots become progressively dimmer until there seems to be a dot missing. If you progress still farther from the midpoint, the dots appear again and get brighter, but are much less bright than the central dots. If you progress still farther from the midpoint, the dots get dimmer again and then disappear again, which is where the dotted line stops.

Figure 3. Double slits produce two coherent sources of waves that interfere. (a) Light spreads out (diffracts) from each slit, because the slits are narrow. These waves overlap and interfere constructively (bright lines) and destructively (dark regions). We can only see this if the light falls onto a screen and is scattered into our eyes. (b) Double slit interference pattern for water waves are nearly identical to that for light. Wave action is greatest in regions of constructive interference and least in regions of destructive interference. (c) When light that has passed through double slits falls on a screen, we see a pattern such as this. (credit: PASCO)

To understand the double slit interference pattern, we consider how two waves travel from the slits to the screen, as illustrated in Figure 4. Each slit is a different distance from a given point on the screen. Thus different numbers of wavelengths fit into each path. Waves start out from the slits in phase (crest to crest), but they may end up out of phase (crest to trough) at the screen if the paths differ in length by half a wavelength, interfering destructively as shown in Figure 4a. If the paths differ by a whole wavelength, then the waves arrive in phase (crest to crest) at the screen, interfering constructively as shown in Figure 4b. More generally, if the paths taken by the two waves differ by any half-integral number of wavelengths [(1/2) λ , (3/2) λ , (5/2) λ , etc.], then destructive interference occurs. Similarly, if the paths taken by the two waves differ by any integral number of wavelengths ( λ , 2 λ , 3 λ , etc.), then constructive interference occurs.

Both parts of the figure show a schematic of a double slit experiment. Two waves, each of which is emitted from a different slit, propagate from the slits to the screen. In the first schematic, when the waves meet on the screen, one of the waves is at a maximum whereas the other is at a minimum. This schematic is labeled dark (destructive interference). In the second schematic, when the waves meet on the screen, both waves are at a minimum.. This schematic is labeled bright (constructive interference).

Figure 4. Waves follow different paths from the slits to a common point on a screen. (a) Destructive interference occurs here, because one path is a half wavelength longer than the other. The waves start in phase but arrive out of phase. (b) Constructive interference occurs here because one path is a whole wavelength longer than the other. The waves start out and arrive in phase.

Take-Home Experiment: Using Fingers as Slits

Look at a light, such as a street lamp or incandescent bulb, through the narrow gap between two fingers held close together. What type of pattern do you see? How does it change when you allow the fingers to move a little farther apart? Is it more distinct for a monochromatic source, such as the yellow light from a sodium vapor lamp, than for an incandescent bulb?

The figure is a schematic of a double slit experiment, with the scale of the slits enlarged to show the detail. The two slits are on the left, and the screen is on the right. The slits are represented by a thick vertical line with two gaps cut through it a distance d apart. Two rays, one from each slit, angle up and to the right at an angle theta above the horizontal. At the screen, these rays are shown to converge at a common point. The ray from the upper slit is labeled l sub one, and the ray from the lower slit is labeled l sub two. At the slits, a right triangle is drawn, with the thick line between the slits forming the hypotenuse. The hypotenuse is labeled d, which is the distance between the slits. A short piece of the ray from the lower slit is labeled delta l and forms the short side of the right triangle. The long side of the right triangle is formed by a line segment that goes downward and to the right from the upper slit to the lower ray. This line segment is perpendicular to the lower ray, and the angle it makes with the hypotenuse is labeled theta. Beneath this triangle is the formula delta l equals d sine theta.

Figure 5. The paths from each slit to a common point on the screen differ by an amount dsinθ, assuming the distance to the screen is much greater than the distance between slits (not to scale here).

Figure 5 shows how to determine the path length difference for waves traveling from two slits to a common point on a screen. If the screen is a large distance away compared with the distance between the slits, then the angle θ between the path and a line from the slits to the screen (see the figure) is nearly the same for each path. The difference between the paths is shown in the figure; simple trigonometry shows it to be d sin θ , where d is the distance between the slits. To obtain constructive interference for a double slit , the path length difference must be an integral multiple of the wavelength, or d sin θ = mλ, for m = 0, 1, −1, 2, −2, . . . (constructive).

Similarly, to obtain destructive interference for a double slit , the path length difference must be a half-integral multiple of the wavelength, or

[latex]d\sin\theta=\left(m+\frac{1}{2}\right)\lambda\text{, for }m=0,1,-1,2,-2,\dots\text{ (destructive)}\\[/latex],

where λ is the wavelength of the light, d is the distance between slits, and θ is the angle from the original direction of the beam as discussed above. We call m the order of the interference. For example, m = 4 is fourth-order interference.

The equations for double slit interference imply that a series of bright and dark lines are formed. For vertical slits, the light spreads out horizontally on either side of the incident beam into a pattern called interference fringes, illustrated in Figure 6. The intensity of the bright fringes falls off on either side, being brightest at the center. The closer the slits are, the more is the spreading of the bright fringes. We can see this by examining the equation d sin θ = mλ, for m = 0, 1, −1, 2, −2, . . . .

For fixed λ and m , the smaller d is, the larger θ must be, since [latex]\sin\theta=\frac{m\lambda}{d}\\[/latex]. This is consistent with our contention that wave effects are most noticeable when the object the wave encounters (here, slits a distance d apart) is small. Small d gives large θ , hence a large effect.

The figure consists of two parts arranged side-by-side. The diagram on the left side shows a double slit arrangement along with a graph of the resultant intensity pattern on a distant screen. The graph is oriented vertically, so that the intensity peaks grow out and to the left from the screen. The maximum intensity peak is at the center of the screen, and some less intense peaks appear on both sides of the center. These peaks become progressively dimmer upon moving away from the center, and are symmetric with respect to the central peak. The distance from the central maximum to the first dimmer peak is labeled y sub one, and the distance from the central maximum to the second dimmer peak is labeled y sub two. The illustration on the right side shows thick bright horizontal bars on a dark background. Each horizontal bar is aligned with one of the intensity peaks from the first figure.

Figure 6. The interference pattern for a double slit has an intensity that falls off with angle. The photograph shows multiple bright and dark lines, or fringes, formed by light passing through a double slit.

Example 1. Finding a Wavelength from an Interference Pattern

Suppose you pass light from a He-Ne laser through two slits separated by 0.0100 mm and find that the third bright line on a screen is formed at an angle of 10.95º relative to the incident beam. What is the wavelength of the light?

The third bright line is due to third-order constructive interference, which means that m = 3. We are given d = 0.0100 mm and θ = 10.95º. The wavelength can thus be found using the equation d sin θ = mλ for constructive interference.

The equation is d sin θ = mλ . Solving for the wavelength λ gives [latex]\lambda=\frac{d\sin\theta}{m}\\[/latex].

Substituting known values yields

[latex]\begin{array}{lll}\lambda&=&\frac{\left(0.0100\text{ nm}\right)\left(\sin10.95^{\circ}\right)}{3}\\\text{ }&=&6.33\times10^{-4}\text{ nm}=633\text{ nm}\end{array}\\[/latex]

To three digits, this is the wavelength of light emitted by the common He-Ne laser. Not by coincidence, this red color is similar to that emitted by neon lights. More important, however, is the fact that interference patterns can be used to measure wavelength. Young did this for visible wavelengths. This analytical technique is still widely used to measure electromagnetic spectra. For a given order, the angle for constructive interference increases with λ , so that spectra (measurements of intensity versus wavelength) can be obtained.

Example 2. Calculating Highest Order Possible

Interference patterns do not have an infinite number of lines, since there is a limit to how big m can be. What is the highest-order constructive interference possible with the system described in the preceding example?

Strategy and Concept

The equation d sin θ = mλ ( for m = 0, 1, −1, 2, −2, . . . ) describes constructive interference. For fixed values of d and λ , the larger m is, the larger sin θ is. However, the maximum value that sin θ can have is 1, for an angle of 90º. (Larger angles imply that light goes backward and does not reach the screen at all.) Let us find which m corresponds to this maximum diffraction angle.

Solving the equation d sin θ = mλ for m gives [latex]\lambda=\frac{d\sin\theta}{m}\\[/latex].

Taking sin θ = 1 and substituting the values of d and λ from the preceding example gives

[latex]\displaystyle{m}=\frac{\left(0.0100\text{ mm}\right)\left(1\right)}{633\text{ nm}}\approx15.8\\[/latex]

Therefore, the largest integer m can be is 15, or m = 15.

The number of fringes depends on the wavelength and slit separation. The number of fringes will be very large for large slit separations. However, if the slit separation becomes much greater than the wavelength, the intensity of the interference pattern changes so that the screen has two bright lines cast by the slits, as expected when light behaves like a ray. We also note that the fringes get fainter further away from the center. Consequently, not all 15 fringes may be observable.

Section Summary

Young’s double slit experiment gave definitive proof of the wave character of light.
An interference pattern is obtained by the superposition of light from two slits.
There is constructive interference when d sin θ = mλ ( for m = 0, 1, −1, 2, −2, . . . ), where d is the distance between the slits, θ is the angle relative to the incident direction, and m is the order of the interference.
There is destructive interference when d sin θ = mλ ( for m = 0, 1, −1, 2, −2, . . . ).

Conceptual Questions

Young’s double slit experiment breaks a single light beam into two sources. Would the same pattern be obtained for two independent sources of light, such as the headlights of a distant car? Explain.
Suppose you use the same double slit to perform Young’s double slit experiment in air and then repeat the experiment in water. Do the angles to the same parts of the interference pattern get larger or smaller? Does the color of the light change? Explain.
Is it possible to create a situation in which there is only destructive interference? Explain.
Figure 7 shows the central part of the interference pattern for a pure wavelength of red light projected onto a double slit. The pattern is actually a combination of single slit and double slit interference. Note that the bright spots are evenly spaced. Is this a double slit or single slit characteristic? Note that some of the bright spots are dim on either side of the center. Is this a single slit or double slit characteristic? Which is smaller, the slit width or the separation between slits? Explain your responses.

The figure shows a photo of a horizontal line of equally spaced red dots of light on a black background. The central dot is the brightest and the dots on either side of center are dimmer. The dot intensity decreases to almost zero after moving six dots to the left or right of center. If you continue to move away from the center, the dot brightness increases slightly, although it does not reach the brightness of the central dot. After moving another six dots, or twelve dots in all, to the left or right of center, there is another nearly invisible dot. If you move even farther from the center, the dot intensity again increases, but it does not reach the level of the previous local maximum. At eighteen dots from the center, there is another nearly invisible dot.

Figure 7. This double slit interference pattern also shows signs of single slit interference. (credit: PASCO)

Problems & Exercises

At what angle is the first-order maximum for 450-nm wavelength blue light falling on double slits separated by 0.0500 mm?
Calculate the angle for the third-order maximum of 580-nm wavelength yellow light falling on double slits separated by 0.100 mm.
What is the separation between two slits for which 610-nm orange light has its first maximum at an angle of 30.0º?
Find the distance between two slits that produces the first minimum for 410-nm violet light at an angle of 45.0º.
Calculate the wavelength of light that has its third minimum at an angle of 30.0º when falling on double slits separated by 3.00 μm.
What is the wavelength of light falling on double slits separated by 2.00 μm if the third-order maximum is at an angle of 60.0º?
At what angle is the fourth-order maximum for the situation in Question 1?
What is the highest-order maximum for 400-nm light falling on double slits separated by 25.0 μm?
Find the largest wavelength of light falling on double slits separated by 1.20 μm for which there is a first-order maximum. Is this in the visible part of the spectrum?
What is the smallest separation between two slits that will produce a second-order maximum for 720-nm red light?
(a) What is the smallest separation between two slits that will produce a second-order maximum for any visible light? (b) For all visible light?
(a) If the first-order maximum for pure-wavelength light falling on a double slit is at an angle of 10.0º, at what angle is the second-order maximum? (b) What is the angle of the first minimum? (c) What is the highest-order maximum possible here?

The figure shows a schematic of a double slit experiment. A double slit is at the left and a screen is at the right. The slits are separated by a distance d. From the midpoint between the slits, a horizontal line labeled x extends to the screen. From the same point, a line angled upward at an angle theta above the horizontal also extends to the screen. The distance between where the horizontal line hits the screen and where the angled line hits the screen is marked y, and the distance between adjacent fringes is given by delta y, which equals x times lambda over d.

Figure 8. The distance between adjacent fringes is [latex]\Delta{y}=\frac{x\lambda}{d}\\[/latex], assuming the slit separation d is large compared with λ .

Using the result of the problem above, calculate the distance between fringes for 633-nm light falling on double slits separated by 0.0800 mm, located 3.00 m from a screen as in Figure 8.
Using the result of the problem two problems prior, find the wavelength of light that produces fringes 7.50 mm apart on a screen 2.00 m from double slits separated by 0.120 mm (see Figure 8).

coherent: waves are in phase or have a definite phase relationship

constructive interference for a double slit: the path length difference must be an integral multiple of the wavelength

destructive interference for a double slit: the path length difference must be a half-integral multiple of the wavelength

incoherent: waves have random phase relationships

order: the integer m used in the equations for constructive and destructive interference for a double slit

Selected Solutions to Problems & Exercises

3. 1.22 × 10 −6 m

9. 1200 nm (not visible)

11. (a) 760 nm; (b) 1520 nm

13. For small angles sin θ − tan θ ≈ θ (in radians).

For two adjacent fringes we have, d sin θ m = mλ and d sin θ m + 1 = ( m + 1) λ

Subtracting these equations gives

[latex]\begin{array}{}d\left(\sin{\theta }_{\text{m}+1}-\sin{\theta }_{\text{m}}\right)=\left[\left(m+1\right)-m\right]\lambda \\ d\left({\theta }_{\text{m}+1}-{\theta }_{\text{m}}\right)=\lambda \\ \text{tan}{\theta }_{\text{m}}=\frac{{y}_{\text{m}}}{x}\approx {\theta }_{\text{m}}\Rightarrow d\left(\frac{{y}_{\text{m}+1}}{x}-\frac{{y}_{\text{m}}}{x}\right)=\lambda \\ d\frac{\Delta y}{x}=\lambda \Rightarrow \Delta y=\frac{\mathrm{x\lambda }}{d}\end{array}\\[/latex]

College Physics. Authored by : OpenStax College. Located at : http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a/College_Physics . License : CC BY: Attribution . License Terms : Located at License

Chapter 27 Wave Optics

27.3 Young’s Double Slit Experiment

Explain the phenomena of interference.
Define constructive interference for a double slit and destructive interference for a double slit.

Why do we not ordinarily observe wave behavior for light, such as observed in Young’s double slit experiment? First, light must interact with something small, such as the closely spaced slits used by Young, to show pronounced wave effects. Furthermore, Young first passed light from a single source (the Sun) through a single slit to make the light somewhat coherent. By coherent , we mean waves are in phase or have a definite phase relationship. Incoherent means the waves have random phase relationships. Why did Young then pass the light through a double slit? The answer to this question is that two slits provide two coherent light sources that then interfere constructively or destructively. Young used sunlight, where each wavelength forms its own pattern, making the effect more difficult to see. We illustrate the double slit experiment with monochromatic (single [latex]{\lambda}[/latex]) light to clarify the effect. Figure 2 shows the pure constructive and destructive interference of two waves having the same wavelength and amplitude.

When light passes through narrow slits, it is diffracted into semicircular waves, as shown in Figure 3 (a). Pure constructive interference occurs where the waves are crest to crest or trough to trough. Pure destructive interference occurs where they are crest to trough. The light must fall on a screen and be scattered into our eyes for us to see the pattern. An analogous pattern for water waves is shown in Figure 3 (b). Note that regions of constructive and destructive interference move out from the slits at well-defined angles to the original beam. These angles depend on wavelength and the distance between the slits, as we shall see below.

To understand the double slit interference pattern, we consider how two waves travel from the slits to the screen, as illustrated in Figure 4 . Each slit is a different distance from a given point on the screen. Thus different numbers of wavelengths fit into each path. Waves start out from the slits in phase (crest to crest), but they may end up out of phase (crest to trough) at the screen if the paths differ in length by half a wavelength, interfering destructively as shown in Figure 4 (a). If the paths differ by a whole wavelength, then the waves arrive in phase (crest to crest) at the screen, interfering constructively as shown in Figure 4 (b). More generally, if the paths taken by the two waves differ by any half-integral number of wavelengths [[latex]{(1/2) \;\lambda}[/latex], [latex]{(3/2) \;\lambda}[/latex], [latex]{(5/2) \;\lambda}[/latex], etc.], then destructive interference occurs. Similarly, if the paths taken by the two waves differ by any integral number of wavelengths ([latex]{\lambda}[/latex], [latex]{2 \lambda}[/latex], [latex]{3 \lambda}[/latex], etc.), then constructive interference occurs.

Take-Home Experiment: Using Fingers as Slits

Figure 5 shows how to determine the path length difference for waves traveling from two slits to a common point on a screen. If the screen is a large distance away compared with the distance between the slits, then the angle [latex]{\theta}[/latex] between the path and a line from the slits to the screen (see the figure) is nearly the same for each path. The difference between the paths is shown in the figure; simple trigonometry shows it to be [latex]{d \;\text{sin} \;\theta}[/latex], where [latex]{d}[/latex] is the distance between the slits. To obtain constructive interference for a double slit , the path length difference must be an integral multiple of the wavelength, or

Similarly, to obtain destructive interference for a double slit , the path length difference must be a half-integral multiple of the wavelength, or

where [latex]{\lambda}[/latex] is the wavelength of the light, [latex]{d}[/latex] is the distance between slits, and [latex]{\theta}[/latex] is the angle from the original direction of the beam as discussed above. We call [latex]{m}[/latex] the order of the interference. For example, [latex]{m = 4}[/latex] is fourth-order interference.

The equations for double slit interference imply that a series of bright and dark lines are formed. For vertical slits, the light spreads out horizontally on either side of the incident beam into a pattern called interference fringes, illustrated in Figure 6 . The intensity of the bright fringes falls off on either side, being brightest at the center. The closer the slits are, the more is the spreading of the bright fringes. We can see this by examining the equation

For fixed [latex]{\lambda}[/latex] and [latex]{m}[/latex], the smaller [latex]{d}[/latex] is, the larger [latex]{\theta}[/latex] must be, since [latex]{\text{sin} \;\theta = m \lambda / d}[/latex].

This is consistent with our contention that wave effects are most noticeable when the object the wave encounters (here, slits a distance [latex]{d}[/latex] apart) is small. Small [latex]{d}[/latex] gives large [latex]{\theta}[/latex], hence a large effect.

Example 1: Finding a Wavelength from an Interference Pattern

Suppose you pass light from a He-Ne laser through two slits separated by 0.0100 mm and find that the third bright line on a screen is formed at an angle of [latex]{10.95 ^{\circ}}[/latex] relative to the incident beam. What is the wavelength of the light?

The third bright line is due to third-order constructive interference, which means that [latex]{m = 3}[/latex]. We are given [latex]{d = 0.0100 \;\text{mm}}[/latex] and [latex]{\theta = 10.95^{\circ}}[/latex]. The wavelength can thus be found using the equation [latex]{d \;\text{sin} \;\theta = m \lambda}[/latex] for constructive interference.

The equation is [latex]{d \;\text{sin} \;\theta = m \lambda}[/latex]. Solving for the wavelength [latex]{\lambda}[/latex] gives

Substituting known values yields

To three digits, this is the wavelength of light emitted by the common He-Ne laser. Not by coincidence, this red color is similar to that emitted by neon lights. More important, however, is the fact that interference patterns can be used to measure wavelength. Young did this for visible wavelengths. This analytical technique is still widely used to measure electromagnetic spectra. For a given order, the angle for constructive interference increases with [latex]{\lambda}[/latex], so that spectra (measurements of intensity versus wavelength) can be obtained.

Example 2: Calculating Highest Order Possible

Interference patterns do not have an infinite number of lines, since there is a limit to how big [latex]{m}[/latex] can be. What is the highest-order constructive interference possible with the system described in the preceding example?

Strategy and Concept

The equation [latex]{d \;\text{sin} \;\theta = m \lambda \; (\text{for} \; m = 0, \; 1, \; -1, \; 2, \; -2, \; \dots)}[/latex] describes constructive interference. For fixed values of [latex]{d}[/latex] and [latex]{\lambda}[/latex], the larger [latex]{m}[/latex] is, the larger [latex]{\text{sin} \;\theta}[/latex] is. However, the maximum value that [latex]{\text{sin} \;\theta}[/latex] can have is 1, for an angle of [latex]{90 ^{\circ}}[/latex]. (Larger angles imply that light goes backward and does not reach the screen at all.) Let us find which [latex]{m}[/latex] corresponds to this maximum diffraction angle.

Solving the equation [latex]{d \;\text{sin} \;\theta = m \lambda}[/latex] for [latex]{m}[/latex] gives

Taking [latex]{\text{sin} \;\theta = 1}[/latex] and substituting the values of [latex]{d}[/latex] and [latex]{\lambda}[/latex] from the preceding example gives

Therefore, the largest integer [latex]{m}[/latex] can be is 15, or

Section Summary

Young’s double slit experiment gave definitive proof of the wave character of light.
An interference pattern is obtained by the superposition of light from two slits.
There is constructive interference when [latex]{d \;\text{sin} \;\theta = m \lambda \;(\text{for} \; m = 0, \; 1, \; -1, \;2, \; -2, \dots)}[/latex], where [latex]{d}[/latex] is the distance between the slits, [latex]{\theta}[/latex] is the angle relative to the incident direction, and [latex]{m}[/latex] is the order of the interference.
There is destructive interference when [latex]{d \;\text{sin} \;\theta = (m+ \frac{1}{2}) \lambda}[/latex] (for [latex]{m = 0, \; 1, \; -1, \; 2, \; -2, \; \dots}[/latex]).

Conceptual Questions

1: Young’s double slit experiment breaks a single light beam into two sources. Would the same pattern be obtained for two independent sources of light, such as the headlights of a distant car? Explain.

2: Suppose you use the same double slit to perform Young’s double slit experiment in air and then repeat the experiment in water. Do the angles to the same parts of the interference pattern get larger or smaller? Does the color of the light change? Explain.

3: Is it possible to create a situation in which there is only destructive interference? Explain.

4: Figure 7 shows the central part of the interference pattern for a pure wavelength of red light projected onto a double slit. The pattern is actually a combination of single slit and double slit interference. Note that the bright spots are evenly spaced. Is this a double slit or single slit characteristic? Note that some of the bright spots are dim on either side of the center. Is this a single slit or double slit characteristic? Which is smaller, the slit width or the separation between slits? Explain your responses.

Problems & Exercises

2: Calculate the angle for the third-order maximum of 580-nm wavelength yellow light falling on double slits separated by 0.100 mm.

3: What is the separation between two slits for which 610-nm orange light has its first maximum at an angle of [latex]{30.0 ^{\circ}}[/latex]?

4: Find the distance between two slits that produces the first minimum for 410-nm violet light at an angle of [latex]{45.0 ^{\circ}}[/latex].

5: Calculate the wavelength of light that has its third minimum at an angle of [latex]{30.0 ^{\circ}}[/latex] when falling on double slits separated by [latex]{3.00 \;\mu \text{m}}[/latex]. Explicitly, show how you follow the steps in Chapter 27.7 Problem-Solving Strategies for Wave Optics .

6: What is the wavelength of light falling on double slits separated by [latex]{2.00 \;\mu \text{m}}[/latex] if the third-order maximum is at an angle of [latex]{60.0 ^{\circ}}[/latex]?

7: At what angle is the fourth-order maximum for the situation in Problems & Exercises 1 ?

8: What is the highest-order maximum for 400-nm light falling on double slits separated by [latex]{25.0 \;\mu \text{m}}[/latex]?

9: Find the largest wavelength of light falling on double slits separated by [latex]{1.20 \;\mu \text{m}}[/latex] for which there is a first-order maximum. Is this in the visible part of the spectrum?

10: What is the smallest separation between two slits that will produce a second-order maximum for 720-nm red light?

11: (a) What is the smallest separation between two slits that will produce a second-order maximum for any visible light? (b) For all visible light?

12: (a) If the first-order maximum for pure-wavelength light falling on a double slit is at an angle of [latex]{10.0^{\circ}}[/latex], at what angle is the second-order maximum? (b) What is the angle of the first minimum? (c) What is the highest-order maximum possible here?

13: Figure 8 shows a double slit located a distance [latex]{x}[/latex] from a screen, with the distance from the center of the screen given by [latex]{y}[/latex]. When the distance [latex]{d}[/latex] between the slits is relatively large, there will be numerous bright spots, called fringes. Show that, for small angles (where [latex]{\text{sin} \;\theta \approx \theta}[/latex], with [latex]{\theta}[/latex] in radians), the distance between fringes is given by [latex]{\Delta y = x \lambda /d}[/latex].

14: Using the result of the problem above, calculate the distance between fringes for 633-nm light falling on double slits separated by 0.0800 mm, located 3.00 m from a screen as in Figure 8 .

15: Using the result of the problem two problems prior, find the wavelength of light that produces fringes 7.50 mm apart on a screen 2.00 m from double slits separated by 0.120 mm (see Figure 8 ).

1: [latex]{0.516 ^{\circ}}[/latex]

3: [latex]{1.22 \times 10^{-6} \;\text{m}}[/latex]

7: [latex]{2.06 ^{\circ}}[/latex]

9: 1200 nm (not visible)

11: (a) 760 nm

(b) 1520 nm

13: For small angles [latex]{\text{sin} \;\theta - \;\text{tan} \;\theta \approx \theta}[/latex] (in radians).

For two adjacent fringes we have,

Subtracting these equations gives

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154 Young’s Double Slit Experiment

[latexpage]

Learning Objectives

Explain the phenomena of interference.
Define constructive interference for a double slit and destructive interference for a double slit.

Why do we not ordinarily observe wave behavior for light, such as observed in Young’s double slit experiment? First, light must interact with something small, such as the closely spaced slits used by Young, to show pronounced wave effects. Furthermore, Young first passed light from a single source (the Sun) through a single slit to make the light somewhat coherent. By coherent , we mean waves are in phase or have a definite phase relationship. Incoherent means the waves have random phase relationships. Why did Young then pass the light through a double slit? The answer to this question is that two slits provide two coherent light sources that then interfere constructively or destructively. Young used sunlight, where each wavelength forms its own pattern, making the effect more difficult to see. We illustrate the double slit experiment with monochromatic (single \(\lambda \)) light to clarify the effect. (Figure) shows the pure constructive and destructive interference of two waves having the same wavelength and amplitude.

When light passes through narrow slits, it is diffracted into semicircular waves, as shown in (Figure) (a). Pure constructive interference occurs where the waves are crest to crest or trough to trough. Pure destructive interference occurs where they are crest to trough. The light must fall on a screen and be scattered into our eyes for us to see the pattern. An analogous pattern for water waves is shown in (Figure) (b). Note that regions of constructive and destructive interference move out from the slits at well-defined angles to the original beam. These angles depend on wavelength and the distance between the slits, as we shall see below.

To understand the double slit interference pattern, we consider how two waves travel from the slits to the screen, as illustrated in (Figure) . Each slit is a different distance from a given point on the screen. Thus different numbers of wavelengths fit into each path. Waves start out from the slits in phase (crest to crest), but they may end up out of phase (crest to trough) at the screen if the paths differ in length by half a wavelength, interfering destructively as shown in (Figure) (a). If the paths differ by a whole wavelength, then the waves arrive in phase (crest to crest) at the screen, interfering constructively as shown in (Figure) (b). More generally, if the paths taken by the two waves differ by any half-integral number of wavelengths [\(\left(1/2\right)\lambda \), \(\left(3/2\right)\lambda \), \(\left(5/2\right)\lambda \), etc.], then destructive interference occurs. Similarly, if the paths taken by the two waves differ by any integral number of wavelengths (\(\lambda \), \(2\lambda \), \(3\lambda \), etc.), then constructive interference occurs.

(Figure) shows how to determine the path length difference for waves traveling from two slits to a common point on a screen. If the screen is a large distance away compared with the distance between the slits, then the angle \(\theta \) between the path and a line from the slits to the screen (see the figure) is nearly the same for each path. The difference between the paths is shown in the figure; simple trigonometry shows it to be \(d\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \), where \(d\) is the distance between the slits. To obtain constructive interference for a double slit , the path length difference must be an integral multiple of the wavelength, or

Similarly, to obtain destructive interference for a double slit , the path length difference must be a half-integral multiple of the wavelength, or

where \(\lambda \) is the wavelength of the light, \(d\) is the distance between slits, and \(\theta \) is the angle from the original direction of the beam as discussed above. We call \(m\) the order of the interference. For example, \(m=4\) is fourth-order interference.

The equations for double slit interference imply that a series of bright and dark lines are formed. For vertical slits, the light spreads out horizontally on either side of the incident beam into a pattern called interference fringes, illustrated in (Figure) . The intensity of the bright fringes falls off on either side, being brightest at the center. The closer the slits are, the more is the spreading of the bright fringes. We can see this by examining the equation

For fixed \(\lambda \) and \(m\), the smaller \(d\) is, the larger \(\theta \) must be, since \(\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\mathrm{m\lambda }/d\). This is consistent with our contention that wave effects are most noticeable when the object the wave encounters (here, slits a distance \(d\) apart) is small. Small \(d\) gives large \(\theta \), hence a large effect.

Suppose you pass light from a He-Ne laser through two slits separated by 0.0100 mm and find that the third bright line on a screen is formed at an angle of \(\text{10}\text{.}\text{95º}\) relative to the incident beam. What is the wavelength of the light?

The third bright line is due to third-order constructive interference, which means that \(m=3\). We are given \(d=0\text{.}\text{0100}\phantom{\rule{0.25em}{0ex}}\text{mm}\) and \(\theta =\text{10}\text{.}\text{95º}\). The wavelength can thus be found using the equation \(d\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\mathrm{m\lambda }\) for constructive interference.

The equation is \(d\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\mathrm{m\lambda }\). Solving for the wavelength \(\lambda \) gives

Substituting known values yields

To three digits, this is the wavelength of light emitted by the common He-Ne laser. Not by coincidence, this red color is similar to that emitted by neon lights. More important, however, is the fact that interference patterns can be used to measure wavelength. Young did this for visible wavelengths. This analytical technique is still widely used to measure electromagnetic spectra. For a given order, the angle for constructive interference increases with \(\lambda \), so that spectra (measurements of intensity versus wavelength) can be obtained.

Interference patterns do not have an infinite number of lines, since there is a limit to how big \(m\) can be. What is the highest-order constructive interference possible with the system described in the preceding example?

Strategy and Concept

The equation \(d\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\mathrm{m\lambda }\phantom{\rule{0.25em}{0ex}}\text{(for}\phantom{\rule{0.25em}{0ex}}m=0,\phantom{\rule{0.25em}{0ex}}1,\phantom{\rule{0.25em}{0ex}}-1,\phantom{\rule{0.25em}{0ex}}2,\phantom{\rule{0.25em}{0ex}}-2,\phantom{\rule{0.25em}{0ex}}\dots \right)\) describes constructive interference. For fixed values of \(d\) and \(\lambda \), the larger \(m\) is, the larger \(\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \) is. However, the maximum value that \(\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \) can have is 1, for an angle of \(\text{90º}\). (Larger angles imply that light goes backward and does not reach the screen at all.) Let us find which \(m\) corresponds to this maximum diffraction angle.

Solving the equation \(d\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\mathrm{m\lambda }\) for \(m\) gives

Taking \(\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =1\) and substituting the values of \(d\) and \(\lambda \) from the preceding example gives

Therefore, the largest integer \(m\) can be is 15, or

Section Summary

Young’s double slit experiment gave definitive proof of the wave character of light.
An interference pattern is obtained by the superposition of light from two slits.
There is constructive interference when \(d\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\mathrm{m\lambda }\phantom{\rule{0.25em}{0ex}}\text{(for}\phantom{\rule{0.25em}{0ex}}m=0,\phantom{\rule{0.25em}{0ex}}1,\phantom{\rule{0.25em}{0ex}}-1,\phantom{\rule{0.25em}{0ex}}2,\phantom{\rule{0.25em}{0ex}}-2,\phantom{\rule{0.25em}{0ex}}\dots \right)\), where \(d\) is the distance between the slits, \(\theta \) is the angle relative to the incident direction, and \(m\) is the order of the interference.
There is destructive interference when \(d\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\left(m+\frac{1}{2}\right)\lambda \phantom{\rule{0.25em}{0ex}}\text{(for}\phantom{\rule{0.5em}{0ex}}m=0,\phantom{\rule{0.25em}{0ex}}1,\phantom{\rule{0.25em}{0ex}}-1,\phantom{\rule{0.25em}{0ex}}2,\phantom{\rule{0.25em}{0ex}}-2,\phantom{\rule{0.25em}{0ex}}\dots \right)\).

Conceptual Questions

Young’s double slit experiment breaks a single light beam into two sources. Would the same pattern be obtained for two independent sources of light, such as the headlights of a distant car? Explain.

Suppose you use the same double slit to perform Young’s double slit experiment in air and then repeat the experiment in water. Do the angles to the same parts of the interference pattern get larger or smaller? Does the color of the light change? Explain.

Is it possible to create a situation in which there is only destructive interference? Explain.

(Figure) shows the central part of the interference pattern for a pure wavelength of red light projected onto a double slit. The pattern is actually a combination of single slit and double slit interference. Note that the bright spots are evenly spaced. Is this a double slit or single slit characteristic? Note that some of the bright spots are dim on either side of the center. Is this a single slit or double slit characteristic? Which is smaller, the slit width or the separation between slits? Explain your responses.

Problems & Exercises

At what angle is the first-order maximum for 450-nm wavelength blue light falling on double slits separated by 0.0500 mm?

\(0\text{.}\text{516º}\)

Calculate the angle for the third-order maximum of 580-nm wavelength yellow light falling on double slits separated by 0.100 mm.

What is the separation between two slits for which 610-nm orange light has its first maximum at an angle of \(\text{30}\text{.}0º\)?

\(1\text{.}\text{22}×{\text{10}}^{-6}\phantom{\rule{0.25em}{0ex}}\text{m}\)

Find the distance between two slits that produces the first minimum for 410-nm violet light at an angle of \(\text{45}\text{.}0º\).

Calculate the wavelength of light that has its third minimum at an angle of \(\text{30}\text{.}0º\) when falling on double slits separated by \(3\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{μm}\). Explicitly, show how you follow the steps in Problem-Solving Strategies for Wave Optics .

What is the wavelength of light falling on double slits separated by \(2\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{μm}\) if the third-order maximum is at an angle of \(\text{60}\text{.}0º\)?

At what angle is the fourth-order maximum for the situation in (Figure) ?

\(2\text{.}\text{06º}\)

What is the highest-order maximum for 400-nm light falling on double slits separated by \(\text{25}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{μm}\)?

Find the largest wavelength of light falling on double slits separated by \(1\text{.}\text{20}\phantom{\rule{0.25em}{0ex}}\text{μm}\) for which there is a first-order maximum. Is this in the visible part of the spectrum?

1200 nm (not visible)

What is the smallest separation between two slits that will produce a second-order maximum for 720-nm red light?

(a) What is the smallest separation between two slits that will produce a second-order maximum for any visible light? (b) For all visible light?

(b) 1520 nm

(a) If the first-order maximum for pure-wavelength light falling on a double slit is at an angle of \(\text{10}\text{.}0º\), at what angle is the second-order maximum? (b) What is the angle of the first minimum? (c) What is the highest-order maximum possible here?

(Figure) shows a double slit located a distance \(x\) from a screen, with the distance from the center of the screen given by \(y\). When the distance \(d\) between the slits is relatively large, there will be numerous bright spots, called fringes. Show that, for small angles (where \(\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \approx \theta \), with \(\theta \) in radians), the distance between fringes is given by \(\text{Δ}y=\mathrm{x\lambda }/d\).

For small angles \(\text{sin}\phantom{\rule{0.25em}{0ex}}\theta -\text{tan}\phantom{\rule{0.25em}{0ex}}\theta \approx \theta \phantom{\rule{0.25em}{0ex}}\left(\text{in radians}\right)\).

For two adjacent fringes we have,

Subtracting these equations gives

Using the result of the problem above, calculate the distance between fringes for 633-nm light falling on double slits separated by 0.0800 mm, located 3.00 m from a screen as in (Figure) .

Using the result of the problem two problems prior, find the wavelength of light that produces fringes 7.50 mm apart on a screen 2.00 m from double slits separated by 0.120 mm (see (Figure) ).

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Derivation of Physics Formula
Youngs Double Slit Experiment Derivation

Young’s Double Slits Experiment Derivation

Introduction to young’s double slits experiment.

During the year 1801, Thomas Young carried out an experiment where the wave and particle nature of light and matter were demonstrated. The schematic diagram of the experimental setup is shown below-

young double slits experiment derivation

A beam of monochromatic light is made incident on the first screen, which contains the slit S 0 . The emerging light then incident on the second screen which consists of two slits namely, S 1 , S 2 . These two slits serve as a source of coherent light. The emerging light waves from these slits interfere to produce an interference pattern on the screen. The interference pattern consists of consecutive bright and dark fringes. The dark fringes are the result of destructive interference and bright fringes are the result of constructive interference.

You may also want to check out these topics given below!

Superposition of waves
Diffraction
Polarisation by scattering

young double slits experiment derivation-1

The light falls on the screen at the point P. which is at a distance y from the centre O.
The distance between the double-slit system and the screen is L
The two slits are separated by the distance d
Distance travelled by the light ray from slit 1 to point P on the screen is r 1
Distance travelled by the light ray from slit 2 to point P on the screen is r 2
Thus, the light ray from slit 2 travels an extra distance of ẟ = r 2 -r 1 than light ray from slit 1.
This extra distance is termed as Path difference.

Refer to Figure(3) Applying laws of cosines; we can write –

∵ Cos(90° – θ) = Sinθ

∵ Cos(90° + θ) = -Sinθ

Subtracting equation (2) from (1) we get-

Let us impose the limit that the distance between the double slit system and the screen is much greater than the distance between the slits. That is L >> d. The sum of r 1 and r 2 can be approximated to r 1 + r 2 ≅2r. Thus, the path difference becomes –

In this limit, the two rays r 1 and r 2 are essentially treated as parallel. (See Figure(4))

young double slits experiment derivation-3

Figure(4): Assuming L >> d, The path difference between two rays.

To compare the phase of two waves, the value of path difference (ẟ) plays a crucial role.

Constructive interference is seen when path difference (𝛿) is zero or integral multiple of the wavelength (λ).

(Constructive interference) 𝛿 = dSin𝜃 = mλ —-(5), m = 0, ±⁤1, ±2, ±3, ±4, ±5, ……

Where m is order number. The Zeroth order maximum (m=0)corresponds to the central bright fringe, here 𝜃=0. The first order maxima(m=±⁤1)(bright fringe) are on either side the central fringe.

Similarly, when 𝛿 is an odd integral multiple of λ/2, the resultant fringes will be 180 0 out of phase, thus, forming a dark fringe.

(Destructive interference) 𝛿 = dSin𝜃 = λ—(6), m = 0, ±⁤1, ±2, ±3, ±4, ±5, ……

Assuming the distance between the slits are much greater than the wavelength of the incident light, we get-

Substituting it in the constructive and destructive interference condition we can get the position of bright and dark fringes, respectively. The equation is as follows-

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Young's double slit experiment

This experiment to use the effects of interference to measure the wavelength of light was devised by Thomas Young in 1801, although the original idea was due to Grimaldi. The method produces non-localised interference fringes by division of wavefront, and a sketch of the experimental arrangement is shown in Figure 1.

Light from a monochromatic line source passes through a lens and is focused on to a single slit S. It then falls on a double slit (S 1 and S 2 ) and this produces two wave trains that interfere with each other in the region on the right of the diagram. The interference pattern at any distance from the double slit may be observed with a micrometer eyepiece or by placing a screen in the path of the waves. The separation across double slit should be less than 1 mm, the width of each slit about 0.3 mm, and the distance between the double slit and the screen between 50 cm and 1 m. The single slit, the source and the double slit must be parallel to produce the optimum interference pattern. Alternatively a laser may be used and the fringes viewed on a screen some metres away without the need for a micrometer eyepiece or a single slit. The formula relating the dimensions of the apparatus and the wavelength of light may be proved as follows.

Note that the fringe width is directly proportional to the wavelength, and so light with a longer wavelength will give wider fringes. Although the diagram shows distinct light and dark fringes, the intensity actually varies as the cos 2 of angle from the centre. If white light is used a white centre fringe is observed, but all the other fringes have coloured edges, the blue edge being nearer the centre. Eventually the fringes overlap and a uniform white light is produced.

What Is The Double-Slit Experiment?

The double-slit experiment, observation affects reality, the various interpretations:.

The double-slit experiment shows the duality of the quantum world. A photon’s wave/particle duality is affected when it is observed.

Light has been one of the major areas of inquiry for physicists since we first began questioning the world around us. Understandably so, as it is the medium by which we see, measure and understand the world. It holds a powerful symbolism in our imaginations, is reflected in our religions and is famously quoted in our scriptures.

Rigorous science has enlightened our ignorance about Light. Until the 1800s, light was thought to be made up of particles, attested by Newtonian physics.

This came rather intuitively, as we see light traveling in a straight line, like bullets coming out of a gun.

Prison cell interior , sunrays coming through a barred window - Illustration(nobeastsofierce)S

However, nature is often weirder than our expectations and light’s weird behavior was first shown by Thomas Young in his now heavily worked upon and immortalized double-slit experiment. This experiment provides some fascinating insights into the minute workings of nature and has challenged everything we know about light, matter, and reality itself. Let’s revisit the experiment that has baffled legendary scientists – including Einstein!

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Double-slit Experiment

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Plane wave representing a particle passing through two slits, resulting in an interference pattern on a screen some distance away from the slits. [1] .

The double-slit experiment is an experiment in quantum mechanics and optics demonstrating the wave-particle duality of electrons , photons , and other fundamental objects in physics. When streams of particles such as electrons or photons pass through two narrow adjacent slits to hit a detector screen on the other side, they don't form clusters based on whether they passed through one slit or the other. Instead, they interfere: simultaneously passing through both slits, and producing a pattern of interference bands on the screen. This phenomenon occurs even if the particles are fired one at a time, showing that the particles demonstrate some wave behavior by interfering with themselves as if they were a wave passing through both slits.

Niels Bohr proposed the idea of wave-particle duality to explain the results of the double-slit experiment. The idea is that all fundamental particles behave in some ways like waves and in other ways like particles, depending on what properties are being observed. These insights led to the development of quantum mechanics and quantum field theory , the current basis behind the Standard Model of particle physics , which is our most accurate understanding of how particles work.

The original double-slit experiment was performed using light/photons around the turn of the nineteenth century by Thomas Young, so the original experiment is often called Young's double-slit experiment. The idea of using particles other than photons in the experiment did not come until after the ideas of de Broglie and the advent of quantum mechanics, when it was proposed that fundamental particles might also behave as waves with characteristic wavelengths depending on their momenta. The single-electron version of the experiment was in fact not performed until 1974. A more recent version of the experiment successfully demonstrating wave-particle duality used buckminsterfullerene or buckyballs , the \(C_{60}\) allotrope of carbon.

Waves vs. Particles

Double-slit experiment with electrons, modeling the double-slit experiment.

To understand why the double-slit experiment is important, it is useful to understand the strong distinctions between wave and particles that make wave-particle duality so intriguing.

Waves describe oscillating values of a physical quantity that obey the wave equation . They are usually described by sums of sine and cosine functions, since any periodic (oscillating) function may be decomposed into a Fourier series . When two waves pass through each other, the resulting wave is the sum of the two original waves. This is called a superposition since the waves are placed ("-position") on top of each other ("super-"). Superposition is one of the most fundamental principles of quantum mechanics. A general quantum system need not be in one state or another but can reside in a superposition of two where there is some probability of measuring the quantum wavefunction in one state or another.

Left: example of superposed waves constructively interfering. Right: superposed waves destructively interfering. [2]

If one wave is \(A(x) = \sin (2x)\) and the other is \(B(x) = \sin (2x)\), then they add together to make \(A + B = 2 \sin (2x)\). The addition of two waves to form a wave of larger amplitude is in general known as constructive interference since the interference results in a larger wave.

If one wave is \(A(x) = \sin (2x)\) and the other is \(B(x) = \sin (2x + \pi)\), then they add together to make \(A + B = 0\) \(\big(\)since \(\sin (2x + \pi) = - \sin (2x)\big).\) This is known as destructive interference in general, when adding two waves results in a wave of smaller amplitude. See the figure above for examples of both constructive and destructive interference.

Two speakers are generating sounds with the same phase, amplitude, and wavelength. The two sound waves can make constructive interference, as above left. Or they can make destructive interference, as above right. If we want to find out the exact position where the two sounds make destructive interference, which of the following do we need to know?

a) the wavelength of the sound waves b) the distances from the two speakers c) the speed of sound generated by the two speakers

This wave behavior is quite unlike the behavior of particles. Classically, particles are objects with a single definite position and a single definite momentum. Particles do not make interference patterns with other particles in detectors whether or not they pass through slits. They only interact by colliding elastically , i.e., via electromagnetic forces at short distances. Before the discovery of quantum mechanics, it was assumed that waves and particles were two distinct models for objects, and that any real physical thing could only be described as a particle or as a wave, but not both.

In the more modern version of the double slit experiment using electrons, electrons with the same momentum are shot from an "electron gun" like the ones inside CRT televisions towards a screen with two slits in it. After each electron goes through one of the slits, it is observed hitting a single point on a detecting screen at an apparently random location. As more and more electrons pass through, one at a time, they form an overall pattern of light and dark interference bands. If each electron was truly just a point particle, then there would only be two clusters of observations: one for the electrons passing through the left slit, and one for the right. However, if electrons are made of waves, they interfere with themselves and pass through both slits simultaneously. Indeed, this is what is observed when the double-slit experiment is performed using electrons. It must therefore be true that the electron is interfering with itself since each electron was only sent through one at a time—there were no other electrons to interfere with it!

When the double-slit experiment is performed using electrons instead of photons, the relevant wavelength is the de Broglie wavelength \(\lambda:\)

\[\lambda = \frac{h}{p},\]

where \(h\) is Planck's constant and \(p\) is the electron's momentum.

Calculate the de Broglie wavelength of an electron moving with velocity \(1.0 \times 10^{7} \text{ m/s}.\)

Usain Bolt, the world champion sprinter, hit a top speed of 27.79 miles per hour at the Olympics. If he has a mass of 94 kg, what was his de Broglie wavelength?

Express your answer as an order of magnitude in units of the Bohr radius \(r_{B} = 5.29 \times 10^{-11} \text{m}\). For instance, if your answer was \(4 \times 10^{-5} r_{B}\), your should give \(-5.\)

Image Credit: Flickr drcliffordchoi.

While the de Broglie relation was postulated for massive matter, the equation applies equally well to light. Given light of a certain wavelength, the momentum and energy of that light can be found using de Broglie's formula. This generalizes the naive formula \(p = m v\), which can't be applied to light since light has no mass and always moves at a constant velocity of \(c\) regardless of wavelength.

The below is reproduced from the Amplitude, Frequency, Wave Number, Phase Shift wiki.

In Young's double-slit experiment, photons corresponding to light of wavelength \(\lambda\) are fired at a barrier with two thin slits separated by a distance \(d,\) as shown in the diagram below. After passing through the slits, they hit a screen at a distance of \(D\) away with \(D \gg d,\) and the point of impact is measured. Remarkably, both the experiment and theory of quantum mechanics predict that the number of photons measured at each point along the screen follows a complicated series of peaks and troughs called an interference pattern as below. The photons must exhibit the wave behavior of a relative phase shift somehow to be responsible for this phenomenon. Below, the condition for which maxima of the interference pattern occur on the screen is derived.

Left: actual experimental two-slit interference pattern of photons, exhibiting many small peaks and troughs. Right: schematic diagram of the experiment as described above. [3]

Since \(D \gg d\), the angle from each of the slits is approximately the same and equal to \(\theta\). If \(y\) is the vertical displacement to an interference peak from the midpoint between the slits, it is therefore true that

\[D\tan \theta \approx D\sin \theta \approx D\theta = y.\]

Furthermore, there is a path difference \(\Delta L\) between the two slits and the interference peak. Light from the lower slit must travel \(\Delta L\) further to reach any particular spot on the screen, as in the diagram below:

Light from the lower slit must travel further to reach the screen at any given point above the midpoint, causing the interference pattern.

The condition for constructive interference is that the path difference \(\Delta L\) is exactly equal to an integer number of wavelengths. The phase shift of light traveling over an integer \(n\) number of wavelengths is exactly \(2\pi n\), which is the same as no phase shift and therefore constructive interference. From the above diagram and basic trigonometry, one can write

\[\Delta L = d\sin \theta \approx d\theta = n\lambda.\]

The first equality is always true; the second is the condition for constructive interference.

Now using \(\theta = \frac{y}{D}\), one can see that the condition for maxima of the interference pattern, corresponding to constructive interference, is

\[n\lambda = \frac{dy}{D},\]

i.e. the maxima occur at the vertical displacements of

\[y = \frac{n\lambda D}{d}.\]

The analogous experimental setup and mathematical modeling using electrons instead of photons is identical except that the de Broglie wavelength of the electrons \(\lambda = \frac{h}{p}\) is used instead of the literal wavelength of light.

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The discovery of light's wave-particle duality

The observation of interference effects definitively indicates the presence of overlapping waves. Thomas Young postulated that light is a wave and is subject to the superposition principle; his great experimental achievement was to demonstrate the constructive and destructive interference of light (c. 1801). In a modern version of Young’s experiment, differing in its essentials only in the source of light, a laser equally illuminates two parallel slits in an otherwise opaque surface. The light passing through the two slits is observed on a distant screen. When the widths of the slits are significantly greater than the wavelength of the light, the rules of geometrical optics hold—the light casts two shadows, and there are two illuminated regions on the screen. However, as the slits are narrowed in width, the light diffracts into the geometrical shadow, and the light waves overlap on the screen. (Diffraction is itself caused by the wave nature of light, being another example of an interference effect—it is discussed in more detail below.)

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The superposition principle determines the resulting intensity pattern on the illuminated screen. Constructive interference occurs whenever the difference in paths from the two slits to a point on the screen equals an integral number of wavelengths (0, λ, 2λ,…). This path difference guarantees that crests from the two waves arrive simultaneously. Destructive interference arises from path differences that equal a half-integral number of wavelengths (λ/2, 3λ/2,…). Young used geometrical arguments to show that the superposition of the two waves results in a series of equally spaced bands, or fringes, of high intensity, corresponding to regions of constructive interference, separated by dark regions of complete destructive interference.

An important parameter in the double-slit geometry is the ratio of the wavelength of the light λ to the spacing of the slits d . If λ/ d is much smaller than 1, the spacing between consecutive interference fringes will be small, and the interference effects may not be observable. Using narrowly separated slits, Young was able to separate the interference fringes. In this way he determined the wavelengths of the colours of visible light. The very short wavelengths of visible light explain why interference effects are observed only in special circumstances—the spacing between the sources of the interfering light waves must be very small to separate regions of constructive and destructive interference.

Observing interference effects is challenging because of two other difficulties. Most light sources emit a continuous range of wavelengths, which result in many overlapping interference patterns, each with a different fringe spacing. The multiple interference patterns wash out the most pronounced interference effects, such as the regions of complete darkness. Second, for an interference pattern to be observable over any extended period of time, the two sources of light must be coherent with respect to each other. This means that the light sources must maintain a constant phase relationship. For example, two harmonic waves of the same frequency always have a fixed phase relationship at every point in space, being either in phase, out of phase, or in some intermediate relationship. However, most light sources do not emit true harmonic waves; instead, they emit waves that undergo random phase changes millions of times per second. Such light is called incoherent . Interference still occurs when light waves from two incoherent sources overlap in space, but the interference pattern fluctuates randomly as the phases of the waves shift randomly. Detectors of light, including the eye, cannot register the quickly shifting interference patterns, and only a time-averaged intensity is observed. Laser light is approximately monochromatic (consisting of a single wavelength) and is highly coherent; it is thus an ideal source for revealing interference effects.

After 1802, Young’s measurements of the wavelengths of visible light could be combined with the relatively crude determinations of the speed of light available at the time in order to calculate the approximate frequencies of light. For example, the frequency of green light is about 6 × 10 14 Hz ( hertz , or cycles per second). This frequency is many orders of magnitude larger than the frequencies of common mechanical waves. For comparison, humans can hear sound waves with frequencies up to about 2 × 10 4 Hz. Exactly what was oscillating at such a high rate remained a mystery for another 60 years.

Young's Double Slit Experiment

The Original Experiment

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M.S., Mathematics Education, Indiana University
B.A., Physics, Wabash College

Throughout the nineteenth century, physicists had a consensus that light behaved like a wave, in large part thanks to the famous double slit experiment performed by Thomas Young. Driven by the insights from the experiment, and the wave properties it demonstrated, a century of physicists sought out the medium through which light was waving, the luminous ether . Though the experiment is most notable with light, the fact is that this sort of experiment can be performed with any type of wave, such as water. For the moment, however, we'll focus on the behavior of light.

What Was the Experiment?

In the early 1800s (1801 to 1805, depending on the source), Thomas Young conducted his experiment. He allowed light to pass through a slit in a barrier so it expanded out in wave fronts from that slit as a light source (under Huygens' Principle ). That light, in turn, passed through the pair of slits in another barrier (carefully placed the right distance from the original slit). Each slit, in turn, diffracted the light as if they were also individual sources of light. The light impacted an observation screen. This is shown to the right.

When a single slit was open, it merely impacted the observation screen with greater intensity at the center and then faded as you moved away from the center. There are two possible results of this experiment:

Particle interpretation: If light exists as particles, the intensity of both slits will be the sum of the intensity from the individual slits.

Wave interpretation: If light exists as waves, the light waves will have interference under the principle of superposition , creating bands of light (constructive interference) and dark (destructive interference).

When the experiment was conducted, the light waves did indeed show these interference patterns. A third image that you can view is a graph of the intensity in terms of position, which matches with the predictions from interference.

Impact of Young's Experiment

At the time, this seemed to conclusively prove that light traveled in waves, causing a revitalization in Huygen's wave theory of light, which included an invisible medium, ether , through which the waves propagated. Several experiments throughout the 1800s, most notably the famed Michelson-Morley experiment , attempted to detect the ether or its effects directly.

They all failed and a century later, Einstein's work in the photoelectric effect and relativity resulted in the ether no longer being necessary to explain the behavior of light. Again a particle theory of light took dominance.

Expanding the Double Slit Experiment

Still, once the photon theory of light came about, saying the light moved only in discrete quanta, the question became how these results were possible. Over the years, physicists have taken this basic experiment and explored it in a number of ways.

In the early 1900s, the question remained how light — which was now recognized to travel in particle-like "bundles" of quantized energy, called photons, thanks to Einstein's explanation of the photoelectric effect — could also exhibit the behavior of waves. Certainly, a bunch of water atoms (particles) when acting together form waves. Maybe this was something similar.

One Photon at a Time

It became possible to have a light source that was set up so that it emitted one photon at a time. This would be, literally, like hurling microscopic ball bearings through the slits. By setting up a screen that was sensitive enough to detect a single photon, you could determine whether there were or were not interference patterns in this case.

One way to do this is to have a sensitive film set up and run the experiment over a period of time, then look at the film to see what the pattern of light on the screen is. Just such an experiment was performed and, in fact, it matched Young's version identically — alternating light and dark bands, seemingly resulting from wave interference.

This result both confirms and bewilders the wave theory. In this case, photons are being emitted individually. There is literally no way for wave interference to take place because each photon can only go through a single slit at a time. But the wave interference is observed. How is this possible? Well, the attempt to answer that question has spawned many intriguing interpretations of quantum physics , from the Copenhagen interpretation to the many-worlds interpretation.

It Gets Even Stranger

Now assume that you conduct the same experiment, with one change. You place a detector that can tell whether or not the photon passes through a given slit. If we know the photon passes through one slit, then it cannot pass through the other slit to interfere with itself.

It turns out that when you add the detector, the bands disappear. You perform the exact same experiment, but only add a simple measurement at an earlier phase, and the result of the experiment changes drastically.

Something about the act of measuring which slit is used removed the wave element completely. At this point, the photons acted exactly as we'd expect a particle to behave. The very uncertainty in position is related, somehow, to the manifestation of wave effects.

More Particles

Over the years, the experiment has been conducted in a number of different ways. In 1961, Claus Jonsson performed the experiment with electrons, and it conformed with Young's behavior, creating interference patterns on the observation screen. Jonsson's version of the experiment was voted "the most beautiful experiment" by Physics World readers in 2002.

In 1974, technology became able to perform the experiment by releasing a single electron at a time. Again, the interference patterns showed up. But when a detector is placed at the slit, the interference once again disappears. The experiment was again performed in 1989 by a Japanese team that was able to use much more refined equipment.

The experiment has been performed with photons, electrons, and atoms, and each time the same result becomes obvious — something about measuring the position of the particle at the slit removes the wave behavior. Many theories exist to explain why, but so far much of it is still conjecture.

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Young’s double slit experiment derivation

One of the first demonstration of the intererference of light waves was given by Young – an English physicist in 1801. We have learnt that two essential conditions to obtain an interference phenomenon are :

Two sources should be coherent and
Two coherent sources must be placed close to each other as the wavelength of light is very small.
1 Young’s double slit experiment derivation
2 Theory of the Experiment
3.1 Maxima or Bright fringes
3.2 Minima or Dark fringes
4.1 Double slit experiment formula?
4.2 Fringe width formula in Young’s experiment?

Young placed a monochromatic source (S) of light in front of a narrow slit S 0 and arranged two very narrow slits S₁ and S₂ close to each other in front of slit S 0 young’s double slit experiment derivation diagram below. Slits S₁ and S₂ are equidistant from S 0 , so the spherical wavefronts emitted by slit S 0 reach the slits S₁ and S₂ in equal time.

These wavefronts after arriving at S₁ and S₂ spread out of these slits. Thus the emerging waves are of the same amplitude and wavelength and are in phase. Hence slits S₁ and S₂ behave as coherent sources.

The wavefronts emitted by coherent sources S₁ and S₂ superpose and give rise to interference . When these wavefronts are received on the screen, interference fringes are seen as shown in young’s double slit experiment diagram below.

The points where the destructive interference takes place, we get minima or dark fringe and where the constructive interference takes place, maxima or bright fringe is obtained. The pattern of these dark and bright fringes obtained on the screen is called interference pattern.

Young had used sun light as source of light and circular slits in his experiment.

Theory of the Experiment

Suppose S is the monochromatic source of light. S 0 is the slit through which the light passes and illuminates the slits S₁ and S₂. The waves emitted by slits S₁ and S₂ are the part of the same wavefront, so these waves have the same frequency and the same phase.

Hence slits S 1 and S 2 behave as two coherent sources. Interference takes place on the screen. If we consider a point O on the perpendicular bisector of S₁S 2 , the waves traveling along S₁O and S₂O have traveled equal distances. Hence they will arrive at O in phase and interfere constructively to make O the centre of a bright fringe or maxima.

Derivation of Young’s double slit experiment

To locate the position of the maxima and minima on both sides of O, consider any point P at a distance x from O. Join S 1 P and S 2 P. Now draw S 1 N normal on S 2 P. Then the path difference between S 2 P and S 1 P

Now from △ S 1 PL,

and from △ S 2 PM,

Since the distance of screen from slits S 1 and S 2 is very large, so S 2 P ≈S 1 P ≈D

Path difference,

Maxima or Bright fringes

If the path difference (S 2 P-S 1 P) = xd/D is an integral multiple of λ, then the point P will be the position of bright fringe or maxima.

That is for bright fringe,

Eqn. (1) gives the position of different bright fringes.

P = 0, x =0 , i.e., the central fringe at O will be bright.

This is the position of first bright fringe w.r.t. point O.

This is the position of second bright fringe w.r.t. point O.

………………………………………………………………………………………

This is the position of pth bright fringe w.r.t. point O.

This is the position of (p+1) bright fringe w.r.t. point O.

The distance between two successive bright fringes is called fringe width and is given by

Minima or Dark fringes

If the path difference (S 2 P-S 1 P)=xd/D is an odd multiple of λ/2 , then the point P will be the position of dark fringes or minima.

Thus for dark fringes,

Eqn. (3) gives the position of different dark fringes.

This is the position of first dark fringe w.r.t. point O.

This is the position of second dark fringe w.r.t. point O.

This is the position of third dark fringe w.r.t. point O.

…………………………………………………………………………….

This is the position of pth dark fringe w.r.t. point O.

This is the position of (p+1) dark fringe w.r.t. point O.

The distance between two successive dark fringes is called fringe width (β) of the dark fringes which is given by

This eqn. (4), ‘ β = λD/d’ is called Fringe width formula in Young’s experiment .

From eqns. (2) and (4), it is evident that the fringes width of bright fringe and dark fringe is the same.

If we know the value of “D” and “d” then the measurement of the fringe width ( β ) gives a direct determination of the wavelength of light.

FAQ on Young’s double slit experiment derivation

Double slit experiment formula.

In a double-slit experiment, λ= xd / L is the formula for the calculation of wavelength.

Fringe width formula in Young’s experiment?

If we know the value of “D” and “d” then the measurement of the fringe width ( β ) gives a direct determination of the wavelength of light. Fringe width formula in Young’s experiment is given by: β = λD/d

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Young's double slit experiment diagram. Interference of light waves
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Young's Double Slit Experiment
Learn about the wave theory of light using Young's double slit experiment, which involves two coherent sources of light passing through two slits and forming interference patterns on a screen. Find the position, shape, intensity and fringe width of bright and dark fringes, and the special cases of this experiment.
Young's Double Slit Experiment
Learn how Thomas Young proved that light is a wave by passing it through two slits and observing interference patterns. Explore the conditions for constructive and destructive interference, the role of coherence and wavelength, and the analogy with water waves.
Physics Tutorial: Young's Experiment
Learn how to perform and analyze Young's experiment to measure the wavelength of light using a laser and a double-slit apparatus. See sample data, interactive, and check your understanding questions.
Double-slit experiment
Learn how light and matter exhibit wave-particle duality in this classic quantum mechanics experiment. Explore the history, variations, and interpretations of the double-slit experiment with examples and diagrams.
27.3 Young's Double Slit Experiment
Learn how light shows wave behavior when passing through two slits and forming interference patterns on a screen. The figure shows a double slit located a distance x = 3.2 m from a screen, with the distance from the center of the screen to a point on the screen being y.
PDF Young's Double Slit Experiment
Learn how to measure the wavelength of light using a double slit apparatus and a spectrometer. The web page explains the theory, procedure and analysis of the experiment with graphs and equations.
27.3 Young's Double Slit Experiment
Learn how light shows wave behavior in a double slit experiment and how to calculate the angles of constructive and destructive interference. See examples, diagrams, and a take-home experiment with fingers as slits.
154 Young's Double Slit Experiment
We illustrate the double slit experiment with monochromatic (single \ (\lambda \)) light to clarify the effect. (Figure) shows the pure constructive and destructive interference of two waves having the same wavelength and amplitude. The amplitudes of waves add. (a) Pure constructive interference is obtained when identical waves are in phase.
Physics Video Tutorial
Learn how to use Young's equation and perform Young's experiment to measure the wavelength of light. The video lesson provides examples, illustrations, lesson notes, and additional resources.
Young's Double Slits Experiment Derivation
Learn how to derive the formula for the position of bright and dark fringes in Young's double slit experiment using the laws of cosines and interference. See the diagram, the conditions for constructive and destructive interference, and the FAQs.
Young's double slit experiment
Learn how to measure the wavelength of light using interference fringes produced by a double slit. Find the formula for fringe width and see an example problem.
Young's Double Slit Experiment: A Simple Explanation
Learn how light and matter behave as waves and particles in the double-slit experiment. Discover the quantum weirdness of observation, interference and wave-particle duality.
Double-slit Experiment
Learn about the quantum phenomenon of wave-particle duality using the double-slit experiment. Find out how electrons, photons, and other particles behave like waves and interfere with themselves when passing through two slits.
Light
Learn how Thomas Young demonstrated the interference of light waves by passing them through two parallel slits and observing the fringes on a screen. Find out the conditions and parameters for constructive and destructive interference, and the role of coherence and wavelength in light.
Thomas Young's Double Slit Experiment
Learn how Thomas Young performed the original experiment in the early 1800s and how it proved that light behaved like a wave. Explore how the experiment has been extended to other particles and how it challenges the wave-particle duality of quantum physics.
Young's interference experiment
Learn about the original double-slit experiment performed by Thomas Young in the early 19th century, which supported the wave theory of light. See how he used a paper card to split a beam of light into two parts and observed the interference patterns on a screen.
PDF Chapter 14 Interference and Diffraction
Learn about the superposition of waves, Young's double-slit experiment, intensity distribution, and diffraction patterns in this chapter from MIT's electromagnetics course. See examples, problems, and conceptual questions on interference and diffraction of light.
Young's double slit experiment derivation
Learn how to derive the formula for the position of dark fringes in Young's double slit experiment. The formula is given by x = (p+1/2)λD/d, where p is the order of the dark fringe and D and d are the distance and separation of the slits.
Young's Double Slit Experiment, Chapter 10, Wave Optics ...
Class 12 Physics https://www.youtube.com/@DynamicVidyapeeth/playlists?view=50&sort=dd&shelf_id=2Chapter 1, Electric Charges and Fields https://youtube.com/pl...
Physics Simulation: Young's Experiment
Turn it on. Shine it at a couple of closely-spaced double slits and project the diffraction pattern onto a distant screen. Make some measurements and use Young's Equation to determine the wavelength of light. Repeat your measurements and calculations for all three lasers and check your answers.
Wave Interference
Explore wave interference with a dripping faucet, audio speaker, or laser. Adjust the sources, barriers, and apertures to create and observe different patterns of interference and diffraction.

Wave Optics

Take-Home Experiment: Using Fingers as Slits

Example 1. Finding a Wavelength from an Interference Pattern

Example 2. Calculating Highest Order Possible

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Section Summary

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Selected Solutions to Problems & Exercises

27.3 Young’s Double Slit Experiment

Take-Home Experiment: Using Fingers as Slits

Example 1: Finding a Wavelength from an Interference Pattern

Example 2: Calculating Highest Order Possible

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154 Young’s Double Slit Experiment

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Double-slit Experiment

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