young's 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]

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27.3 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.

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 λ ) light to clarify the effect. Figure 27.11 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 27.12 (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 27.12 (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 27.13 . 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 27.13 (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 27.13 (b). More generally, if the paths taken by the two waves differ by any half-integral number of wavelengths [ ( 1 / 2 ) λ ( 1 / 2 ) λ , ( 3 / 2 ) λ ( 3 / 2 ) λ , ( 5 / 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 λ , 2 λ 2 λ , 3 λ 3 λ , etc.), then constructive interference occurs.

Take-Home Experiment: Using Fingers as Slits

Figure 27.14 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 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 d sin θ , where d 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 λ λ is the wavelength of the light, d d is the distance between slits, and θ θ is the angle from the original direction of the beam as discussed above. We call m m the order of the interference. For example, m = 4 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 27.15 . 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 λ λ and m m , the smaller d d is, the larger θ θ must be, since sin θ = mλ / d sin θ = mλ / d . This is consistent with our contention that wave effects are most noticeable when the object the wave encounters (here, slits a distance d d apart) is small. Small d d gives large θ θ , hence a large effect.

Example 27.1

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 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 m = 3 . We are given d = 0 . 0100 mm d = 0 . 0100 mm and θ = 10 . 95º θ = 10 . 95º . The wavelength can thus be found using the equation d sin θ = mλ d sin θ = mλ for constructive interference.

The equation is d sin θ = mλ d sin θ = mλ . Solving for the wavelength λ λ 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 λ λ , so that spectra (measurements of intensity versus wavelength) can be obtained.

Example 27.2

Calculating highest order possible.

Interference patterns do not have an infinite number of lines, since there is a limit to how big m 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, … ) d sin θ = mλ (for m = 0, 1, − 1, 2, − 2, … ) describes constructive interference. For fixed values of d d and λ λ , the larger m m is, the larger sin θ sin θ is. However, the maximum value that sin θ sin θ can have is 1, for an angle of 90º 90º . (Larger angles imply that light goes backward and does not reach the screen at all.) Let us find which m m corresponds to this maximum diffraction angle.

Solving the equation d sin θ = mλ d sin θ = mλ for m m gives

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

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

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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.

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.

Take-Home Experiment: Using Fingers as Slits

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

$\boldsymbol{d \;\textbf{sin} \;\theta = m +}$

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

$\boldsymbol{d \;\textbf{sin} \;\theta = m \theta , \;\textbf{for} \; m = 0, \;1, \; -1, \; 2, \; -2, \; \dots}.$

Example 1: Finding a Wavelength from an Interference Pattern

$\boldsymbol{10.95 ^{\circ}}$

Substituting known values yields

$$\begin{array}{r @{{}={}}l} \boldsymbol{\lambda} & \boldsymbol{\frac{(0.0100 \;\textbf{mm})(\textbf{sin} 10.95^{\circ})}{3}} \\[1em] & \boldsymbol{6.33 \times 10^{-4} \;\textbf{mm} = 633 \;\textbf{nm}}. \end{array}$

Example 2: Calculating Highest Order Possible

Strategy and Concept

$\boldsymbol{d \;\textbf{sin} \;\theta = m \lambda \; (\textbf{for} \; m = 0, \; 1, \; -1, \; 2, \; -2, \; \dots)}$

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.

$\boldsymbol{d \;\textbf{sin} \;\theta = m \lambda \;(\textbf{for} \; m = 0, \; 1, \; -1, \;2, \; -2, \dots)}$

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.

$\boldsymbol{30.0 ^{\circ}}$

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

$\boldsymbol{25.0 \;\mu \textbf{m}}$

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?

$\boldsymbol{10.0^{\circ}}$

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 ).

$\boldsymbol{0.516 ^{\circ}}$

9: 1200 nm (not visible)

11: (a) 760 nm

(b) 1520 nm

$\boldsymbol{\textbf{sin} \;\theta - \;\textbf{tan} \;\theta \approx \theta}$

For two adjacent fringes we have,

$\boldsymbol{d \;\textbf{sin} \;\theta _{\textbf{m}} = m \lambda}$

Subtracting these equations gives

$\begin{array}{r @{{}={}}l} \boldsymbol{d (\textbf{sin} \; \theta _{\textbf{m} + 1} - \textbf{sin} \; \theta _{\textbf{m}})} & \boldsymbol{[(m + 1) - m] \lambda} \\[1em] \boldsymbol{d(\theta _{{\textbf{m}} + 1} - \theta _{\textbf{m}})} & \boldsymbol{\lambda} \end{array}$

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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.

Lookang, . CC-3.0 Licensing . Retrieved from https://commons.wikimedia.org/w/index.php?curid=17014507
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Why is light important for life on Earth?

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Table Of Contents

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.)

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 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

PHYS102: Introduction to Electromagnetism

Young's double slit experiment.

As the reading illustrates, Huygens' principle is not just a philosophical interpretation – it is also a computational tool . In particular, the idea of circular (or spherical ) elementary waves makes it relatively easy to explain how a wave can bend around corners and spread out after passing through a constriction. This is called diffraction because it allows wave energy to go around corners in directions that the rays of geometric optics (or the trajectories of classical particles) would not be permitted to go.

Read about the proof that light is a wave in this experiment Thomas Young gave using diffraction by a pair of closely spaced slits.

Figure 27.10 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.

By coherent , we mean waves are in phase or have a definite phase relationship. Incoherent means the waves have random phase relationships.

Figure 27.11 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.

Figure 27.12 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)

Take-Home Experiment: Using Fingers as Slits

Figure 27.13 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.

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

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 27.15 . 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

Figure 27.15 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 27.1 Finding a Wavelength from an Interference Pattern

Substituting known values yields

Example 27.2 Calculating Highest Order Possible

Strategy and concept.

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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and .

(separate page)

Young, a contemporary of Newton, performed his celebrated experiment with light, to demonstrate its wave nature. Here, we'll look first at a similar experiment using water waves, for which the displacements are visible. Two pencils attached to a frame are being sinusoidally vibrated in the vertical direction. They touch the water and create waves that spread out radially.

In the upper views, on the axis of symmetry, we can see constructive interference: along this line, the combined waves from the two sources has maximum amplitude. In the left images, this is marked by a red line on both the upper and lower views. This is called constructive interference and it creates an antinode in the wave pattern.

A little to the left of that line, we can see a line where the wave combination hardly disturbs the water at all: destructive interference or a node. This is marked by a blue line on both views. Along this blue line, the distance from the two sources differs by half a wavelength, hence the destructive interference: the waves arrive there half a cycle out of phase.

To the left of both lines, there is another line of antinode, again marked with a red line. Along this line, the distance from the two sources differs by one wavelength. The pattern of nodal and antinodal lines continues all the way around the two sources.

The two different views are of the same apparatus, taken from different angles. In the upper shot, we see the waves on the water surface. On the lower, we see the distribution of light intensity due to the refaction of light by the waves. In the experiment above, the clip is cycling over seven frames. For this frequency, the lower view is not very clear. For that reason, we show below a slightly higher frequency.

In the clip at right, the frequency is about 15% higher. This time the lower view is clear, but the upper view is less clear. By the way: for waves of this size, both the surface tension of water and gravity contribute to the restoring force, so the wave speed is not constant, but is a complicated function of the wavelength, so wavelength and period are not proportional.

Young's experiment with laser light

We can see an analogy with the water experiment above: on the axis of symmetry, we see a bright spot, where light from the two sources interferes constructively. A little to the left, a node (black in the pattern). Then an antinode (bright red) where the distance from the two slits must differ by one wavelength. Let's look at the geometry in another diagram.

Young's experiment: the geometry

It's interesting to note that the photographed pattern doesn't 'look like' the intensity graph plotted beside it. We return to this below.

Comparison: Young's experiment with water waves and with light

So both water waves and light exhibit interference – a property of waves. But does this explain how light casts shadows? Go to this page about Shadows, particles and waves . This link will return you to the multimedia tutorial The Nature of Light .

Pathlengths and interference If we look down on a Young's experiment from above, what do we see? From the two sources (two slits for light) wave fronts spread out symmetrically in all directions. The animation shows this for the water waves, which radiate in all directions. For light, the radiation spreads over only a small angle, as we'll see later.

In the animation above, grey circles are wavefronts. The pink lines show loci of constructive interference. The black lines show points where the path difference from the two slits differs by an integral number of wavelengths. Grey straight lines are loci of destructive interference.

Small angle approximation

For Young's experiment with light, the distance L to the screen is typically very much greater the distance between slits, so lines drawn from a point on the screen to each of the slits are almost parallel, an approximation we'll use next. (Further, the interference pattern is usually spread over a very small angle, so the small angle approximation is usually appropriate.)

Finding the angular position of maxima and minima

The distance between the slits is d . Consider paths from the two slits at angle θ . From the (almost) right angle triangle shown, the difference in path length to the screen from the two slits is

Divide this by the wavelength to get the number of cycles out of phase, and it follows that the phase difference at the screen, in radians, is

For de structive interference , the minima in brightness, the phase difference should be π, 3π etc, i.e.

Phasors: calculating the interference intensity as a function of angle

The phasors in our example have the same magnitude A because the two slits are approximately equidistant from the screen. The phase φ varies as we move away from the midpoint of the pattern, as shown in the previous section. At θ = 0, φ = 0 and the resultant has a magnitude of 2 A , as shown at left. The diagram at right shows an arbitrary phase difference φ . Looking at the two similar right angle triangles, each with angle φ /2, we can see that the phasor sum has magnitude 2 A .cos ( φ /2).

(φ/2) vs. the image

Looking at the image, we can see two ways in which it doesn't look like a cos 2 function. First, the maxima become weaker as we go further from the centre of the pattern. This is the result of diffraction, which is due to For a Young's experiment using radio waves, it is easy to make the source dimensions much smaller than the distance between them. This is not at all easy for light and, as we shall see below, the finite size of the slits contributes a broader pattern that modulates the cos 2 function.

Looking at the image, we also notice that the brightness of the maxima does not appear to vary much – it looks as though the brightness saturates over nearly a third of a cycle. The main reason is the nonlinear response of the eye to light intensity: the eye has a dynamic range of about 90 dB, and this cannot be achieved with a linear response. (The camera also has a nonlinear response but the image looks much like this to the naked eye, as well.) So let's check

Measuring the intensity as a function of angle

Here, a phototransistor is illuminated by the interference pattern. Due to the photovoltaic effect, each photon captured by the base emitter junction (used here as a photodiode) provides a conduction electron, so the diode current is proportional to the number of photons. This current is input to an operational amplifier, whose feedback resistor converts it to a voltage, so the voltage shown on the oscilloscope is proportiotnal to the rate of photon capture. The scanning across the pattern is done by my hand, which does not move at exactly constant speed. Consequently, the recorded shape is not precisely cos 2 , but it does show the clear peak with no saturation.

Why does more light make the pattern darker?

So here arise the remarkable questions about interference. The intensity of light is proportional to the rate of photon arrivals. Two slits admit twice as many photons so, on a.verage, the lower pattern must be twice as bright as the upper.

Twice as many photons gives less light. Look, however, at the yellow rectangle at left, located at a minimum in the interference pattern. At this point, doubling the number of photos passing through the slits has decreased the light intensity. In the wave picture, we know why: the electric fields due to the waves coming from the two slits cancel out, giving negligible electric field and so negligible intensity: destructive interference. Nevertheless, it is interesting that, at this point, adding more photons removes light!

Twice as many photons gives four times as much light. Next, look at the yellow rectangle at right, located at a maximum in the interference pattern. At this point, doubling the number of photos passing through the slits increases the light intensity by a factor of four (see equation (1) above ). In the wave picture, we know why: the electric fields due to the waves coming from the two slits are in phase, which doubles the electric field and so gives four times the intensity: constructive interference. Nevertheless, it is interesting that, at this point, twice as many photons give four times as much light!

There is no magic here: the photons are differently distributed between the two patterns: at the maxima, there are four times as many in the interference pattern, at the minima none. So, on average over the pattern, there are twice as many photons from two slits as there are from one. The interference pattern determines the probability of capturing photons . We'll look at this again when we do Young's experiment, one photon at a time .

Combination of interference and diffraction

Further information.

IIT JEE Study Material
Youngs Double Slit Experiment

Young's Double Slit Experiment

What is young’s double slit experiment.

Young’s double slit experiment uses two coherent sources of light placed at a small distance apart. Usually, only a few orders of magnitude greater than the wavelength of light are used. Young’s double slit experiment helped in understanding the wave theory of light , which is explained with the help of a diagram. As shown, a screen or photodetector is placed at a large distance, ‘D’, away from the slits.

Download Complete Chapter Notes of Wave Optics Download Now

JEE Main 2021 LIVE Physics Paper Solutions 24 Feb Shift-1 Memory-based

The original Young’s double slit experiment used diffracted light from a single source passed into two more slits to be used as coherent sources. Lasers are commonly used as coherent sources in modern-day experiments.

Position of Fringes
Shape of Fringes
Intensity of Fringes

Special Cases

Displacement of Fringes

Each source can be considered a source of coherent light waves . At any point on the screen at a distance ‘y’ from the centre, the waves travel distances l 1 and l 2 to create a path difference of Δl at the point. The point approximately subtends an angle of θ at the sources (since the distance D is large, there is only a very small difference between the angles subtended at sources).

Derivation of Young’s Double Slit Experiment

Consider a monochromatic light source ‘S’ kept at a considerable distance from two slits: s 1 and s 2 . S is equidistant from s 1 and s 2 . s 1 and s 2 behave as two coherent sources as both are derived from S.

The light passes through these slits and falls on a screen which is at a distance ‘D’ from the position of slits s 1 and s 2 . ‘d’ is the separation between two slits.

If s 1 is open and s 2 is closed, the screen opposite to s 1 is closed, and only the screen opposite to s 2 is illuminated. The interference patterns appear only when both slits s 1 and s 2 are open.

When the slit separation (d) and the screen distance (D) are kept unchanged, to reach P, the light waves from s 1 and s 2 must travel different distances. It implies that there is a path difference in Young’s double slit experiment between the two light waves from s 1 and s 2 .

Approximations in Young’s double slit experiment

Approximation 1: D > > d: Since D > > d, the two light rays are assumed to be parallel.
Approximation 2: d/λ >> 1: Often, d is a fraction of a millimetre, and λ is a fraction of a micrometre for visible light.

Under these conditions, θ is small. Thus, we can use the approximation sin θ = tan θ ≈ θ = λ/d.

∴ path difference, Δz = λ/d

This is the path difference between two waves meeting at a point on the screen. Due to this path difference in Young’s double slit experiment, some points on the screen are bright, and some points are dark.

Now, we will discuss the position of these light and dark fringes and fringe width.

Position of Fringes in Young’s Double Slit Experiment

Position of bright fringes.

For maximum intensity or bright fringe to be formed at P,

Path difference, Δz = nλ (n = 0, ±1, ±2, . . . .)

i.e., xd/D = nλ

The distance of the n th bright fringe from the centre is

x n = nλD/d

Similarly, the distance of the (n-1) th bright fringe from the centre is

x (n-1) = (n -1)λD/d

Fringe width, β = x n – x (n-1) = nλD/d – (n -1)λD/d = λD/d

(n = 0, ±1, ±2, . . . .)

Position of Dark Fringes

For minimum intensity or dark fringe to be formed at P,

Path difference, Δz = (2n + 1) (λ/2) (n = 0, ±1, ±2, . . . .)

i.e., x = (2n +1)λD/2d

The distance of the n th dark fringe from the centre is

x n = (2n+1)λD/2d

x (n-1) = (2(n-1) +1)λD/2d

Fringe width, β = x n – x (n-1) = (2n + 1) λD/2d – (2(n -1) + 1)λD/2d = λD/d

Fringe Width

The distance between two adjacent bright (or dark) fringes is called the fringe width.

If the apparatus of Young’s double slit experiment is immersed in a liquid of refractive index (μ), then the wavelength of light and fringe width decreases ‘μ’ times.

If white light is used in place of monochromatic light, then coloured fringes are obtained on the screen, with red fringes larger in size than violet.

Angular Width of Fringes

Let the angular position of n th bright fringe is θ n, and because of its small value, tan θ n ≈ θ n

Similarly, the angular position of (n+1) th bright fringe is θ n+1, then

∴ The angular width of a fringe in Young’s double slit experiment is given by,

Angular width is independent of ‘n’, i.e., the angular width of all fringes is the same.

Maximum Order of Interference Fringes

But ‘n’ values cannot take infinitely large values as it would violate the 2 nd approximation.

i.e., θ is small (or) y < < D

When the ‘n’ value becomes comparable to d/ λ, path difference can no longer be given by d γ/D.

Hence for maxima, path difference = nλ

The above represents the box function or greatest integer function.

Similarly, the highest order of interference minima

The Shape of Interference Fringes in YDSE

From the given YDSE diagram, the path difference between the two slits is given by

The above equation represents a hyperbola with its two foci as, s 1 and s 2 .

The interference pattern we get on the screen is a section of a hyperbola when we revolve the hyperbola about the axis s 1 s 2 .

If the screen is a yz plane, fringes are hyperbolic with a straight central section.

If the screen is xy plane , the fringes are hyperbolic with a straight central section.

The Intensity of Fringes in Young’s Double Slit Experiment

For two coherent sources, s 1 and s 2 , the resultant intensity at point p is given by

I = I 1 + I 2 + 2 √(I 1 . I 2 ) cos φ

Putting I 1 = I 2 = I 0 (Since, d<<<D)

I = I 0 + I 0 + 2 √(I 0 .I 0 ) cos φ

I = 2I 0 + 2 (I 0 ) cos φ

I = 2I 0 (1 + cos φ)

For maximum intensity

phase difference φ = 2nπ

Then, path difference $\begin{array}{l}\Delta x=\frac{\lambda }{{2}{\pi }}\left( {2}n{\pi } \right)\end{array} $ = nλ

The intensity of bright points is maximum and given by

I max = 4I 0

For minimum intensity

φ = (2n – 1) π

Phase difference φ = (2n – 1)π

Thus, the intensity of minima is given by

If I 1 ≠ I 2 , I min ≠ 0.

Rays Not Parallel to Principal Axis:

From the above diagram,

Using this, we can calculate different positions of maxima and minima.

Source Placed beyond the Central Line:

If the source is placed a little above or below this centre line, the wave interaction with S 1 and S 2 has a path difference at point P on the screen.

Δ x= (distance of ray 2) – (distance of ray 1)

= bd/a + yd/D → (*)

We know Δx = nλ for maximum

Δx = (2n – 1) λ/2 for minimum

By knowing the value of Δx from (*), we can calculate different positions of maxima and minima .

Displacement of Fringes in YDSE

When a thin transparent plate of thickness ‘t’ is introduced in front of one of the slits in Young’s double slit experiment, the fringe pattern shifts toward the side where the plate is present.

The dotted lines denote the path of the light before introducing the transparent plate. The solid lines denote the path of the light after introducing a transparent plate.

Where μt is the optical path.

Then, we get,

Term (1) defines the position of a bright or dark fringe; term (2) defines the shift that occurred in the particular fringe due to the introduction of a transparent plate.

Constructive and Destructive Interference

For constructive interference, the path difference must be an integral multiple of the wavelength.

Thus, for a bright fringe to be at ‘y’,

Or, y = nλD/d

Where n = ±0,1,2,3…..

The 0th fringe represents the central bright fringe.

Similarly, the expression for a dark fringe in Young’s double slit experiment can be found by setting the path difference as

Δl = (2n+1)λ/2

This simplifies to

(2n+1)λ/2 = y d/D

y = (2n+1)λD/2d

Young’s double slit experiment was a watershed moment in scientific history because it firmly established that light behaved like a wave.

The double slit experiment was later conducted using electrons , and to everyone’s surprise, the pattern generated was similar as expected with light. This would forever change our understanding of matter and particles, forcing us to accept that matter, like light, also behaves like a wave.

Wave Optics

Young’s double slit experiment.

Frequently Asked Questions on Young’s Double Slit Experiment

What was the concept explained by young’s double slit experiment.

Young’s double slit experiment helps in understanding the wave theory of light.

What are the formulas derived from Young’s double slit experiment?

For constructive interference, dsinθ = mλ , for m = 0,1,-1,2,-2

For destructive interference, dsinθ = (m+½)λ, for m = 0,1,-1,2,-2 Here, d is the distance between the slits. λ is the wavelength of the light waves.

What is called a fringe width?

The distance between consecutive bright or dark fringe is called the fringe width.

What kind of source is used in Young’s double slit experiment?

A coherent source is used in Young’s double slit experiment.

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz

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Young's Double Slit Experiment
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 ...
Physics Tutorial: Young's Experiment
The previous section of Lesson 3 discussed Thomas Young's effort to derive an equation relating the wavelength of a light source to reliably measured distances associated with a two-point source light interference pattern. The equation, known as Young's equation is: λ = y • d / (m • L). In 1801, Young devised and performed an experiment to measure the wavelength of light.
27.3 Young's Double Slit Experiment
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 27.10).
Double-slit experiment
A diagram of Wheeler's delayed choice experiment, showing the principle of determining the path of the photon after it passes through the slit ... In 1991, Carnal and Mlynek performed the classic Young's double slit experiment with metastable helium atoms passing through micrometer-scale slits in gold foil. [62] [63] In 1999, a quantum ...
Physics Video Tutorial
Young's Experiment Video Tutorial. The Young's Experiment Video Tutorial introduces and explains Young's equation and discusses its use in Young's experiment to determine the wavelength of light. Numerous examples and illustrations assist in the explanations. The video lesson answers the following questions:
27.3 Young's Double Slit Experiment
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). Figure 1. Young's double slit experiment. Here pure-wavelength light sent through a pair of vertical slits is diffracted into a pattern on ...
Double-slit Experiment
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.
Young's interference experiment
Unlike the modern double-slit experiment, Young's experiment reflects sunlight (using a steering mirror) through a small hole, and splits the thin beam in half using a paper card. [6] [8] [9] He also mentions the possibility of passing light through two slits in his description of the experiment: Modern illustration of the double-slit experiment
PDF Young's Double Slit Experiment
slit spacing, d (mm) Number of fringes(N) distance be-tween N fringes (cm) fringe spacing, y(cm) L(m) (nm) 0.176 0.35 0.70 1.40 Analysis 1 ...
Young's double-slit experiment
Young's double-slit experiment When monochromatic light passing through two narrow slits illuminates a distant screen, a characteristic pattern of bright and dark fringes is observed. This interference pattern is caused by the superposition of overlapping light waves originating from the two slits.
Young's Double Slits Experiment Derivation
The schematic diagram of the experimental setup is shown below-. Figure (1): Young double slit experimental set up along with the fringe pattern. 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.
Young's Double-Slit Experiment
Past Papers. Edexcel. Spanish. Past Papers. CIE. Spanish Language & Literature. Past Papers. Other Subjects. Revision notes on 3.3.3 Young's Double-Slit Experiment for the AQA A Level Physics syllabus, written by the Physics experts at Save My Exams.
PDF Chapter 14 Interference and Diffraction
14.2 Young's Double-Slit Experiment In 1801 Thomas Young carried out an experiment in which the wave nature of light was demonstrated. The schematic diagram of the double-slit experiment is shown in Figure 14.2.1. Figure 14.2.1 Young's double-slit experiment. A monochromatic light source is incident on the first screen which contains a slit .
Young's Experiment Notes
Notes: The Young's Experiment Interactive is an adjustable-size file that displays nicely on smart phones, on tablets such as the iPad, on Chromebooks, and on laptops and desktops. The size of the Interactive can be scaled to fit the device that it is displayed on. The compatibility with smart phones, iPads, other tablets, and Chromebooks make ...
Young's double slit experiment derivation
Young's double slit experiment derivation. 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 ...
PHYS102: Young's Double Slit Experiment
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 27.10). Figure 27.10 Young's double slit experiment. Here pure-wavelength light sent through a pair of vertical slits is diffracted into a ...
Thomas Young's Double Slit Experiment
Young's Double Slit Experiment. The Original Experiment. 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 ...
Young's experiment
Young's experiment demonstrates interference of waves from two similar sources. It is a classic demonstration of the interference and of the nature of waves. Here we look first at Young's experiment using water waves, where the displacements due to the waves can be seen directly, as at right. ... The diagram at right shows an arbitrary phase ...
Young's Double Slit Experiment
Young's double slit experiment uses two coherent sources of light placed at a small distance apart. Usually, only a few orders of magnitude greater than the wavelength of light are used. Young's double slit experiment helped in understanding the wave theory of light, which is explained with the help of a diagram. As shown, a screen or ...
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...
Young's Experiment Interactive
Using this Interactive. The Young's Double Slit Experiment Interactive is shown in the iFrame below. There is a small hot-spot in the lower-right corner of the iFrame. Dragging this hot-spot allows you to change the size of iFrame to whatever dimensions you prefer. Now available with Task Tracker compatibility.

young's experiment diagram

Measuring the intensity as a function of angle

Why does more light make the pattern darker?

Combination of interference and diffraction

Young's Double Slit Experiment

JEE Main 2021 LIVE Physics Paper Solutions 24 Feb Shift-1 Memory-based

Table of Contents

Special Cases

Derivation of Young’s Double Slit Experiment

Position of Fringes in Young’s Double Slit Experiment

Position of Dark Fringes

Fringe Width

Angular Width of Fringes

Maximum Order of Interference Fringes

The Shape of Interference Fringes in YDSE

The Intensity of Fringes in Young’s Double Slit Experiment

For minimum intensity

Rays Not Parallel to Principal Axis:

Source Placed beyond the Central Line:

Displacement of Fringes in YDSE

Constructive and Destructive Interference

Wave Optics

Frequently Asked Questions on Young’s Double Slit Experiment

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