young's experiment light waves

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.

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 27.10 ).

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

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?

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

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.

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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).

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 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.

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.

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?

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.

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.

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?

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]

Candela Citations

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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.

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 ).

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

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?

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

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 [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

College Physics Copyright © August 22, 2016 by OpenStax is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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Young's Experiment

• Anatomy of a Two-Point Source Interference Pattern
• The Path Difference
• Young's Equation
• Young's Experiment
• Other Applications of Two-Point Source Interference

Today's classroom version of the same experiment is typically performed using a laser beam as the source. Rather than using a note card to split the single beam into two coherent beams, a carbon-coated glass slide with two closely spaced etched slits is used. The slide with its slits is most commonly purchased from a manufacturer who provides a measured value for the slit separation distance - the d value in Young's equation. Light from the laser beam diffracts through the slits and emerges as two separate coherent waves. The interference pattern is then projected onto a screen where reliable measurements can be made of L and y for a given bright spot with order value m . Knowing these four values allows a student to determine the value of the wavelength of the original light source.

To illustrate some typical results from this experiment and the subsequent analysis, consider the sample data provided below for d, y, L and m.

(Note: AN 0 = central antinode and AN 4 = fourth antinode)  

The determination of the wavelength demands that the above values for d, y, L and m be substituted into Young's equation.

Careful inspection of the units of measurement is always advisable. The sample data here reveal that each measured quantity is recorded with a different unit. Before substituting these measured values into the above equation, it is important to give some thought to the treatment of units. One means of resolving the issue of nonuniform units is to simply pick a unit of length and to convert all quantities to that unit. If doing so, one might want to pick a unit that one of the data values already has so that there is one less conversion. A wise choice is to choose the meter as the unit to which all other measured values are converted. Since there are 1000 millimeters in 1 meter, the 0.250 mm is equivalent to 0.000250 meter. And since there are 100 centimeters in 1 meter, the 10.2 cm is equivalent to 0.102 m. Thus, the new values of d, y and L are:

While the conversion of all the data to the same unit is not the only means of treating such measured values, it might be the most advisable - particularly for those students who are less at ease with such conversions.

Now that the issue regarding the units of measurement has been resolved, substitution of the measured values into Young's equation can be performed.

λ  = 6.52 x 10 -7 m

As is evident here, the wavelength of visible light is rather small. For this reason wavelength is often expressed using the unit nanometer, where 1 meter is equivalent to 10 9 nanometers. Multiplying by 10 9 will convert the wavelength from meters to nanometers (abbreviated nm).

We Would Like to Suggest ...

Check Your Understanding

1. The diagram below depicts the results of Young's Experiment. The appropriate measurements are listed on the diagram. Use these measurements to determine the wavelength of light in nanometers. (GIVEN: 1 meter = 10 9 nanometers)

Answer: 657 nm

First, identify known values in terms of their corresponding variable symbol:

L = 10.2 m = 1020 cm y = 22.5 cm m = 10 d = 0.298 mm = 0.0298 cm

(Note: m was chosen as 10 since the y distance corresponds to the distance from the 5th bright band on one side of the central band and the 5th bright band on the other side of the central band.)

Then convert all known values to an identical unit. In this case, cm has been chosen as the unit to use. The converted values are listed in the table above.

Substitute all values into Young's equation and perform calculation of the wavelength. The unit of wavelength is cm.

λ = y • d / ( m • L) λ = ( 22.5 cm ) • ( 0.0298 cm ) / [ ( 10 ) • ( 1020 cm ) ] λ = 6.57 x 10 -5 cm

Finally convert to nanometers using a conversion factor. If there are 10 9 nm in 1 meter, then there must be 10 7 nm in the smaller centimeter.

λ = ( 6.57 x 10 -5 cm ) • ( 10 7 nm / 1 cm ) = 657 nm

2. A student uses a laser and a double-slit apparatus to project a two-point source light interference pattern onto a whiteboard located 5.87 meters away. The distance measured between the central bright band and the fourth bright band is 8.21 cm. The slits are separated by a distance of 0.150 mm. What would be the measured wavelength of light?

Answer: 524 nm

L = 5.87 m = 587 cm y = 8.21 cm m = 4 d = 0.150 mm = 0.0150 cm

λ = y • d / ( m • L) λ = ( 8.21 cm ) • ( 0.0150 cm ) / [ ( 4 ) • ( 587 cm ) ] λ = 5.24 x 10 -5 cm

λ = ( 5.24 x 10 -5 cm ) • ( 10 7 nm / 1 cm ) = 524 nm

3. The analysis of any two-point source interference pattern and a successful determination of wavelength demands an ability to sort through the measured information and equating the values with the symbols in Young's equation. Apply your understanding by interpreting the following statements and identifying the values of y, d, m and L. Finally, perform some conversions of the given information such that all information share the same unit.

This question simply asks to equate the stated information with the variables of Young's equation and to perform conversions such that all information is in the same unit.

(Note that m = 4.5 represents the fifth nodal position or dark band from the central bright band. Also note that the given values have been converted to cm.)

b. An interference pattern is produced when light is incident upon two slits that are 50.0 micrometers apart. The perpendicular distance from the midpoint between the slits to the screen is 7.65 m. The distance between the two third-order antinodes on opposite sides of the pattern is 32.9 cm.

This question simply asks to equate the stated information with the variables of Young's equation and to perform conversions such that all information is in the same unit. 

(Note that m = 6 corresponds to six spacings. There are three spacings between the central antinode and the third antinode. The stated distance is twice as far so the m value must be doubled. Also note that the given values have been converted to cm. There are 10 6 µm in one meter; so there are 10 4 µm in one centimeter.)

c. The fourth nodal line on an interference pattern is 8.4 cm from the first antinodal line when the screen is placed 235 cm from the slits. The slits are separated by 0.25 mm.

 ( Note that the fourth nodal line is assigned the order value of 3.5. Also note that the given values have been converted to cm.)

d. Two sources separated by 0.500 mm produce an interference pattern 525 cm away. The fifth and the second antinodal line on the same side of the pattern are separated by 98 mm.

  ( Note that there are three spacings between the second and the fifth bright bands. Since all spacings are the same distance apart, the distance between the second and the fifth bright bands would be the same as the distance between the central and the third bright bands. Thus, m = 3. Also note that the given values have been converted to cm.)

e. Two slits that are 0.200 mm apart produce an interference pattern on a screen such that the central maximum and the 10th bright band are distanced by an amount equal to one-tenth the distance from the slits to the screen.

 ( Note that there are 10 spacings between the central anti-node and the tenth bright band or tenth anti-node. And observe that they do not state the actual values of L and y; the value of y is expressed in terms of L. )

f. The fifth antinodal line and the second nodal line on the opposite side of an interference pattern are separated by a distance of 32.1 cm when the slits are 6.5 m from the screen. The slits are separated by 25.0 micrometers.

 ( Note that there are five spacings between the central anti-node and the fifth anti-node. And there are 1.5 spacings from the central anti-node in the opposite direction out to the second nodal line. Thus, m = 6.5. Also note that the given values have been converted to cm. There are 10 6 µm in one meter; so there are 10 4 µm in one centimeter.)

g. If two slits 0.100 mm apart are separated from a screen by a distance of 300 mm, then the first-order minimum will be 1 cm from the central maximum.

  ( Note that a the first-order minimum is a point of minimum brightness or a nodal position. The first-order minimum is the first nodal position and is thus the m = 0.5 node. Also note that the given values have been converted to cm. )

h. Consecutive bright bands on an interference pattern are 3.5 cm apart when the slide containing the slits is 10.0 m from the screen. The slit separation distance is 0.050 mm.

   ( Note that the spacing between adjacent bands is given. This distance is equivalent with the distance from the central bright band to the first antinode. Thus, m = 1. Also note that the given values have been converted to cm. )