83 Single Slit Diffraction

Single-Slit Diffraction

So far, we have examined interference patterns produced by two slits and by diffraction gratings. Surprisingly, even a single narrow slit can produce a complex pattern of bright and dark regions. Although only one opening is present, light emerging from different parts of the slit interferes with itself. This phenomenon is known as single-slit diffraction. The diffraction pattern produced by a single slit differs noticeably from those produced by a double slit or a diffraction grating. Instead of many equally bright interference maxima, a single slit produces one broad, intense central maximum surrounded by much weaker secondary maxima. The intensity decreases rapidly as the angle from the center increases. This behavior illustrates an important property of waves: every portion of a wavefront can act as a source of new waves. Even though there is only one physical opening, different regions within that opening emit wavelets that interfere with one another after passing through the slit.

How a Single Slit Produces Interference

To understand this pattern, we apply Huygens's principle. Every point across the width of the slit acts as a source of tiny secondary wavelets. Immediately after passing through the slit, these wavelets spread in many directions. When the observation screen is far from the slit, rays traveling toward a particular point on the screen are nearly parallel. Whether that point appears bright or dark depends on the phase relationship of the wavelets arriving there. Looking straight ahead ([latex]\theta=0[/latex]), every wavelet travels the same distance to the screen. Since all wavelets remain in phase, they interfere constructively, producing the bright central maximum. At other angles, however, wavelets originating from different parts of the slit travel different distances. These path differences determine whether the waves reinforce or cancel one another.

The First Diffraction Minimum

Figure 83.2 illustrates the origin of the first dark fringe. Suppose the path difference between light coming from the top and bottom of the slit equals one wavelength. If we mentally divide the slit into two equal halves, every point in the upper half has a corresponding point in the lower half. The wavelets from each pair travel paths differing by one-half wavelength, [latex]\lambda/2[/latex]. Because waves separated by half a wavelength arrive exactly out of phase, each pair cancels through destructive interference. Since every point has a matching point that cancels it, essentially no light reaches the observation point, producing the first diffraction minimum.

Higher-Order Maxima and Minima

As the observation angle increases further, the path difference between the top and bottom of the slit continues to increase. Between successive minima, cancellation is incomplete. Some wavelets reinforce while others partially cancel, producing secondary maxima. These maxima are much weaker than the central maximum because only part of the light interferes constructively at the same time. When the path difference becomes two wavelengths, every wavelet once again has a partner that is exactly out of phase, producing the second minimum. This process repeats for larger path differences, creating alternating bright and dark regions on either side of the center.

Condition for Destructive Interference

For a slit of width [latex]D[/latex], the path difference between light from opposite edges of the slit is

[latex]\Delta L=D\sin\theta.[/latex]

Whenever this path difference equals an integer multiple of the wavelength, complete pairwise cancellation occurs and a diffraction minimum is produced.   The condition for destructive interference is therefore

[latex]D\sin\theta=m\lambda,\qquad m=\pm1,\pm2,\pm3,\ldots[/latex]

where

• • [latex]D[/latex] is the slit width,

• • [latex]\lambda[/latex] is the wavelength of the light,

• • [latex]\theta[/latex] is the diffraction angle measured from the original direction of the beam, and

• • [latex]m[/latex] is the order of the diffraction minimum.

Notice that there is no minimum corresponding to [latex]m=0[/latex]. When [latex]\theta=0[/latex], every point across the slit is exactly in phase, producing the bright central maximum rather than a minimum.

Key Concept

Single-slit diffraction demonstrates that interference does not require multiple slits. Different parts of the same slit act as coherent wave sources and interfere with one another after the light passes through the opening.   A narrower slit produces a wider diffraction pattern because the waves spread out more strongly after passing through the opening. This inverse relationship between slit width and diffraction angle is one of the fundamental properties of wave behavior.

Example 83.1: Calculating Single-Slit Diffraction

Visible light with a wavelength of 550 nm passes through a single slit. The second diffraction minimum occurs at an angle of [latex]45.0^\circ[/latex] relative to the original direction of the light.

1. What is the width of the slit?
2. At what angle does the first minimum occur?

Strategy

For a single slit, diffraction minima occur when

[latex]D\sin\theta=m\lambda.[/latex]

In part (a), the wavelength, angle, and order of the minimum are known, so we solve for the slit width [latex]D[/latex]. In part (b), we use the calculated slit width and set [latex]m=1[/latex] to find the angle of the first minimum.

Solution for Part (a)

The second minimum corresponds to

[latex]m=2.[/latex]

The known quantities are

[latex]\lambda=550\ \text{nm}=5.50\times10^{-7}\ \text{m}[/latex]

and

[latex]\theta_2=45.0^\circ.[/latex]

Solving the minimum condition for [latex]D[/latex] gives

[latex]D=\frac{m\lambda}{\sin\theta_2}.[/latex]

Substituting the known values,

[latex]D=\frac{2(5.50\times10^{-7}\ \text{m})}{\sin(45.0^\circ)}.[/latex]

[latex]D=1.56\times10^{-6}\ \text{m}.[/latex]

Therefore, the slit width is

[latex]\boxed{D=1.56\ \mu\text{m}}.[/latex]

Solution for Part (b)

For the first minimum, [latex]m=1[/latex]. The minimum condition becomes

[latex]D\sin\theta_1=\lambda.[/latex]

Solving for [latex]\sin\theta_1[/latex],

[latex]\sin\theta_1=\frac{\lambda}{D}.[/latex]

Substituting the wavelength and slit width,

[latex]\sin\theta_1=\frac{5.50\times10^{-7}\ \text{m}}{1.56\times10^{-6}\ \text{m}}=0.354.[/latex]

Therefore,

[latex]\theta_1=\sin^{-1}(0.354)=20.7^\circ.[/latex]

The first minimum occurs at

[latex]\boxed{\theta_1=20.7^\circ}.[/latex]

Discussion

The slit width is only about three times the wavelength of the light:

[latex]\frac{D}{\lambda}=\frac{1.56\ \mu\text{m}}{0.550\ \mu\text{m}}\approx2.84.[/latex]

Because the slit width is comparable to the wavelength, the light diffracts through relatively large angles. The central maximum extends from the first minimum on one side to the first minimum on the other side. Its total angular width is therefore approximately

[latex]2\theta_1=2(20.7^\circ)=41.4^\circ.[/latex]

The next bright region lies between the first and second minima. Its angular width on one side is approximately

[latex]\theta_2-\theta_1=45.0^\circ-20.7^\circ=24.3^\circ.[/latex]

This confirms that the central maximum is considerably wider than the secondary maxima. It is also much brighter because all parts of the slit interfere constructively at [latex]\theta=0[/latex].

Section Summary

• A single slit produces a diffraction pattern consisting of one broad, bright central maximum surrounded by much narrower and dimmer secondary maxima.
• The central maximum is approximately twice as wide as the neighboring bright fringes because it extends from the first minimum on one side of the pattern to the first minimum on the other.
• Dark fringes (diffraction minima) occur when light from different parts of the slit interferes destructively.
• The condition for destructive interference is

• Here, [latex]D[/latex] is the slit width, [latex]\lambda[/latex] is the wavelength of the light, [latex]\theta[/latex] is the angle measured from the original direction of the beam, and [latex]m[/latex] is the order of the diffraction minimum.
• There is no minimum corresponding to [latex]m=0[/latex]. At [latex]\theta=0[/latex], all portions of the wavefront arrive in phase, producing the bright central maximum.
• Narrower slits produce wider diffraction patterns, while wider slits produce narrower diffraction patterns.

Problems & Exercises

43. (a) At what angle does the first diffraction minimum occur for 550-nm light passing through a single slit that is [latex]1.00~\mu\text{m}[/latex] wide? (b) Is a second minimum possible for this slit? Explain.
44. (a) Calculate the angle of the first diffraction minimum for 410-nm violet light passing through a slit that is [latex]2.00~\mu\text{m}[/latex] wide. (b) At what angle does the first minimum occur for 700-nm red light passing through the same slit?
45. (a) How wide must a single slit be to produce its first diffraction minimum for 633-nm light at an angle of [latex]28.0^\circ[/latex]? (b) At what angle would the second minimum occur?
46. (a) Determine the width of a single slit that produces its first diffraction minimum at [latex]60.0^\circ[/latex] for 600-nm light. (b) Find the wavelength of light that would produce its first minimum at [latex]62.0^\circ[/latex] for the same slit.
47. Find the wavelength of light whose third diffraction minimum occurs at an angle of [latex]48.6^\circ[/latex] when it passes through a single slit that is [latex]3.00~\mu\text{m}[/latex] wide.
48. Calculate the wavelength of light that produces its first diffraction minimum at [latex]36.9^\circ[/latex] when passing through a single slit of width [latex]1.00~\mu\text{m}[/latex].
49. Sodium vapor light with an average wavelength of 589 nm passes through a single slit that is [latex]7.50~\mu\text{m}[/latex] wide.
    1. At what angle does the second diffraction minimum occur?
    2. What is the highest-order diffraction minimum that can exist for this slit?
50. 1. Find the angle of the third diffraction minimum for 633-nm light passing through a slit that is [latex]20.0~\mu\text{m}[/latex] wide.
    2. What slit width would place this third minimum at an angle of [latex]85.0^\circ[/latex]?

    Show your solution using the standard problem-solving strategy for wave optics.

51. Two sodium emission lines have wavelengths of 589.1 nm and 589.6 nm.
    1. Find the angular separation between their first diffraction minima when the light passes through a single slit that is [latex]2.00~\mu\text{m}[/latex] wide.
    2. If the diffraction pattern is projected onto a screen 1.00 m away, what is the distance between these minima?
    3. Comment on whether this separation would be easy to measure experimentally.
52. 1. What is the minimum slit width, expressed as a multiple of [latex]\lambda[/latex], that can produce a first diffraction minimum?
    2. What minimum slit width is required to produce 50 diffraction minima?
    3. What minimum slit width is required to produce 1000 diffraction minima?
53. A single slit produces its first diffraction minimum at an angle of [latex]14.5^\circ[/latex].
    1. At what angle does the second minimum occur?
    2. At what angle does the third minimum occur?
    3. Is a fourth minimum possible? Explain.
    4. Use your answers to show that the central maximum is approximately twice as wide as the next bright fringe.
54. A double-slit experiment also exhibits a single-slit diffraction envelope. Determine the ratio of the slit width to the slit separation if the first minimum of the single-slit diffraction pattern coincides with the fifth interference maximum of the double-slit pattern. Explain why this greatly reduces the intensity of that interference maximum.
55. Integrated Concepts. A breakwater protecting a harbor consists of a rock barrier with a 50.0-m-wide opening. Ocean waves having a wavelength of 20.0 m approach the opening perpendicularly. At what angle relative to the incoming waves are boats inside the harbor best protected from wave action?
56. Integrated Concepts. An aircraft maintenance technician walks past a tall hangar doorway that acts as a single slit for sound waves entering the building. A jet engine outside the hangar produces a 600-Hz sound, and the doorway is 0.800 m wide. Assuming the speed of sound is 340 m/s, at what angle relative to the doorway will the technician encounter the first minimum in sound intensity?