Learning Objectives

Learning Objectives

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

  • Discuss the single slit diffraction pattern

The information presented in this section supports the following AP® learning objectives and science practices:

  • 6.C.2.1 The student is able to make claims about the diffraction pattern produced when a wave passes through a small opening and to qualitatively apply the wave model to quantities that describe the generation of a diffraction pattern when a wave passes through an opening whose dimensions are comparable to the wavelength of the wave.

Light passing through a single slit forms a diffraction pattern somewhat different from those formed by double slits or diffraction gratings. Figure 10.21 shows a single slit diffraction pattern. Note that the central maximum is larger than those on either side, and that the intensity decreases rapidly on either side. In contrast, a diffraction grating produces evenly spaced lines that dim slowly on either side of center.

Part a of the figure shows a slit in a vertical bar. To the right of the bar is a graph of intensity versus height. The graph is turned ninety degrees counterclockwise so that the intensity scale increases to the left and the height increases as you go up the page. Just in front of the gap, a strong central peak extends leftward from the graph’s baseline, and many smaller satellite peaks appear above and below this central peak. Part b of the figure shows a drawing of the two-dimensional intensity patte
Figure 10.21 (a) Single slit diffraction pattern. Monochromatic light passing through a single slit has a central maximum and many smaller and dimmer maxima on either side. The central maximum is six times higher than shown. (b) The drawing shows the bright central maximum and dimmer and thinner maxima on either side.

The analysis of single slit diffraction is illustrated in Figure 10.22. Here, we consider light coming from different parts of the same slit. According to Huygens’s principle, every part of the wavefront in the slit emits wavelets. These are like rays that start out in phase and head in all directions. Each ray is perpendicular to the wavefront of a wavelet. Assuming the screen is very far away compared with the size of the slit, rays heading toward a common destination are nearly parallel. When they travel straight ahead, as in Figure 10.22(a), they remain in phase, and a central maximum is obtained. However, when rays travel at an angle θθ size 12{θ} {} relative to the original direction of the beam, each travels a different distance to a common location, and they can arrive in or out of phase. In Figure 10.22(b), the ray from the bottom travels a distance of one wavelength λλ size 12{λ} {} farther than the ray from the top. Thus, a ray from the center travels a distance λ/2λ/2 size 12{λ/2} {} farther than the one on the left, arrives out of phase, and interferes destructively. A ray from slightly above the center and one from slightly above the bottom will also cancel one another. In fact, each ray from the slit will have another to interfere destructively, and a minimum in intensity will occur at this angle. There will be another minimum at the same angle to the right of the incident direction of the light.

The figure shows four schematics of a ray bundle passing through a single slit. The slit is represented as a gap in a vertical line. In the first schematic, the ray bundle passes horizontally through the slit. This schematic is labeled theta equals zero and bright. The second schematic is labeled dark and shows the ray bundle passing through the slit an angle of roughly fifteen degrees above the horizontal. The path length difference between the top and bottom ray is lambda, and the schematic is labeled s
Figure 10.22 Light passing through a single slit is diffracted in all directions and may interfere constructively or destructively, depending on the angle. The difference in path length for rays from either side of the slit is seen to be Dsin θ.Dsin θ. size 12{D`"sin"`θ} {}

At the larger angle shown in Figure 10.22(c), the path lengths differ by /2/2 size 12{3λ/2} {} for rays from the top and bottom of the slit. One ray travels a distance λλ size 12{λ} {} different from the ray from the bottom and arrives in phase, interfering constructively. Two rays, each from slightly above those two, will also add constructively. Most rays from the slit will have another to interfere with constructively, and a maximum in intensity will occur at this angle. However, all rays do not interfere constructively for this situation, and so the maximum is not as intense as the central maximum. Finally, in Figure 10.22(d), the angle shown is large enough to produce a second minimum. As seen in the figure, the difference in path length for rays from either side of the slit is Dsinθ,Dsinθ, size 12{D`"sin"θ} {} and we see that a destructive minimum is obtained when this distance is an integral multiple of the wavelength.

The graph shows the variation of intensity as a function of sine theta. The curve has a strong peak at sine theta equals zero, then has small oscillations spreading symmetrically to the left and right of this central peak. The oscillations all appear to be of the same height. Between each oscillation, the curve appears to go to zero, and each zero is labeled. The first zero to the left of the main peak is labeled minus lambda over d and the first zero to the right is labeled lambda over d. The second zero
Figure 10.23 A graph of single slit diffraction intensity showing the central maximum to be wider and much more intense than those to the sides. In fact, the central maximum is six times higher than shown here.

Thus, to obtain destructive interference for a single slit,

10.23 Dsinθ=,form=1,–1,2,–2,3, (destructive),Dsinθ=,form=1,–1,2,–2,3, (destructive), size 12{D`"sin"θ= ital "mλ",~m="1,"`"2,"`"3,"` dotslow } {}

where DD size 12{D} {} is the slit width, λλ size 12{λ} {} is the light’s wavelength, θθ size 12{θ} {} is the angle relative to the original direction of the light, and mm size 12{m} {} is the order of the minimum. Figure 10.23 shows a graph of intensity for single slit interference, and it is apparent that the maxima on either side of the central maximum are much less intense and not as wide. This is consistent with the illustration in Figure 10.21(b).

Example 10.4 Calculating Single Slit Diffraction

Visible light of wavelength 550 nm falls on a single slit and produces its second diffraction minimum at an angle of 45.0º45.0º size 12{"45" "." 0°} {} relative to the incident direction of the light. (a) What is the width of the slit? (b) At what angle is the first minimum produced?

The schematic shows a single slit to the left and the resulting intensity pattern on a screen is graphed on the right. The single slit is represented by a gap of size d in a vertical line. A ray of wavelength lambda enters the gap from the left, then five rays leave from the gap center and head to the right. One ray continues on the horizontal centerline of the schematic. Two rays angle upward: the first at an unknown angle theta one above the horizontal and the second at an angle theta two equals forty f
Figure 10.24 A graph of the single slit diffraction pattern is analyzed in this example.

Strategy

From the given information, and assuming the screen is far away from the slit, we can use the equation Dsinθ=Dsinθ= size 12{D`"sin"θ= ital "mλ"} {} first to find D,D, size 12{D} {} and again to find the angle for the first minimum θ1.θ1. size 12{θ rSub { size 8{1} } } {}

Solution for (a)

We are given that λ=550 nm,λ=550 nm, size 12{λ="500"`"nm"} {} m=2,m=2, size 12{m=2} {} and θ2=45.0º.θ2=45.0º. Solving the equation Dsinθ=Dsinθ= size 12{D`"sin"θ= ital "mλ"} {} for DD size 12{D} {} and substituting known values gives

10.24 D = sin θ 2 = 2 ( 550 nm ) sin 45.0º = 1100 × 10 9 0.707 = 1.56 × 10 6 . D = sin θ 2 = 2 ( 550 nm ) sin 45.0º = 1100 × 10 9 0.707 = 1.56 × 10 6 .

Solution for (b)

Solving the equation Dsinθ=Dsinθ= size 12{D`"sin"θ= ital "mλ"} {} for sinθ1sinθ1 size 12{"sin"θ rSub { size 8{1} } } {} and substituting the known values gives

10.25 sinθ1=D=1550×109m1.56×106m.sinθ1=D=1550×109m1.56×106m. size 12{"sin"θ rSub { size 8{1} } = { {mλ} over {D} } = { {1 left ("550" times "10" rSup { size 8{ - 9} } m right )} over {1 "." "56" times "10" rSup { size 8{ - 6} } m} } } {}

Thus the angle θ1θ1 size 12{θ rSub { size 8{1} } } {} is

10.26 θ1=sin10.354=20.7º.θ1=sin10.354=20.7º. size 12{θ rSub { size 8{1} } ="sin" rSup { size 8{ - 1} } 0 "." "354"="20" "." 7°} {}

Discussion

We see that the slit is narrow—it is only a few times greater than the wavelength of light. This is consistent with the fact that light must interact with an object comparable in size to its wavelength to exhibit significant wave effects such as this single slit diffraction pattern. We also see that the central maximum extends 20.7º20.7º on either side of the original beam, for a width of about 41º.41º. The angle between the first and second minima is only about 24º(45.0º−20.7º).24º(45.0º−20.7º). size 12{"24"°` \( "45" "." 0° - "20" "." 7° \) } {} Thus the second maximum is only about half as wide as the central maximum.