Gregg Wolfe; Erika Gasper; John Stoke; Julie Kretchman; David Anderson; Nathan Czuba; Sudhi Oberoi; Liza Pujji; Irina Lyublinskaya; Douglas Ingram

Learning Objectives

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

Discuss the single slit diffraction pattern.

Light passing through a single slit forms a diffraction pattern somewhat different from those formed by double slits or diffraction gratings. Figure 27.21 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 pattern that is observed from single slit diffraction. The central stripe is quite broad compared to the satellite stripes, and there are dark areas between all the stripes.

Figure 27.21 (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 27.22. 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 27.22(a), they remain in phase, and a central maximum is obtained. However, when rays travel at an angle $θ$ 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 27.22(b), the ray from the bottom travels a distance of one wavelength $λ$ farther than the ray from the top. Thus a ray from the center travels a distance $λ / 2$ 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 sine theta equals lambda over d. The third schematic is labeled bright and shows the ray bundle passing through the slit at an angle of about twenty five degrees above the horizontal. The path length difference between the top and bottom rays is three lambda over two d, and the schematic is labeled sine theta equals three lambda over two d. The final schematic is labeled dark and shows the ray bundle passing through the slit at an angle of about forty degrees above the horizontal. The path length difference between the top and bottom rays is two lambda over d, and the schematic is labeled sine theta equals two lambda over d.

Figure 27.22 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

D sin θ

.

At the larger angle shown in Figure 27.22(c), the path lengths differ by $3λ / 2$ for rays from the top and bottom of the slit. One ray travels a distance $λ$ 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 27.22(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 $D sin θ$ , 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 to the left is labeled minus two lambda over d and the second zero to the right is labeled two lambda over d. The third zero to the left is labeled minus three lambda over d and the third zero to the right is labeled three lambda over d.

Figure 27.23 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,

D sin θ = mλ, for m = 1, –1, 2, –2, 3, \dots (destructive),

27.21

where $D$ is the slit width, $λ$ is the light’s wavelength, $θ$ is the angle relative to the original direction of the light, and $m$ is the order of the minimum. Figure 27.23 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 27.21(b).

Example 27.4

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 $45.0º$ 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 five degrees above the horizontal. The final two rays angle downward at the same angles, so that they are symmetric about the horizontal with respect to the two rays that angle upward. The intensity on the screen is a maximum where the central ray hits the screen, whereas it is a minimum where the angled rays hit the screen.

Figure 27.24 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 $D sin θ = mλ$ first to find $D$ , and again to find the angle for the first minimum $θ_{1}$ .

Solution for (a)

We are given that $λ = 550 nm$ , $m = 2$ , and $θ_{2} = 45.0º$ . Solving the equation $D sin θ = mλ$ for $D$ and substituting known values gives

\begin{array}{l} D & = & \frac{mλ}{sin θ_{2}} = \frac{2 (550 nm)}{sin 45.0º} \\ = & \frac{1100 \times 10^{- 9}}{0.707} \\ = & 1.56 \times 10^{- 6} . \end{array}

27.22

Solution for (b)

Solving the equation $D sin θ = mλ$ for $sin θ_{1}$ and substituting the known values gives

sin θ_{1} = \frac{mλ}{D} = \frac{1 (550 \times 10^{- 9} m)}{1 . 56 \times 10^{- 6} m} .

27.23

Thus the angle $θ_{1}$ is

θ_{1} = {sin}^{- 1} 0.354 = 20.7º.

27.24

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 in order to exhibit significant wave effects such as this single slit diffraction pattern. We also see that the central maximum extends $20.7º$ on either side of the original beam, for a width of about $41º$ . The angle between the first and second minima is only about $24º (45.0º - 20.7º)$ . Thus the second maximum is only about half as wide as the central maximum.

27.5 Single Slit Diffraction