Diffraction by a Single Slit: Locating the Minima
Let us now examine the diffraction pattern of plane waves of light of wavelength A that are diffracted by a single. long. narrow slit of width a in an otherwise opaque screen B, as shown in cross section in . (In that figure, the slit’s length extends into and out of the page, and the incoming wavefronts are parallel to screenB.) When the diffracted light reaches viewing screen C, waves from different points within the slit undergo interference and produce a diffraction pattern of bright and dark fringes (interference maxima and minima) on the screen. To locate the fringes. we shall use a procedure somewhat similar to the one we used to locate the fringes a two-slit interference pattern. However, diffraction is more mathematically challenging, and here we shall be able to find equations for only the dark fringes. Before we do that. however, we can justify the central bright fringe seen in Fig.
37-1 by noting that the Huygens wa velvet from all points in the slit travel about the same distance to reach the center of the pattern and thus are in phase there. As for the other bright fringes. we can say only that they are approximately halfway between adjacent dark fringes. To find the dark fringes. we shall use a clever (and simplifying) strategy thai involves pairing up all the rays coming through the slit and then finding what condition causes the wavelets of the rays in each pair to cancel each other. We apply this strategy in-4a to locate the first dark fringe, at point PI’ First, we mentally divide the slit into two zones of equal widths a/2. Then we extend to PI a light ray rl from the top point of the top zone and a light ray r2 from the top point of the bottom zone. A central axis is drawn from the center of the slit to screen C, and P I is located at an angle f) to that axis.The wavelets of the pair of rays rl and r2 are in phase within the slit because they originate from the same wavefront passing through the slit, along the width of the slit. However, to produce the first dark fringe they must be out of phase by A/2 when they reach PI: this phase difference is due to their path length difference. with the wavelet of r2 traveling a longer path to reach PI than the wavelet of rl. To display this path length difference, we find a point b on ray r2 such that the path length from b to PI matches the pah length of ray • Then the path length difference between the two rays is the i stance from the center of the slit to b. hen viewing screen C is near screen B, as in , the diffraction pattern on C is difficult to describe mathematically. However, we can simplify the mathematics considerably if we arrange for the screen separation D to be much larger than the slit width a. Then we can approximate rays rl and r2 as being parallel, at angle8 to the central axis ). We can also approximate the triangle formed by point b, the top point of the slit, and the center point of the slit as being a right triangle, and one of the angles inside that triangle as being 8. The path length difference
between rays rl and r2 (which is still the distance from the center of the slitto point b) is then equal to (al2) sin 8.