Diffraction by a Single Slit: Locating the Minima

Diffraction by a Single Slit: Locating the Minima

We can repeal this analysis for any other pair of rays originating at corresponding points in the two zones (say, at the midpoints of the zones) and extending to point PI_ Each such pair of rays has the same path length difference (a/2) sin O.  Setting this common path length difference equal to Al2 (our condition for the first dark fringe), we have

a A
– sin 0 =-
2 2′

 

Fig.37-5(0) Waves from the top points of four zones of width 0/4 undergo totally destructive interference at point Pl. (h) For D :l> a. we can approximate rays ,." r2' "3' and "4 as being parallel, at angle 0 to the central axis
Fig.37-5(0) Waves from the top points
of four zones of width 0/4 undergo totally
destructive interference at point
Pl. (h) For D :l> a. we can approximate
rays ,.” r2′ “3′ and “4 as being
parallel, at angle 0 to the central axis

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Given slit width a and wavelength A, Eq. 37-1 tells us the angle 0 of the first dark fringe above and (by symmetry) below the central axis. Note that if we begin with a > A and then narrow the slit while holding the wavelength constant, we increase the angle at which the first dark fringes appear: that is, the extent of the diffraction (the extent of the flaring and the width of the pattern) is greater for a narrower slit. When we have reduced the slit width to the
wavelength (that is, a = A), the an gle of the first dark fringes is 90°. Since the firs dark fringes mark the two edges of the central bright fringe, that bright fringe must then cover the entire viewings been. We find the second dark fringes above and below the central axis as we found the first dark fringes, except that we now divide the slit into four zones of equal widths a/4, as shown in Fig. 37-5a. We then extend rays rl, r2′ r3′ and r4 from the top points of the zones to point P2′ the location of the second dark fringe above the central axis. To produce that fringe, the path length  difference between rl and r2,  that between r2 and r3′ and that between r3 and r4 must all be equal to Al2.For D ~ a, we can approximate these four rays as being parallel, at angle 0 to the central axis. To display their path length differences, we extend a perpendicular  line through each adjacent pair of rays, as shown in b, to form a  of right triangles, each of which has a path length difference as one side. We see from te top tr angle that the path length difference between rl and /”2 is (a/4) sin O. Similarly, from the bottom triangle, the path length difference between r3 and /”4 is  also (a/4) sin O. In fact, the path length difference for any two rays that originate at corresponding points in two adjacent zones is (a/4) sin O. Since in each such case the path length difference is equal to Al2, we have

 

– sin 0 =-
4 2′

We could now continue to locate dark fringes in the diffraction pattern by splitting up the slit into more zones of equal width. We would always choose an even number of zones so that the zones (and their waves) could be paired as we have been doing. We would find that the dark fringes above and below the central axis can be located with the following general equation:

a sin 0 = mA, for m = I, 2, 3, …

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