Chapter 46: Problem 32

Question: Suppose that, as in sample problem 7 (Capter 46) the envelope of the central peak contains 11 fringes. How many fringes lie between the first and second minima of the envelope?

Solution: The general expressin for the intensity pattern for two slits of width a and separation d is

$$I(\theta)~=~ I(0)\ \cos^2\beta\ \left({\sin\alpha\over\alpha}\right)^2 \ \ \ ,$$

where

$$\beta~=~{\pi d\sin\theta\over\lambda} \ \ \ ,$$

and

$$\alpha~=~{\pi a\sin\theta\over\lambda} \ \ \ .$$

If there are 11 fringes, then we know that the interference fringes (corresponding to $d\sin\theta~=~m\lambda$) for $m~=~0,\pm 1,\pm 2,\pm 3,\pm 4, \pm5$ all lie in the diffraction central maximum. Therefore, we know that the intereference fringes corresponding to m = +6, +7, +8, +9, +10, (or alternately m = -6, -7, -8, -9, -10) will lie between the first and second diffraction minima. Thus we have 5 fringes.