Chapter 35, Problem 087 In Figure (a), the waves along rays 1 and 2 are initiall
ID: 2078412 • Letter: C
Question
Chapter 35, Problem 087 In Figure (a), the waves along rays 1 and 2 are initially in phase, with the same wavelength in air. Ray 2 goes through a material with length L and index of refraction n. The rays are then reflected by mirrors to a common point P on a screen. Suppose that we can vary L from 0 to 214.8 nm. Suppose also that, from L = 0 to L = 189.6 nm, the intensity I of the light at point P varies with L as given in Figure (b) below, where the horizontal scale is set by Ls 189.6.0 nm. At what values of L greater than 189.6 nm is intensity 1 a maximum and b zero? c What multiple of gives the phase difference between the ra sat pointP when 10,4 nm Screen Ray 2 Ray l 0 Ls L(nm) (a) Number (b) Number (c) Number Click if you would like to Show Work for this question: Open Show Work Units Units Units UnitsExplanation / Answer
When the interference between two waves is completely destructive, their phase difference is given by
= (2m+1) m=0,1,2,
The equivalent condition is that their path-length difference is an odd multiple of / 2, where is the wavelength of the light.
(a) Looking at the figure (where a portion of a periodic pattern is shown) we see that half of the periodic pattern is of length L = 160 nm (judging from the maximum at x = 0 to the minimum at x = 160nm);
this suggests that the wavelength (the full length of the periodic pattern) is = 2 L = 320 nm.
A maximum should be reached again at x = 320nm (and at x = 640 nm, x = 960nm, …).
(b) From our discussion in part (b), we expect a minimum to be reached at each value x = 160m + n(320nm), where n = 1, 2, 3, … .
For instance, for n = 1 we would find the minimum at x = 480 nm.
(c) With = 320 nm (found in part (a)), we can express x = 107.4 nm as x = 107.4/320 = 0.335 wavelength.
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