In figure a, the waves along rays 1 and 2 are initially in phase, with the same

ID: 1895889 • Letter: I

Question

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 n from n = 1.0 to n = 2.6. Suppose also that, from n = 1.0 to n = ns = 1.75, the intensity I of the light at point P varies with n as given in figure b. At what values of n greater than 1.60 is intensity I (a) maximum and (b) zero? (c) What multiple of ? gives the phase difference between the rays at point P when n = 2.50?

fig. A

fig. B

(a) Number_________Unit________

(b) Number_________Unit________

Explanation / Answer

So from the plot we can see that there is a minimum at n=1.6
so there the path length difference is half a wavelength

so path length difference= (# wavelengths ray 2 - # wavelength ray 1)
= (L/(/n) - L/)=nL-L=(n-1)L

so we have (n-1)L=PLD

now we know PLD=/2 when n=1.6

so .6 L = /2

so /L=1.2

a) so now we want the max after 1.6 so we wnat PLD = 1

so (n-1)L= so n-1=/L=1.2

n=2.2

b) next min is when PLD=3/2

(n-1)L = 3/2

so n-1=3/2 * 1.2 = 1.8

so n=2.8

c) at n=2.5

PLD=(2.5-1)L= 1.5 L but L= /1.2

so PLD = 1.5/1.2 = 1.25

so the factor is 1.25

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