Two Slabs Two beams of light start together and then hit a slab of two different

Question

Two Slabs

Two beams of light start together and then hit a slab of two different kinds of material. This will cause one of the beams to get "ahead" of the other; that is, one will emerge from the slab sooner than the other. The beams have a wavelength of 550 nm outside the slabs, and the slab is d = 2.10 microns thick. If the top half of the slab has index of refraction 1.62 and the bottom has index 1.45, by what time interval will one of the beams be ahead of the other once they've gone through the slab?

Explanation / Answer

The wavelength of the light will vary according to the medium in which it travels, and we are not told what the medium is in which its wavelength is 500nm. I will answer the question as though the given wavelength refers to a vacuum, and that the required phase difference is that which exists after the 2 beams have re-emerged into vacuum again, or a the point at which they are about to do so. I will also assume that the 2 beams are in-phase when they enter the two media mentioned. Let the media be m1 & m2.

Speed of light in m1 is c/n1, speed in m2 is c/n2.

If d is the thickness, the difference in time required to traverse the 2 slabs is t = (d/c)*(n1 - n2) = 119*10*-17s

The period (T) of the waves is L/c where L is the vacuum wavelength and c is the speed of light

T = 1.66*10^-15s

The phase shift in terms of number of periods is (1.66*10^-15)/119*10^-17 = 1.394

The phase difference is therefore 1.394*2*pi = 8.75rad . Phase differences are usually quoted as being grater than pi.

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