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