Two beams of light start together and then hit a slab of two different kinds of
ID: 1403688 • Letter: T
Question
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 660 nm outside the slabs, and the slab is d = 1.60 microns thick. If the top half of the slab has index of refraction 1.78 and the bottom has index 1.46, by what time interval will one of the beams be ahead of the other once they've gone through the slab?
Explanation / Answer
We know from snell law that
nV = n1V1
where n and V is the refractive index of air and velocity in air
and suffix 1 represent the media 1
Now
nV = 1.78V1
Similarly
nV = 1.46V2
so 1.78V1 = 1.46V2
V2 = 1.219 V1
therefore the ray emerges sooner from second media
Now we know that velocity of light in air is c = 3*108 m/s
so
V1 = 1.685*108 m/s
V2 = 2.054*108 m/s
Time taken through media 1 = (1.6*10-6)/(1.685*108) = 0.9495*10-14 s
Time taken throgh media 2 = (1.6*10-6)/(2.054*108) = 0.77896*10-14 s
So time interval = 0.17054*10-14 s