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A 1-cm-thick layer of water stands on a horizontal slab of glass. A light ray in...

A 1-cm-thick layer of water stands on a horizontal slab of glass. A light ray in the air is incident on the water 62 degrees from the normal.

After entering the glass, what is the ray's angle from the normal?
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Answer #1
u(glass)/u(air) = sini/sinr

1.5 = sin(62)/sin r

sin r = 0.8829/1.5

sin r = 0.5886
r = 36.05 answer

introducing water in between doesn't make any difference as you can apply eq on both surfaces , terms related to water cancel out
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