A slab of glass with an index of refraction of 1.53 is submerged in a liquid wit

Question

A slab of glass with an index of refraction of 1.53 is submerged in a liquid with an index of refraction of 1.33. Light in the glass is incident on the glass-liquid interface. Find the angle of refraction for the following angles of incidence.

A slab of glass with an index of refraction of 1.53 is submerged in a liquid with an index of refraction of 1.33. Light in the glass is incident on the glass-liquid interface. Find the angle of refraction for the following angles of incidence. (a) 600 0 (b) 45 0 (c) 30° 0

Explanation / Answer

using snell's law for the glass -water interface

a) for 60 degrees

n1*sin(i) = n2*sin(r)

sin(r) = (n1/n2)*sin(i)

sin(r) = (1.53/1.33)*sin(60) = 0.996

r = sin^(-1)(0.996) = 85 degrees

b) for i = 45 degrees

sin(r) = (n1/n2)*sin(i)

sin(r) = (1.53/1.33)*sin(45) =0.813

r= sin^(-1)(0.813) = 54.4 degrees

c) for i = 30 degrees

sin(r)= (n1/n2)*sin(i) = (1.53/1.33)*sin(30) = 0.575

r= sin^(-1)(0.575) = 35 degrees

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