Please help A material with index of refraction n = 1.5 is in the shape of a qua

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A material with index of refraction n = 1.5 is in the shape of a quarter circle of radius R = 15 cm and is surrounded by a vacuum. A light ray parallel to the base of the material is incident from the left at a distance L = 10 cm above the base and emerges from the material at the angle theta. Determine an expression for theta in terms of n, R and L and then find the value of theta for given values of n, R and L. (Answer 23.5 degree) A glass beaker (n = 1.5) is filled with water (n = 1.33). A ray is incident at 75 degree to the normal to the water surface emerges from the side of the beaker, as shown in the figure. What angle theta does this ray make with the horizontal? (n = 1 for air). (Answer: 66.1 degree)

Explanation / Answer

using snells law

sin(i) /sin(r) = refractive index of media 2 / refractive index of media 1

for water surface

sin(75)/sin(r) = 1.33/1

r = 46.57 degree

now again for glass

sin(43.43)/sin(r) = 1.5/1.33

r for glass = 37.55

now again for air

sin(37.55)/sin(theta) = 1/1.5

from this we will get

theta = 66.08 degree

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