The 45-45-90 triangular prism shown in the figure has an index of refraction of

Question

The 45-45-90 triangular prism shown in the figure has an index of refraction of 1.43, and is surrounded by air. A ray of light is incident on the left face at an angle of ? =54.0°. The point of incidence is high enough that the refracted ray hits the opposite sloping side.

1. Calculate the angle at which the ray exits the prism, relative to the normal of that surface. Express your answer as a negative number if the angle is clockwise relative to the normal, and a positive number for a counterclockwise angle.

2. What is the maximum index of refraction for which the ray exits the prism from the right-hand side? (It may help to remember that sin(90° - ?) = cos ?.)

Explanation / Answer

for the refraction at the first face using snell's law -

n1*sin i = n2*sin r

1*sin 54 = 1.43*sin r

sin r = 0.565

r = 34.40 degrees

angle between the normal and incident rays for the refraction at surface 2 can be find as -

r' = 90 - 34.40

r' = 55.59 degrees

AGAIN USING SNELL'S LAW -

1.43*sin 55.59 = 1*sin t

sin t = 1.18

which is not possible so light will not emerge.

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