Question

A light ray passes through from a glass, with an index of refraction of 1.58, to water, with an index of refraction of 1.33. The angle of the refracted ray in the water is 58°. Calculate the angle of incident ray at the glass-water interface.

Answer 1

Answer:

ANGLE OF INCIDENT RAY→ 45.5°

Explanation:

Using snells law

n1 sin(θ1) = n2 sin(θ2)

Here,

n1= 1.58

n2= 1.33

θ2 = 58°

1.58× sin(θ1) = 1.33× sin(58°)

Sin(θ1) = {1.33× sin(58°)} / 1.58

Sin(θ1) = (0.71)

θ1 = 45.5°