Question

A light ray travels inside a block of sodium fluoride that has index of refraction n = 1.33 .The ray strikes the vertical wall at the critical angle, totally reflects, and then emerges into the air above the block. What is the angle ?2 at which the ray emerges?

Answer 1

Concepts and reason

The concept used to solve the problem is the Snell’s law using the variables of angle of incidence and angle of refraction

Initially, express the relation using Snell’s law using the variables of index of refraction of first medium, index of refraction of second medium, angle of incidence and refraction. Then, rearrange the expression to calculate the critical angle. Then, express the relation using Snell’s law in terms of angle of incidence and angle of refraction. Finally, rearrange the expression and substitute the values of angle of incidence to calculate the angle of emerging ray

Fundamentals

The expression for refractive index using Snell’s law,

Here, is the critical angle, is the angle of refraction, is the refractive index of first medium and is the refractive index of second medium.

The expression for refractive index using Snell’s law,

Rearrange the expression in terms of

Substitute for , for and for .

The expression for refractive index,

Rearrange the expression in terms of

Substitute for , for and for

Ans:

The angle of emerging ray is .