Question

A light ray in glass (n=1.5) hits the air-glass interface at an angle of 10 degrees from the normal. What angle from the normal is the light ray in the air (n=1.0)? (You can use the small angle approximation.)

Answer 1

Answer:

The angle from the normal is 15.1°.

Explanation:

We can find the angle by using Snell's law:

$n_{1}sin(\theta_{1}) = n_{2}sin(\theta_{2})$

Where:

n₁: is the first medium (glass) = 1.5

n₂: is the second medium (air) = 1.0

θ₁: is the first angle (in the glass) = 10°

θ₂: is the second angle (in the air) =?

$\theta_{2} = arcsin(\frac{n_{1}sin(\theta_{1})}{n_{2}}) = arcsin(\frac{1.5*sin(10)}{1.0}) = 15.1 ^{\circ}$

Therefore, the angle from the normal is 15.1°.

I hope it helps you!