## Introduction

To understand how to calculate critical angle, we will apply the formula below:

C = sin^{-1} (1 / n)

or C = sin^{-1} (n_{a} / n_{g})

Where

C = critical angle

n = refractive index of the medium

Also

n_{a} = refractive index of air

and

n_{g} = refractive index of glass

*You may also like to read:*

What is Critical Angle and its Formula?

And what is the critical angle for light traveling from crown glass (n = 1.52) into water (n = 1.33)?

## Problem 1

If the refractive indices of glass and water are 1.5 and 1.3 respectively. What will be the critical angle when the angle of refraction in the water medium is 90 degrees?

**Data **

Refractive index of glass, n_{g} = 1.5

The refractive index of water, n_{w} = 1.3

Angle of refraction in the water medium, θ_{w} = 90^{0}

**Unknown**

Critical angle, C = ?

**Formula**

We will apply the formula

C = sin^{-1} (n_{w} sinθ_{w} / n_{g})

Solution

We can now apply our data into the formula to obtain

C = sin^{-1} (n_{w} sinθ_{w} / n_{g}) = sin^{-1} (1.3 x sin90^{0} / 1.5)

The above expression will become

C = sin^{-1} (1.3 x 1 / 1.5)

We will now have

C = = sin^{-1} 0.9 = 64^{0}

**Therefore, the critical angle is 64 degrees.**

## Problem 2

What is the critical angle for a light ray traveling in water with a refractive index of 1.33 that is incident on the surface of the water above which air with a refractive index of 1.00? Answer to the nearest degree.

**Data**

Refractive index of water, n_{w} = 1.33

The refractive index of air, n_{a} = 1.00

**Unknown**

Critical angle, C = ?

**Formula**

We will apply the formula below to solve the problem

C = sin^{-1} (n_{a} / n_{w})

### Solution

We will insert our data into the formula to get

C = sin^{-1} (n_{a} / n_{w}) = sin^{-1} (1 / 1.33)

After dividing 1 by 1.33 we will obtain

C = sin^{-1} 0.75 = 48.6^{0} = 49^{0}

**Therefore, the critical angle for the light ray is 49 degrees.**

## Problem 3

Light rays travel through a layer of kerosene floating on the surface of water that has a refractive index of 1.33. Light rays that are incident on the interface of kerosene and water at angles of 16.9° from the surface or less are totally internally reflected. What is the refractive index of the kerosene? Give your answer to two decimal places.

**Data**

The refractive index of water, n_{w} = 1.33

Critical angle, C = 90^{0} – 16.9^{0} = 73.1^{0}

**Unknown**

The refractive index of the kerosene, n_{k} = ?

**Formula**

We will use the formula

n_{k} = n_{w} / sinC

### Solution

Insert your data into the above formula to obtain

n_{k} = n_{w} / sinC = 1.33 / sin73.1^{0}

We will now have

n_{k} = 1.33 / 0.96 = 1.385 = 1.39

**Therefore, the refractive index of the kerosene, n _{k} is 1.39 in two decimal places.**