How to Calculate Critical Angle

Introduction

To understand how to calculate critical angle, we use any of the two formulas 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)?

How to Calculate Critical Angle

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 = 90⁰

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 sin90⁰ / 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 = 64⁰

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 = 48.6⁰ = 49⁰

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 = 90⁰ – 16.9⁰ = 73.1⁰

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 / sin73.1⁰

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.

Sources

Google Books