# Refraction of Light

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12.3Refraction of Light

In this Physics tutorial, you will learn:

• What is refraction of waves?
• How refraction is applied in light waves?
• What do the Laws of Refraction say about light rays travelling through various media?
• What is the critical angle? When it is observed?
• What happens when the angle of incidence is greater than the critical angle?
• What is the index of refraction? How it is calculated?
• What does the Snell's Law says about the angles during refraction?
• How is the total internal reflection applied in technology?

## Introduction to Refraction of Light

Have you ever watched through a clear river water? Why the river bottom look much closer than it actually is?

When you partially insert your index finger on water, the fingers seems broken. Why does this occur?

Do you think light moves at the same speed in air and water? Explain your opinion.

In this tutorial, we will discuss about diffraction - a phenomenon that occurs in all types of waves, but which is particularly important in light waves. Also, the mathematical apparatus of light refraction, including the laws of refraction will be explained.

## Refraction in Water Waves as an Easier Way to Understand the Phenomenon of Light Refraction

If we have a tank with two different levels of depth as shown in the figure below

and some linear waves are generated from right to left after filling the tank with water, the following pattern is observed.

During this simple experiment, we observe two things:

1. Wavelength changes magnitude (in the specific case this magnitude increases) when moving from shallow to deep water; this can be observed by the increasing difference between two waves shown through the wave lines.
2. Waves also change direction when the depth of water changes. In the specific case, the direction changes in such a way to fit the direction of the border between the two depth levels.

However, if accurate measurement are made, we will observe that frequency (the number of waves passing through a given point of the container in one second) doesn't change. From the equation of waves

v = λ × f

we find that the speed of wave changes (here it increases) when moving in deeper water.

Hence, the only quantity which does not change during this process is the frequency; none from the other quantities (including amplitude) remains the same when any the medium feature (here the depth) change. Such a phenomenon is known as refraction of waves.

By definition, refraction is the change in direction of a wave passing from one medium to another or from a gradual change in the medium.

Our focus in this tutorial is in the refraction of light waves, which are discussed in the next paragraph.

## Refraction of Light Waves

The phenomenon of refraction is more obvious in light waves. Thus, if you insert a finger or a straw in water, it will seem as broken due to the phenomenon of light refraction when it changes medium (here from air to water). Look at the figure.

There is no change in the colour of the straw; this is a confirmation of the fact that frequency of waves during refraction does not change (frequency of light is related to its colour).

The question arisen at this point is: Is there any regularity in the light bending when it enters into a new medium?

The answer is YES. We can find the direction of a light ray when it enters into a new medium if the initial quantities are known. For this, the Laws of Refraction explained in the next paragraph comes to help.

## Laws of Refraction

Like in reflection, we must first introduce the concept of normal line to avoid issues caused by any possible non-regular interface between the two media in which light travels. Thus, normal line in refraction is the line that is perpendicular to the interface of the two media and which passes through the same point in which light touches the second medium as shown below.

Based on this description, the two Laws of Refraction are:

### 1. Incident ray, refracted ray and the normal line are all at the same plane

This law is very similar to the first law of reflection. Look at the figure.

### 2. When the light passes from a less dense medium to a denser one, it approaches the normal line, while when light passes from a denser medium to a less dense one, it moves away from the normal line

For example, when light passes from air into water, it bends towards the normal line as shown in the above figure. On the other hand, when light passes from glass to air for example, light bends away from the normal line.

We can use the concepts of the angle of incidence and angle of refraction to express the 2nd Law of Refraction. This, if we denote the angle of incidence by î, the angle of refraction by ȓ, the density of the first medium by p1 and that of the second medium by p2, we can write:

If p1 < p2, then î > ȓ
If p1 > p2, then î < ȓ

The explanation for this phenomenon is as follows:

Since the frequency of light waves does not change during refraction, the time of motion does not change as well, because time is a multiple of period (t = N × T) and period is the inverse of frequency (T = 1/f). Therefore, we can write for frequency of light waves

f = 1/T
= N/t

Since the speed of light decreases when it enters in a denser medium, the wavelength decreases as well because

c=λ × f

Hence, the light ray shortens its path when it enters a denser medium (for example from air to water). Therefore, it approaches the normal line (the angle of refraction decreases).

The opposite occurs when light passes from a denser medium to a less dense one. In this case, the wavelength increases and therefore the path becomes longer. As a result, the light ray moves away from the normal line, i.e. the angle of refraction increases.

Only in one case the light rays do not bend when they change medium. This occurs when light is incident at 00 to the normal line (or at 900 to the interface of two media). This is because the angle of refraction cannot be smaller than 0. The only thing we can observe in such cases is that objects look in a different distance from the surface than they actually are because light shortens the path when it enters into a denser medium and it elongates the path when moving into a less dense medium. That's why rocks at bottom of the river look closer than they actually are.

## Critical Angle. Total Internal Reflection

When light passes from a denser medium to a less dense one, the angle of refraction is greater than the angle of incidence, as seen in the previous paragraph. If we increase the angle of incidence, the angle of refraction increases too. Therefore, it comes a moment in which the angle of refraction becomes 900 meanwhile the angle of incidence is smaller as shown in the figure.

When the refracted ray 2 is at 900 to the normal line, it does not enter in the econ medium but passes through the interface between the two media as shown in the above figure. In such a case, the corresponding incident angle is known as "critical angle". Thus, by definition, the critical angle is defined as the angle of incidence that provides an angle of refraction of 90-degrees. Make particular note that the critical angle is an angle of incidence value.

For example, for the water-air boundary, the critical angle is 48.6-degrees. This means for incidence angles smaller than 48.60, light will pass from water to air, but when the incidence angle is 48.60 the light ray emitted from inside water does not passes in air but it moves through the interface between water and air.

## The Index of Refraction

As explained earlier, light moves at different speeds in different media. This causes diffraction. To express this phenomenon mathematically, a concept known as Index of Refraction is introduced. The index of refraction is denoted by n and it is calculated by

n = speed of light in vacuum/speed of light in the given medium

For example, since the light speed in air is about 300 000 km/s while in water it is 225 000 km/s, the refractive indexes in air and water are:

nair cvacuum/cair = 300 000 km/s/300 000 km/s = 1

and

nwater cvacuum/cwater = 300 000 km/s/225 000 km/s = 4/3 = 1.33

### Example 1

The refractive index in glycerine is 1.46 and in diamond 2.4. Calculate the speed of light in glycerine and diamond.

### Solution 1

From the formula of refractive index and giving that the speed of light in vacuum is 300 000 km/s, we obtain

nglycerine = cvacuum/cglycerine ⟹ cglycerine = cvacuum/nglycerine
= 300 000 km/s/1.46
= 205 000 km/s
ndiamond = cvacuum/cdiamond ⟹ cdiamond = cvacuum/ndiamond
= 300 000 km/s/2.4
= 125 000 km/s

Remark! Critical angle exists only when the incident medium is denser than the refractive medium as only in this case the refracted ray moves away from the normal line and so, the refraction angle becomes 900 before the incident angle.

## Snell's Law Equation on Refraction of Light

The path of light inside a medium is somehow related to the angle in which the ray enters to that medium. Therefore, there is a relationship between the refractive indexes and the sines of the incident and refraction angles given by an equation known as Snell's Law. The formula used to express mathematically the Snell's Law is

n1 × sinθ1 = n2 × sinθ2

where

• n1 is the refractive index in the incident medium,
• n2 is the refractive index in the refraction medium,
• θ1 is the angle of incidence, and
• θ2 is the angle of refraction

### Example 2

A light ray enters from air to water at 300 (sin 300 = 0.5). Calculate the refraction angle of the light ray in water in these conditions. Take the values for the refractive indexes from the previous paragraph.

### Solution 2

We have the following clues:

θ1 = 300 (sin θ1 = 0.5)
n1 = 1
n2 = 1.33
θ2 = ?

From the Snell's Law, we have

n1 × sinθ1 = n2 × sinθ2

Substituting the values, we obtain

1 × 0.5 = 1.33 × sinθ2
sinθ2 = 0.5/1.33
= 0.3759

Therefore, we obtain for the angle of refraction θ2:

θ2 = sin-1 0.3759
= 220

This is obvious; the angle of refraction is smaller than the angle of incidence because (as stated by the 2nd Law of Refraction) when light passes from a less dense medium to a denser one it approaches the normal line.

### How to Find the Values of Critical Angle in Various Media?

We can use Snell's Law to find the values of critical angles in various media. We just have to know the refractive indexes (or the light's speed in those media) and then apply the Snell's Law for θ2 = 900 (sin 900 = 1). Let's consider an example.

#### Example 3

What is the critical angle when light moves from glass to air? Take the speed of light in glass equal to 200 000 km/s.

#### Solution 3

First, we must calculate the refractive index in glass. We have:

nglass = cvacuum/cglass = 300 000 km/s/200 000 km/s = 1.5

Now, using the Snell's Law, we obtain

n1 × sinθ1 = n2 × sinθ2

and giving that n1 = nglass = 1.5, n2 = nair = 1, sinθ2 = 1 (because θ2 = 900), we obtain for the sine of the critical angle:

sinθ1 = n2 × sinθ2/n1
= 1 × 1/1.5
= 0.6667

Therefore, we obtain for the critical angle θ1 of glass-to-air refraction:

θ1= sin-1 (0.6667)
= 41.80

## Total Internal Reflection

When the angle of incidence is greater than the critical angle, the light ray is reflected back into the first medium after touching the interface between the two media. This means the interface acts like a plane mirror. As a result, a phenomenon known as total internal reflection does occur. During this process, the laws of reflection are applied. However, you can also use the laws of refraction to study this process by considering the refraction angle greater than 900 as shown in the figure.

The ray 3 is incident at a larger angle than the critical. As a result, it is reflected back into the first medium. The angle of refraction is r3 while the angle of reflection is the supplementary of r3 (1800 - r3).

### Example 4

A light ray moves out from glass to air at 500 to the normal line.

1. What phenomenon does occur in this case?
2. What is the refraction angle?

Take the critical angle of glass-to-air equal to 41.80.

### Solution 4

1. Since the angle of incidence is greater than the critical angle, a total internal reflection will take place. Therefore, the angle of reflection will be equal to the angle of incidence, i.e. 500.
2. The angle of refraction is the supplementary of the angle of total internal reflection, i.e. 1800 - 500 = 1300 as shown in the figure below.

A practical application of total internal reflection is the submarine. It cannot be detected from above the water as it emits light at angles that are greater than the critical angle of water-to-air (41.80).

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