# Physics Tutorial: Interference and Diffraction of Light

In this Physics tutorial, you will learn:

• Is there any diffraction in light waves?
• Why diffraction and interference of light are related phenomena?
• How to obtain light diffraction on a screen using a single slit?
• What Young Double-Slit Experiment is used for?
• Why diffraction gratings are used in technology?
• How to find wavelength of a light wave using diffraction gratings?

## Introduction to Interference and Diffraction of Light

Do you think light rays can bend around obstacles? What dimensions these obstacles must have to observe this phenomenon?

Is it possible for a light ray to pass through a small hole? Does it bend during this process?

When you turn on two identical bulbs placed at different positions inside the same room, you can observe some dark and bright concentric circles on the wall. Why do they occur?

Do you think two light waves can overlap? What conditions must they meet in order to make this phenomenon occur?

In this tutorial, we will discuss about interference and diffraction of light - two phenomena that require very specific conditions to take place. Also, the tools used to obtain the interference and diffraction of light waves are discussed and explained. Hence, this is a very special topic - one of the most important in the Physical Optics.

## Does the Diffraction of Light Waves Exist?

It is a known fact that light is an electromagnetic transverse wave. As such, it possesses all properties of waves, including diffraction.

As explained in our Physics tutorial on the Diffraction of Waves, diffraction is the property of a wave bending around a small obstacle, or when it passes through a narrow opening (gap). However, this phenomenon can be observed only when the dimensions of the obstacle or gap are comparable to the wavelength. This is the reason why diffraction of light waves can hardly be observed without using special techniques.

One of techniques we have discussed in our Physics tutorial on our Physics tutorial covering the Interference of Waves is by using a small gap in which the light waves are allowed to pass through. As a result, a number of bright and dark regions are produced on a screen placed at a distance from the gap. These regions form because of waves' overlap (interference). When dealing with light waves, the above gap or opening is known as slit. Hence, we say that the phenomenon described above is called "single slit diffraction" of light waves.

When light passes through a single slit whose width a is on the order of the wavelength of the light, then we can observe a single slit diffraction pattern on a screen that is a distance D >> a away from the slit. The light intensity is a function of angle, i.e the intensity of light weakens when the angle increases as shown in the figure below.

Huygens' Principle tells us that each part of the slit can be thought of as an emitter of waves. All these waves interfere to produce the diffraction pattern In places where crest meets crest we have constructive interference (as in the above figure) and where crest meets trough we have destructive interference.

## Single Slit Diffraction Formula

As explained in the Interference of Waves, the path difference between the two waves after passing through the slit must be a whole multiple of wavelength in order to observe a constructive interference (bright region) or a half multiple of wavelength for destructive interference (dark region). The equations for each case are:

d2 - d1 = N × λ

for bright regions (constructive interference) and

d2 - d1 = 2N - 1/2 × λ

for dark regions (destructive interference), where N is an integer.

The above equations represent the conditions for the interference to occur in single slit diffraction.

Now, let's consider the following figure.

The path difference d2 - d1 represents the opposite side of the angle θ in the small right triangle in the figure, where the hypotenuse is the slit's width a. This means the path difference d2 - d1 is equal to a × sin θ where θ is the angle formed by the rays' direction and the horizontal direction (in reality we consider the small angle inside the triangle which is equal to θ as well (because angles with perpendicular corresponding sides are equal). Therefore, we obtain the equation of single slit diffraction:

d2-d1 = a × sinθ

This equation helps us determine the wavelength if we know the slit's width and the angle formed by the rays to the horizontal direction, because we have

d2 - d1 = a × sinθ = N × λ

for constructive interference and

d2 - d1 = a × sinθ = 2N-1/2 × λ

for destructive interference.

At this point, the drawback is the number N that gives the multiple of wavelength in the path difference, which is impossible to know in these conditions. Therefore, we use the triangles' similarity rules to find indirectly the number N of extra wavelengths of the longest ray because the number of wavelengths caused by this difference in light paths causes a number of bright and dark strips on the screen.

In other words, the difference in light paths

d2 - d1 = a × sinθ = N × λ

brings as a result a displacement y of the point in which the rays converge which is the same multiple of the thickness of a pair of strips (one bright and one dark. Look at the figure.

From the triangles similarity rules we have:

a × sinθ/y = a/d2

Given that

d2 = D/cosθ

and

a × sinθ = N × λ

we obtain

N × λ/y = a/D/cosθ

Hence,

N = a × y × cosθ/D × λ

However, it is not necessary to use such a long approach to find the number of extra cycles N travelled by the wave of the longest path as this number can be obtained by counting the strips. This is because the screen acts like a projector or a magnifier, so the number of strips corresponds to the number of extra waves. Therefore, it is enough the equation

a × sinθ = N × λ

for constructive interference and

a × sinθ = 2N - 1/2 × λ

for destructive interference to find the wavelength λ which is the ultimate goal in these situation.

### Example 1

A light beam containing two coherent and unicolour light waves, passes through a 0.4 μm thin slit and forms the 5-th maximum 2 cm above the central maximum on a screen that is placed 2 m away from the slit. What is the wavelength of these light waves?

### Solution 1

In this problem we have the following clues:

D = 2 m
a = 8 μm = 8 × 10 - 6 m
y = 20 cm = 2 × 10 - 1 m
λ = ?

First, let's determine the angle θ. From the figure, we can see that

tanθ = y/D
= 20 cm/2m
= 20 cm/200 cm
= 0.1

Therefore,

θ = tan-1 0.1
=5.710

Now, since we have a maximum in the given point of the screen, we use the equation

a × sinθ = N × λ

to determine the wavelength λ. Thus, substituting the known values, we obtain

8 × 10-6 × sin5.710 = 5 × λ
8 × 10-6 × 0.01 = 5 × λ
8 × 10-8 = 5 × λ
λ = 8 × 10-8/5 m
= 1.6 ×10-8 m

As you see from the result, these light rays belong to UV spectrum.

## Young's Double-Slit Experiment

The famous scientist Thomas Young developed a technique that is similar to the one described above to calculate the wavelength of a light wave. The only difference is that Young used two slits instead of a single slit to produce interference on the screen through diffraction of light waves. The reason for this decision was because it is very difficult to find point sources to produce light. It is much easier to use a large source such as a lamp or the Sun than a point source with very small dimensions. The two small slits are very close to each other, so the rays coming from the common source are coherent. As a result, the bright and dark fringes produced on the screen are more visible.

If the slits are circular, bright and dark circular fringes are produced on the screen and when slits are linear, bright and dark linear fringes are obtained. The only difference with the single slit diffraction lies in the circular slits, as unlike in the single slit diffraction a net of circular fringes is obtained in the Young Double slit experiment.

To determine the wavelength λ of the two coherent light waves passing through the slits, we must know three quantities:

1. The distance (separation) between the two slits, a
2. The distance (separation) between two fringes, i.e. the distance between the centres of two adjacent bright (or dark) regions, x and
3. The distance from the slits to the screen, D.

Thus, the wavelength λ is calculated by the equation

λ = a × x/D

The above equation is obtained by applying the triangles' similarity rules discussed earlier.

### Example 2

Two coherent yellow light rays of wavelength 580 nm produced by the same source are incident in parallel through two thin slits of separation equal to 0.2 mm. What is the distance from the central maximum in which the seventh maximum is observed, if the screen is 4 m away from the slits? Take the figure above as a reference.

### Solution 2

In this problem, we have the following clues:

λ = 580 nm = 5.8 × 10-7 m
a = 0.2 mm = 2 × 10-4 m
N = 7
D = 4 m
y = ?

First, the equation of Young's Double Slit Experiment is applied to calculate the separation x between two adjacent fringes. We have

λ = a × x/D
x = λ × D/a
= 5.8 × 10-7 × 4/2 × 10-4
= 1.16 × 10-2 m
= 1.16 cm

Therefore, the 7th maximum is displaced by

y = N × x
= 7 × 1.16 cm
= 8.12 cm

in either direction (up or down) from the central maximum.

## Diffraction Gratings

A diffraction grating tool is very similar in concept to the double-slit system discussed earlier but with more openings. It has a very large number of parallel lines on a glass or plastics obtained through grating process as shown in the figure.

The figure below shows an educational diffraction grating tool whose parameters (number of gratings per mm) are written under each glass.

Obviously, we cannot see the gratings on naked eye, as they are very close and thin.

A diffraction grating is often (but not only) used to split the white light into colours. It is also used to determine the wavelength of a light wave similarly to the methods mentioned earlier in single-slit diffraction and Young Double-Slit Experiment.

If we enlarge a part of a diffraction grating and see it in profile, the view obtained will be as shown below.

We can use the equation for the condition of constructive interference

a × sinθ = N × λ

to determine the wavelength of a certain light beam. Let's consider an example.

### Example 3

Monochromatic (one-coloured) light is incident at right angle to a diffraction grating having 2000 lines per centimetre. The angular separation between the zeroth and the fourth-order maxima is 300. Calculate the wavelength of the incident light.

### Solution 3

First, let's work out the separation a between two adjacent gratings. We have

a = 1 cm/2000 gratings/cm
= 0.0005 cm
= 5 × 10-4 cm
= 5 × 10-6 m

Given that N = 4 and sin 300 = 0.5, we obtain

a × sinθ = 4 × λ
5 × 10-6 × 0.5 = 4 × λ
2.5 × 10-6 = 4 × λ
λ = 2.5 × 10-6/4
= 0.625 × 10-6 m
= 6.25 × 10-7 m

## Physics Revision: Interference and Diffraction of Light Summary

Light is an electromagnetic transverse wave. As such, it possesses all properties of waves, including interference and diffraction. Since diffraction is observed only when the dimensions of the obstacle or gap are comparable to the wavelength, the diffraction of light waves can hardly be observed without using special techniques due to the very small wavelengths of light waves.

When light passes through a single slit whose width a is on the order of the wavelength of the light, then we can observe a single slit diffraction pattern on a screen that is a distance D >> a away from the slit. The light intensity is a function of angle, i.e the intensity of light weakens when the angle increases.

This known equation

d2-d1 = a × sinθ

helps us determine the wavelength λ if we know the slit's width a and the angle θ formed by the rays to the horizontal direction, because we have

d2-d1=a × sinθ = N × λ

for constructive interference and

d2-d1=a × sinθ = 2N - 1/2 × λ

for destructive interference.

The famous scientist Thomas Young developed a technique that is similar to the one described above to calculate the wavelength of a light wave. The only difference is that Young used two slits instead of a single slit to produce interference on the screen through diffraction of light waves.

To determine the wavelength λ of the two coherent light waves passing through the slits, we must know three quantities:

1. The distance (separation) between the two slits, a
2. The distance (separation) between two fringes, i.e. the distance between the centres of two adjacent bright (or dark) regions, x and
3. The distance from the slits to the screen, D.

Thus, the wavelength λ is calculated by the equation

λ = a × x/D

A diffraction grating tool is very similar in concept to the double-slit system discussed earlier but with more openings. It has a very large number of parallel lines on a glass or plastics obtained through grating process.

A diffraction grating is often (but not only) used to split the white light into colours. It is also used to determine the wavelength of a light wave similarly to the methods mentioned earlier in single-slit diffraction and Young Double-Slit Experiment.

## Physics Revision Questions for Interference and Diffraction of Light

1. Two yellow coherent light rays (λ = 580 nm) are incident normally to a single slit of width equal to 0.02 mm. What is the angle formed by the fourth minimum to the horizontal direction?

1. 580
2. 5.80
3. 0.580
4. 330

2. The slits' separation in a Young Double Slit Experiment is 0.05 mm and the third maximum is formed on the screen at 2 cm distance from the central maximum. If the light waves used for this experiment are green (λ = 550 nm), calculate the distance D of the slits from the screen.

1. 60 cm
2. 6 m
3. 1.8 m
4. 5.4 m

3. A diffraction grating with 1000 gratings/cm is used to diffract a monochromatic light beam. As a result, the fifth maximum is formed at 150 to the horizontal direction as shown in the figure.

Calculate the wavelength of the light beam in nm.

1. 517.6 nm
2. 105 nm
3. 2588 nm
4. 571 nm