## Interference of Light Waves and Young’s Experiment

### Table of Contents

- Interference
- History
- Types of Interference

- Interference Pattern
- Intensity of light at the point of Interference
- Path difference for constructive and destructive interference
- Young’s double slit experiment
- Summary

## Interference

**Interference** of Light Waves is defined as the modification in the distribution of light energy when two or more waves superimpose each other. For Interference the waves emitted by sources should be with zero phase difference or no phase difference. These sources should emit continuous waves of same wave length and same time period. These sources should be very close to each other. The sources which emit light waves should be coherent sources. If the waves are coherent, the interference pattern is observable and is stable. If the waves are incoherent, the pattern is not visible.

## History

The wave theory of light was proposed by **Christiaan Huygens**. Huygens stated that light is made up of waves that vibrates up and down and is perpendicular to the direction of light. But this concept was not readily accepted and was in contradiction to the particle theory or the corpuscular theory of light proposed by **Sir Isaac Newton**. Isaac Newton described that light is made up of tiny particles. In the beginning, wave theory of light was not accepted.

Later, another scientist named **Thomas Young** performed the famous Interference experiment, which is called the **Young’s Double Slit Experiment (YDSE)**. This experiment proved the wave nature of light. After this, many other experiments were carried out which included the **Interference**, diffraction and polarization patterns. These experiments were only explained by considering the wave nature of light. Thus wave theory came into existence. Corpuscular theory could not explain the various optical phenomena like interference, diffraction, polarization, dispersion and many others.

## Types of Interference

There are two types of Interference:

### Constructive Interference

When crest of one wave falls on the crest of other wave, the resultant amplitude will be maximum. The case in which the resultant amplitude is maximum is called constructive Interference. Here the two interfering waves have a displacement which is in the same direction.

Consider two waves which travels in same direction with same frequency. Now the amplitude of the waves gets add up when the light waves are at same place and time. The new wave looks similar to the original waves, but the amplitude of the wave is higher. In constructive interference, the waves are in phase.

### Destructive Interference

When crest of one wave falls on the trough of other wave, the resultant amplitude will be minimum. The case in which resultant amplitude is minimum is called **Destructive Interference.** Consider the figure given below. Here the first wave is up and the second wave is down. So when they add together the resultant one becomes zero. Thus in destructive interference, the sum of the wave can either be less than the original waves and or it can be zero. Here the two waves will be out of phase.

## Interference Pattern

Constructive Interference results in bright bands and destructive interference results in dark bands. These are called **Fringes**. This formation of bright and dark bands on the screen results in forming an interference pattern. Consider the picture shown below. The two slits present in it act as two coherent sources of light. We can see the pattern of light and dark bands here.

## Intensity of light at the point of Interference

We know that intensity of any light wave is directly proportional to the square of its amplitude. In case of coherent sources, let’s consider the factor of frequency to be a constant. Thus the intensity for the wave is considered as KA^{2}.

I = KA^{2}, where K is the constant which depends on what medium the wave is in.

Consider the resultant amplitude as ‘R’ at the point of interference. Now the resultant intensity at this point can be written as

I_{R }= KR^{2}

= K [A_{1}^{2 }+ A_{2}^{2 }+ 2A_{1 }A_{2} Cos ϕ]

= KA_{1}^{2 }+ KA_{2}^{2}+ 2KA_{1 }A_{2 }Cos ϕ

Substitute KA_{1}^{2} = I_{1 }and KA_{2}^{2 } = I_{2}

Thus I_{R }= I_{1 }+ I_{2 }+ 2 √(I_{1 }I_{2 }Cos ϕ)

So the resultant intensity of two waves after the interference phenomenon is

**I _{R }= I_{1 }+ I_{2 }+ 2 √(I_{1 }I_{2 }Cos ϕ)**

In case of constructive interference, the value of ϕ =0 and so Cos ϕ =1.Then I_{R }= I_{1 }+ I_{2 }+ 2 (_{√}I_{1 }I_{2 }= (√ I_{1 }+ _{√}I_{2})^{2} where the waves are superposed in same phase. Here the resultant intensity is maximum. For destructive interference, the waves superpose in opposite direction. Then Cos ϕ = -1 Thus the resultant intensity is minimum. So, I_{R }= I_{1 }+ I_{2 }- 2 √( I_{1 }I_{2 }= (√I_{1 }- √I_{2})^{2}.

**Interference of equal Intensity waves**

As mentioned earlier, the resultant intensity of waves at the point of interference

I_{R }= (I_{1 }+ I_{2 }+ 2 √(I_{1 }I_{2 }Cos ϕ). Let us consider that the two waves have same and equal intensity, which means I_{1 }= I_{2}= I_{0}.

Thus I_{R }= 2I_{0 }+ 2I_{0} Cos ϕ

= 2I_{0} (1+ Cos ϕ)

1+ Cos ϕ = 2cos^{2} ϕ/2 (Cos ϕ in half angle form)

Then I_{R }= 4I_{0} cos^{2} ϕ/2.

Hence for constructive interference, intensity will be maximum, I_{R }= ( √I_{1 }+ √I_{2})^{2} = 4 I_{0.}

For destructive interference, the intensity is minimum and I_{R }= (√I_{1 }- √I_{2})^{2} = 0.

## Path difference for constructive and destructive interference

In constructive interference, the phase difference is considered as 2nӅ, where n being an integer. Now to find the path difference between the waves,

Δ = /2Ӆ* ϕ

= /2Ӆ*2nӅ

Δ = n λ (n =0, 1, 2……)

Hence, we can say that two waves interfere constructively, when their path difference Δ = λ, 2λ.... nλ. .

Consider two light sources S_{1 }and S_{2}, which are coherent. Assume that the light sources are switched on at the same time. Consider a point P where the two light waves emitted from the coherent sources are superposed. Now the point P is located at a distance a_{1} from source S_{1 }and a distance a_{2} from source a_{2}. If the two sources are in same phase, then the path difference, Δ = a_{2}- a_{1.} If the path difference at point P with respect to the two sources is multiple of λ, then the intensity at point p is maximum and will be equal to 4I_{0}, if the two waves are of equal intensity. Otherwise it is shown as (√I_{1 }+ √I_{2})^{2}. In case of destructive interference, we know that Cos ϕ = -1 and it happens when the phase difference, ϕ = (2n+1) Ӆ which corresponds to opposite phase. To calculate the path difference between the waves in destructive interference, Δ= /2Ӆ* ϕ

= λ/2π*(2n+1) π

= (2n + 1) λ/2

Hence, we can say that two waves interfere destructively, when their path difference Δ = λ/2, 3λ/2....(2n-1) λ/2. (n = 1, 2……)

For destructive interference, the waves superpose in opposite direction. Then Cos ϕ = -1 Thus the resultant intensity is minimum. So, I_{R }= I_{1 }+ I_{2 }- 2 √( I_{1 }I_{2 }= √( I_{1 }- I_{2})^{2}.

## Young’s double slit experiment

As mentioned earlier, Thomas young performed the famous interference experiment which is the Ydse. It was by this experiment; the wave theory of light came into existence. In this experiment, Thomas Young used one single light source and it was the passed through two slits to obtain two coherent sources. Thus each slit act as a light source.

Consider two slits S1 and S2 on a screen which is kept parallel and very close to each other. These slits are illuminated by other narrow slit S. Light spreads out fromS1 and S2 ad falls on a screen. When both slits are open, there is a formation of interference fringes.

Young’s double slit Experiment

We know that I = I_{1} + I_{2} + 2 √ (I_{1} I_{2} Cos ϕ).

**Condition for bright fringes/maxima**, ϕ = 2nΠ or Δ = nλ, path difference

Where n = 0, 1, 2. Φ is the phase difference.

**Condition for dark fringes/minima**, ϕ = (2n-1) Π or Δ = (n-1)/2λ, path difference.

Where n = 1, 2, 3.

**Maxima and Minima**

Path difference between two waves

Δ = S_{1}P- S_{2}P= d sin θ

We know that D >> d, so the angle θ is small

Sin θ = Y_{n }/D

Δ = d sin θ = d (Y_{n}/D) ⇒ Y_{n} = Δ D /d

For n^{th} maxima, Δ** = nλ**

**Y _{n} = nλD/d,** where n = 0, 1, 2, 3. This denotes bright fringe.

n = 0, Y_{n} = 0

n = 1, Y_{n }= λ D/d

n = 2, Y_{n }= 2λD/d

1 ane

For n^{th} minima Δ = (n-1/2) λ

**Y _{n} = **

**(n - ½)**

**λ D/d**where n = 1, 2, 3. This denotes dark fringe

n = 1, Y_{n }= λ D/2d

n = 2, Y_{n }= 3λD/2d

## Summary

**Interference**is the combination of two or more waves and as a result forms a wave of greater or lower amplitude. There are two types of interference- constructive Interference and destructive Interference. Bright bands are formed by constructive interference and dark bands are formed by destructive interference. They are called**Fringes**.- The resultant intensity of two waves after the interference phenomenon is I
_{R }= I_{1 }+I_{2 }+ 2 √(I_{1 }I_{2 }Cos ϕ), where ϕ is the phase difference between two waves. - For constructive interference, intensity will be maximum, I
_{R }= (√I_{1 }+ √I_{2})^{2}= 4 I_{0.}For destructive interference, the intensity is minimum and I_{R }= (√I_{1 }- √I_{2})^{2}= 0 if the light waves have equal intensity. - Two waves interfere constructively, when their path difference Δ = λ, 2λ.... nλ.
- Two waves interfere destructively, when their path difference Δ = λ/2, 3λ/2 …. (2n+1) λ/2.
- Thomas young performed the famous Ydse interference experiment. Condition for bright fringes/maxima, Δ = nλ, path difference. This denotes the bright fringe. For n
^{th}minima Δ = (n-1/2) λ. This denotes the dark fringe.

**Watch this Video for more reference**

**More Readings**

Interference of light waves and Young’s Experiment