Interference Of Light - Condition And Types

Interference of light is a fascinating optical phenomenon that occurs when two or more light waves overlap, resulting in a pattern of alternating bright and dark fringes. This phenomenon is crucial in various scientific and technological applications, from creating anti-reflective coatings on lenses to enhancing the precision of optical instruments. In everyday life, interference can be observed in the vibrant colours of soap bubbles and oil films on water, which are caused by the constructive and destructive interference of light waves. By understanding the conditions and types of interference, we can better grasp how light behaves and harness its properties for practical uses. This article delves into the conditions necessary for interference to occur and explores the different types of interference, providing a comprehensive insight into this intriguing aspect of wave optics. In this article, we will discuss coherent sources, phase difference, path difference, the principle of superposition, resultant intensity, types of Interference and solved examples for better understanding.

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Interference Of Light - Condition And Types

Interference of light is a fascinating optical phenomenon that occurs when two or more light waves overlap, resulting in a pattern of alternating bright and dark fringes. This phenomenon is crucial in various scientific and technological applications, from creating anti-reflective coatings on lenses to enhancing the precision of optical instruments. In order to observe interference in light waves, the following conditions must be met:

• The sources must be coherent.
• The source should be monochromatic (that is, of a single wavelength).

What is Coherent Sources?

Coherent sources are crucial for producing stable and observable interference patterns. These sources emit light waves that maintain a constant phase difference and have the same frequency and wavelength

Two sources are said to be coherent if they produce waves of the same frequency with a constant phase difference. The relation between Phase difference $(\Delta \phi)$ and Path difference $(\Delta x)$

What is the Phase Difference?

Phase difference refers to the difference in the phase angle between two periodic signals, such as light waves or sound waves. It is a measure of how much one wave is shifted in time or space relative to another.

The difference between the phases of two waves at a point is called phase difference.

$
\text { i.e. if } y_1=a_1 \sin \omega t \text { and } y_2=a_2 \sin (\omega t+\phi) \text { so phase difference }=\phi
$

What is Path Difference?

Path difference refers to the difference in the distance travelled by two coherent waves from their respective sources to a common point. It is a critical concept in understanding interference patterns, as the path difference determines the type of interference (constructive or destructive) that occurs at that point. The difference in path lengths of two waves meeting at a point is called the path difference between the waves at that point.

The relation between Phase difference $(\Delta \phi)$ and Path difference $(\Delta x)$ is given as
$
\Delta \phi=\frac{2 \pi}{\lambda} \Delta x=k \Delta x
$
where $\lambda=$ wavelength of waves

Principle of Super Position

According to the principle of Super Position of waves, when two or more waves meet at a point, then the resultant wave has a displacement $(y)$ which is the algebraic sum of the displacements ( $y_1$ and $y_2$ ) of each wave.
i.e $y=y_1+y_2$
consider two waves with the equations as
$
\begin{aligned}
& y_1=A_1 \sin (k x-w t) \\
& y_2=A_2 \sin (k x-w t+\phi)
\end{aligned}
$

where $\phi$ is the phase difference between waves $y_1$ and $y_2$.
According to the principle of Super Position of waves
$
\begin{aligned}
y= & y_1+y_2=A_1 \sin (k x-w t)+A_2 \sin (k x-w t+\phi) \\
& =A_1 \sin (k x-w t)+A_2[\sin (k x-w t) \cos \phi+\sin \phi \cos (k x-\omega t)] \\
\Rightarrow y & =\sin (k x-w t)\left[A_1+A_2 \cos \phi\right]+A_2 \sin \phi \cos (k x-w t) \ldots(1)
\end{aligned}
$

Now let
$
\begin{aligned}
A \cos \theta & =A_1+A_2 \cos \phi \\
\text { and } A \sin \theta & =A_2 \sin \phi
\end{aligned}
$

Putting this in equation (1) we get
$
y=A \sin (k x-\omega t) \cos \theta+A \sin \theta \cos (k x-\omega t)
$
thus we get the equation of the resultant wave as
$
y=A \sin (k x-\omega t+\theta)
$

where $A=$ Resultant amplitude of two waves
$
\begin{aligned}
& \text { and } A=\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \phi} \\
& \text { and } \theta=\tan ^{-1}\left(\frac{A_2 \sin \phi}{A_1+A_2 \cos \phi}\right) \\
&
\end{aligned}
$
where
$A_1=$ the amplitude of wave 1
$A_2=$ the amplitude of wave 2
$
A_{\max }=A_1+A_2 \text { and } A_{\min }=A_1-A_2
$

Resultant Intensity of Two Waves

When two waves interfere, the resultant intensity at any point depends on the amplitude and phase relationship between the waves. The intensity of a wave is proportional to the square of its amplitude. The resultant intensity can be found by considering the amplitudes of the individual waves and the phase difference between them.

Using $I \propto A^2$
we get $I=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi$
where
$I_1=$ The intensity of Wave 1
$I_2=$ The intensity of Wave 2
- $I_{\max }=I_1+I_2+2 \sqrt{I_1 I_2} \Rightarrow I_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2$
- $I_{\min }=I_1+I_2-2 \sqrt{I_1 I_2} \Rightarrow I_{\min }=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2$
- For identical sources
$
I_1=I_2=I_0 \Rightarrow I=I_0+I_0+2 \sqrt{I_0 I_0} \cos \phi=4 I_0 \cos ^2 \frac{\phi}{2}
$

Average intensity : $I_{a v}=\frac{I_{\max }+I_{\min }}{2}=I_1+I_2$

The ratio of maximum and minimum intensities

$
\frac{I_{\max }}{I_{\min }}=\left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)^2=\left(\frac{\sqrt{I_1 / I_2}+1}{\sqrt{I_1 / I_2}-1}\right)^2=\left(\frac{a_1+a_2}{a_1-a_2}\right)^2=\left(\frac{a_1 / a_2+1}{a_1 / a_2-1}\right)^2
$
or
$
\sqrt{\frac{I_1}{I_2}}=\frac{a_1}{a_2}=\left(\frac{\sqrt{\frac{I_{\max }}{I_{\min }}}+1}{\sqrt{\frac{I_{\max }}{I_{\min }}-1}}\right)
$

Interference of Light

Interference of light is a phenomenon that occurs when two or more coherent light waves overlap and combine to form a new wave pattern. This new pattern results from the principle of superposition, where the resultant wave is the sum of the individual waves' amplitudes at each point. The interference can be constructive or destructive, depending on the phase relationship between the waves. It is of the following two types.

1. Constructive interference

• When the waves meet a point with the same phase, constructive interference is obtained at that point.

i.e. we will see a bright fringe/spot.

• The phase difference between the waves at the point of observation is $\phi=0^{\circ}$ or $2 n \pi$

• Path difference between the waves at the point of observation is $\Delta x=n \lambda($ i.e. even multiple of $\lambda / 2)$

• The resultant amplitude at the point of observation will be the maximum

$
\text { i.e } A_{\max }=a_1+a_2
$

If $a_1=a_2=a_0 \Rightarrow A_{\max }=2 a_0$

• The resultant intensity at the point of observation will be the maximum

i.e $\begin{aligned} I_{\max } & =I_1+I_2+2 \sqrt{I_1 I_2} \\ I_{\max } & =\left(\sqrt{I_1}+\sqrt{I_2}\right)^2 \\ \text { If } \quad I_1 & =I_2=I_0 \Rightarrow I_{\max }=4 I_0\end{aligned}$

2. Destructive interference

• When the waves meet a point with the opposite phase, Destructive interference is obtained at that point. i.e we will see dark fringe/spot.

• The phase difference between the waves at the point of observation is

$\begin{aligned} & \phi=180^{\circ} \text { or }(2 n-1) \pi ; n=1,2, \ldots \\ & \text { or }(2 n+1) \pi ; n=0,1,2 \ldots\end{aligned}$

• Path difference between the waves at the point of observation is $\Delta x=(2 n-1) \frac{\lambda}{2}($ i.e. an odd multiple of $\lambda / 2)$

• The resultant amplitude at the point of observation will be a minimum

i.e $A_{\min }=A_1-A_2$

If $A_1=A_2 \Rightarrow A_{\min }=0$

• Resultant intensity at the point of observation will be minimum

$\begin{gathered}I_{\min }=I_1+I_2-2 \sqrt{I_1 I_2} \\ I_{\min }=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2 \\ \text { If } I_1=I_2=I_0 \Rightarrow I_{\min }=0\end{gathered}$

Solved Examples Based on Interference Of Light - Condition And Types

Example 1: Light of wavelength $\lambda_0$ in air enters a medium of refractive index $\mathbf{n}$. If two points $\mathbf{A}$ and $\mathbf{B}$ in this medium lie along the path of this light at a distance $x_1$ then the phase difference $\phi_0$ between these two points is :

$
\text { 1) } \phi_0=\frac{1}{n} \cdot\left(\frac{2 \pi}{\lambda_0}\right) \cdot x
$
2) 0
3) $\phi_0=n \cdot\left(\frac{2 \pi}{\lambda_0}\right) \cdot x$
$
\text { 4) } \phi_0=\frac{1}{n-1} \cdot\left(\frac{2 \pi}{\lambda_0}\right) \cdot x
$

Solution:

Relation between phase & path difference

$
\Delta \phi=\frac{2 \pi}{\lambda} \times \Delta x
$
wherein
$\Delta \phi=$ Phase difference
$\Delta x=$ Path Difference
$\lambda=$ Wavelength
Phase difference $=\frac{2 \pi}{\lambda_0}$ (optical path difference $)$
Optical path difference $=n \cdot x$
$
\phi_0=\frac{2 \pi}{\lambda_0} \cdot(n x)
$

Hence, the answer is the option (3).

Example 2: What is the necessary condition for the interference of lightwave?

1) The sources of the waves must be coherent

2) The waves should be monochromatic

3) Both of these

4) none of these

Solution:

To observe interference in light waves, the following conditions must be met:

The sources must be coherent i.e. Two sources are said to be coherent if they produce waves of the same frequency with a constant phase difference.

The source should be monochromatic (that is, of a single wavelength).

Hence, the answer is the option (3).

Example 3: Two light beams of intensities in the ratio of 9: 4 are allowed to interfere. The ratio of the intensity of maxima and minima will be:

1) 2:3

2) 16:81

3) 25:169

4) 25:1

Solution:

$\begin{aligned} & \frac{I_1}{I_2}=\frac{9}{4} \\ & \frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2}{\left(\sqrt{I_1}-\sqrt{I_2}\right)^2}=\frac{25}{1}\end{aligned}$
Hence, the answer is the option (4)

Example 4: In Young's experiment the interfering has amplitudes in the ratio 3:2, and then ratios of amplitudes between bright and dark fringes are:

1) 5:1

2) 9:4

3) 7:1

4) 49:1

Solution:

The resultant amplitude of two waves

$
A=\sqrt{A_1{ }^2+A_2{ }^2+2 A_1 A_2 \cos \theta}
$
wherein
$A_1=$ amplitude of wave 1
$A_2=$ amplitude of wave 2
$\theta=$ phase difference
We have to obtain the ratio
$
\frac{A_{\max }}{A_{\min }}=\frac{A_1+A_2}{A_1-A_2}
$
and also the corresponding ratio of intensities
$
\begin{aligned}
\frac{I_{\max }}{I_{\min }} & =\frac{\left(A_1+A_2\right)^2}{\left(A_1-A_2\right)^2} \\
\frac{A_1}{A_2} & =\frac{3}{2}
\end{aligned}
$

By correspond and divide

$
\frac{A_1+A_2}{A_1-A_2}=\frac{3+2}{3-2}=5
$

Hence,
$
\frac{A_{\max }}{A_{\min }}=5
$

And,
$
\frac{I_{\max }}{I_{\min }}=25
$

Hence, the answer is the option (1).

Example 5: Consider a YDSE that has different slit widths. As a result, the amplitude of waves from the two silts is A and 2A respectively. If $I_0$ be the maximum intensity of the interference pattern, then the intensity of the pattern at a point where the phase difference between waves is $\phi$ is:

1) $I_{0 \cos ^2 \phi}$
2) $\frac{I_0}{3} \sin ^2 \frac{\phi}{2}$
3) $\frac{I_0}{9}(5+4 \cos \phi)$
4) $\frac{I_0}{9}(5+8 \cos \phi)$

Solution:

The resultant Intensity of two waves

$
I=I_1+I_2+2 \sqrt{I_1 I_2} \cos \theta
$
wherein
$I_1=$ Intensity of Wave 1
$I_2=$ Intensity of Wave 2
$\theta=$ Phase difference
As amplitudes are A and 2A, so intensities would be in ratio 1:4,
Let us say they are I and 41 respectively. For maximum intensity, their phase difference is equal to $\underline{0}^{\circ}$

The maximum intensity of the resultant wave is
$
\begin{aligned}
& I_{\max }=I_0=I+4 I+2 \sqrt{I \times 4 I}=9 I \\
& I=\frac{I_0}{9}
\end{aligned}
$

Intensity at any point,

$\begin{aligned} & I^{\prime}=I_1+I_2+2 \sqrt{I_1 I_2} \cos \theta=I+4 I+2 \sqrt{4 I^2} \cos \theta \\ & I^{\prime}=5 I+4 I \cos \phi=\frac{I_0}{9}(5+4 \cos \phi)\end{aligned}$

Hence, the answer is the option (3).

Summary

Interference of light occurs when overlapping light waves create alternating bright and dark fringes. This phenomenon requires coherent and monochromatic sources. Constructive interference results in bright fringes, while destructive interference leads to dark fringes. Understanding interference is essential for applications like anti-reflective coatings and precise optical instruments. Examples and solved problems illustrate how phase and path differences influence the interference patterns.