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Interference Of Light - Condition And Types - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

Interference of light waves- 1, Interference of light waves- 2 is considered one of the most asked concept.
35 Questions around this concept.

Solve by difficulty

An initially parallel cylindrical beam travels in a medium of refractive index $\mu (I)=\mu _{0}+\mu _{2}I,where\;\mu _{0}\; and\; \mu _{2}$ are positive constant and $I$ is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius. As the beam enters the medium, it will

Travel as a cylindrical beam

Diverge

Converge

Diverge near the axis and converge near the periphery

View Solution

Concepts Covered - 2

Interference of light waves- 1

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).

Coherent sources-

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 $(\Delta x )$

Phase difference $(\Delta \phi )$ :

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

$\text { i.e. if } y_{1}=a_{1} \sin \omega t \text { and } y_{2}=a_{2} \sin (\omega t+\phi) \text { so phase difference }=\phi$

Path difference $(\Delta x )$ :

The difference in path lengths of two waves meeting at a point is called path difference between the waves at that point.

And 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 $\lambda =wavelength \ of \ waves$

Interference of light waves- 2

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

$\begin{array}{l}{y_{1}=A_{1} \sin (k x-w t)} \\ {y_{2}=A_{2} \sin (k x-w t+\phi)}\end{array}$

where $\phi$ is the phase difference between waves $y_1 \ \ and \ \ y_2$ .

And According to the principle of Super Position of waves

$\begin{aligned} y=& y_{1}+y_{2}=A_{1} \sin (k x-w t)+A_{2} \sin (k x-w t+\phi) \\ &=A_{1} \sin (k x-w t)+A_{2}[\sin (k x-w t) \cos \phi+\sin \phi \cos (k x-\omega t)] \\ \Rightarrow y &=\sin (k x-w t)\left[A_{1}+A_{2} \cos \phi\right]+A_{2} \sin \phi \cos (k x-w t) \dots (1) \end{aligned}$

Now let

$Acos\theta =A_{1}+A_{2} \cos \phi \\ and \ \ Asin\theta=A_{2} sin \phi$

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

and $A= \sqrt{{A_{1}}^{2}+{A_{2}}^{2}+2A_{1}A_{2}\cos \phi}$

and $\theta=\tan^{-1} \left (\frac{A_{2} \sin \phi }{A_{1}+A_{2} \cos \phi} \right )$

where

$A_{1}=$ the amplitude of wave 1

$A_{2}=$ the amplitude of wave 2

$\begin{array}{c} \ \ A_{\max }={A_{1}+A_{2}} \ \ and \ \ A_{\min }={A_{1}-A_{2}}\end{array}$

Resultant Intensity of two waves (I)-

Using $I \ \ \alpha \ \ A^2$

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

$\begin{aligned} \ \ I_{\max } &=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}} \Rightarrow I_{\max } &=(\sqrt{I_{1}}+\sqrt{I_{2}})^{2} \end{aligned}$
$\begin{array}{c}{I_{\min }=I_{1}+I_{2}-2 \sqrt{I_{1} I_{2}}} \Rightarrow {I_{\min }=(\sqrt{I_{1}}-\sqrt{I_{2}})^{2}} \end{array}$
For identical sources-

$I_{1}=I_{2}=I_{0} \Rightarrow I=I_{0}+I_{0}+2 \sqrt{I_{0} I_{0}} \cos \phi=4 I_{0} \cos ^{2} \frac{\phi}{2}$

$\text { Average intensity : } I_{a v}=\frac{I_{\max }+I_{\min }}{2}=I_{1}+I_{2}$
The ratio of maximum and minimum intensities

$\frac{I_{\max }}{I_{\min }}=\left(\frac{\sqrt{I_{1}}+\sqrt{I_{2}}}{\sqrt{I_{1}}-\sqrt{I_{2}}}\right)^{2}=\left(\frac{\sqrt{I_{1} / I_{2}}+1}{\sqrt{I_{1} / I_{2}}-1}\right)^2=\left(\frac{a_{1}+a_{2}}{a_{1}-a_{2}}\right)^{2}=\left(\frac{a_{1} / a_{2}+1}{a_{1} / a_{2}-1}\right)^{2}$

$\sqrt{\frac{I_{1}}{I_{2}}}=\frac{a_{1}}{a_{2}}=\left(\frac{\sqrt{\frac{I_{\max }}{I_{\min }}}+1}{\sqrt{\frac{I_{\max }}{I_{\min }}-1}}\right)$

Interference of Light-

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 bright fringe/spot.

The phase difference between the waves at the point of observation is $\phi=0^{\circ} \text { or } 2 n \pi$
Path difference between the waves at the point of observation is $\Delta x=n \lambda(i . e . \text { even multiple of } \lambda / 2)$
The resultant amplitude at the point of observation will be maximum

$\begin{array}{c} i.e \ \ A_{\max }={a_{1}+a_{2}} \\ {\text { If } a_{1}=a_{2}=a_{0} \Rightarrow A_{\max }=2 a_{0}}\end{array}$

Resultant intensity at the point of observation will be maximum

$\begin{aligned} \ \ i.e \ \ I_{\max } &=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}} \\ I_{\max } &=(\sqrt{I_{1}}+\sqrt{I_{2}})^{2} \\ \text { If } \quad I_{1} &=I_{2}=I_{0} \Rightarrow I_{\max }=4 I_{0} \end{aligned}$

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

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

Path difference between the waves at the point of observation is $\Delta x=(2 n-1) \frac{\lambda}{2}(\text { i.e. odd multiple of } \lambda / 2)$
The resultant amplitude at the point of observation will be minimum

$\begin{array}{c} i.e \ \ A_{\min }={A_{1}-A_{2}} \\ {\text { If } A_{1}=A_{2} \Rightarrow A_{\min }=0}\end{array}$

Resultant intensity at the point of observation will be minimum

$\begin{array}{c}{I_{\min }=I_{1}+I_{2}-2 \sqrt{I_{1} I_{2}}} \\ {I_{\min }=(\sqrt{I_{1}}-\sqrt{I_{2}})^{2}} \\ {\text { If } I_{1}=I_{2}=I_{0} \Rightarrow I_{\min }=0}\end{array}$