Question

Consider a YDSE that has different slits width, as a result, amplitude of the waves from two slits are A and 2A, respectively. If ${{I}_{0}}$ be the maximum intensity of the interference pattern, then intensity of the pattern at a point where phase difference between waves is $\phi$ is :
$\text{A}\text{. }{{I}_{0}}{{\cos }^{2}}\phi$
$\text{B}\text{. }\dfrac{{{I}_{0}}}{3}{{\sin }^{2}}\dfrac{\phi }{2}$
$\text{C}\text{. }\dfrac{{{I}_{0}}}{9}\left[ 5+4\cos \phi \right]$
$\text{D}\text{. }\dfrac{{{I}_{0}}}{9}\left[ 5+8\cos \phi \right]$

Answer 1

Hint: Intensity is directly is directly proportional to the square of the amplitude, i.e. $I\propto {{A}^{2}}$. With this, find the relation between the intensities of the waves. The maximum intensity is equal to ${{I}_{\max }}={{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}^{2}}$. Then use the formula, $I={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}}\sqrt{{{I}_{2}}}\cos \phi$ to find the intensity at point where the phase difference is $\phi$.

Complete step by step answer:
Intensity if a wave depends on the amplitude of the wave. The intensity of the wave is directly proportional to the square of the amplitude of the wave.
If the intensity and the amplitude of the wave are I and A respectively, then $I\propto {{A}^{2}}$.
Therefore, we can write an equation as $I=k{{A}^{2}}$…… (i) ,
where k is a proportionality constant.
In YDSE (young’s double slit experiment), when light waves from the two sources (slits) interference at a point, the resultant amplitude is given as $A=\sqrt{A_{1}^{2}+A_{2}^{2}+2{{A}_{1}}{{A}_{2}}\cos \phi }$.
Here, ${{A}_{1}}$ and ${{A}_{2}}$ are the amplitudes of the waves emitted by the two sources respectively. $\phi$ is the phase difference between the waves at that point.
Therefore, we get ${{A}^{2}}=A_{1}^{2}+A_{2}^{2}+2{{A}_{1}}{{A}_{2}}\cos \phi$ ……. (ii).
Let the intensities of the waves with amplitudes ${{A}_{1}}$ and ${{A}_{2}}$ be ${{I}_{1}}$ and ${{I}_{2}}$ respectively. And let the resultant intensity due to the interference of the two waves at that point be I.
Therefore, $I=k{{A}^{2}}$.
From equation (i) we get, ${{I}_{1}}=kA_{1}^{2}$ and ${{I}_{2}}=kA_{2}^{2}$.
Substitute these values of ${{A}_{1}}$ and ${{A}_{2}}$ in equation (ii).
Hence,
$\dfrac{I}{k}=\dfrac{{{I}_{1}}}{k}+\dfrac{{{I}_{2}}}{k}+2\sqrt{\dfrac{{{I}_{1}}}{k}}\sqrt{\dfrac{{{I}_{2}}}{k}}\cos \phi$.
$I={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}}\sqrt{{{I}_{2}}}\cos \phi$ ….. (iii).
With this data, we will be able to solve the given question.
It is given that the amplitudes of the two waves are A and 2A and let their intensities be ${{I}_{1}}$ and ${{I}_{2}}$.
Therefore, we get
${{I}_{1}}=k{{A}^{2}}$ and ${{I}_{2}}=k{{\left( 2A \right)}^{2}}=4k{{A}^{2}}$.
Hence, $\dfrac{{{I}_{1}}}{{{I}_{2}}}=\dfrac{k{{A}^{2}}}{4k{{A}^{2}}}\Rightarrow {{I}_{2}}=4{{I}_{1}}$.
Maximum intensity that can be produced is given as ${{I}_{\max }}={{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}^{2}}$.
It is given that ${{I}_{\max }}={{I}_{0}}$ and substitute the value of ${{I}_{2}}$ in the above equation.
Hence, we get
${{I}_{0}}={{\left( \sqrt{{{I}_{1}}}+\sqrt{4{{I}_{1}}} \right)}^{2}}={{\left( \sqrt{{{I}_{1}}}+2\sqrt{{{I}_{1}}} \right)}^{2}}=9{{I}_{1}}$
$\Rightarrow {{I}_{1}}=\dfrac{{{I}_{0}}}{9}$.
$\Rightarrow {{I}_{2}}=4{{I}_{1}}=\dfrac{4{{I}_{0}}}{9}$
Now let us find the intensity at a point where the phase difference between the waves is $\phi$.
Use equation (iii).
Therefore we get,
$I={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}}\sqrt{{{I}_{2}}}\cos \phi$
But we know that ${{I}_{1}}=\dfrac{{{I}_{0}}}{9}$ and ${{I}_{2}}=\dfrac{4{{I}_{0}}}{9}$.
Therefore,
$I=\dfrac{{{I}_{0}}}{9}+\dfrac{4{{I}_{0}}}{9}+2\sqrt{\dfrac{{{I}_{0}}}{9}}\sqrt{\dfrac{4{{I}_{0}}}{9}}\cos \phi$
$\Rightarrow I=\dfrac{{{I}_{0}}}{9}+\dfrac{4{{I}_{0}}}{9}+2\sqrt{\dfrac{4I_{0}^{2}}{81}}\cos \phi$
$\Rightarrow I=\dfrac{{{I}_{0}}}{9}+\dfrac{4{{I}_{0}}}{9}+2\times \dfrac{2{{I}_{0}}}{9}\cos \phi$
$\Rightarrow I=\dfrac{5{{I}_{0}}}{9}+\dfrac{4{{I}_{0}}}{9}\cos \phi$
$\Rightarrow I=\dfrac{{{I}_{0}}}{9}\left[ 5+4\cos \phi \right]$.
Hence, the correct option is C.

Note: The maximum intensity due the interference of light is when $\cos \phi$ is maximum, i.e. when $\cos \phi =1$. Therefore, the phase difference is equal to $0,2\pi ,4\pi ...$
A phase difference of $\phi$ is equal to a path difference of $\Delta x=\dfrac{\lambda }{2\pi }\left( \phi \right)$.
$\lambda$ is the wavelength of the light.