Question

In Young's double slits experiment, one of the slits is wider than another, so that the amplitude of the light from one slit is double that from the other slit. If ${{I}_{m}}$ is the maximum intensity, the resultant intensity I when they interfere at phase difference $\phi $ is given by.
$\begin{align}
  & \text{A}\text{. }\dfrac{{{I}_{m}}}{9}\left( 4+5\cos \phi \right) \\
 & \text{B}\text{. }\dfrac{{{I}_{m}}}{3}\left( 1+2{{\cos }^{2}}\dfrac{\phi }{2} \right) \\
 & \text{C}\text{. }\dfrac{{{I}_{m}}}{5}\left( 1+4{{\cos }^{2}}\dfrac{\phi }{2} \right) \\
 & \text{D}\text{. }\dfrac{{{I}_{m}}}{9}\left( 1+8{{\cos }^{2}}\dfrac{\phi }{2} \right) \\
\end{align}$

Answer 1

Hint: A Young’s double slit experiment is performed with one of the slits wider than the other and hence the amplitude of a light twice the amplitude of the other light. We know that amplitude is directly proportional to the square of intensity. By using the equation to find the net intensity after interference and to find the maximum intensity, we will get the solution.
Formula used:
$I\propto {{A}^{2}}$
${{I}_{m}}={{I}_{1}}+{{I}_{2}}$
${{I}_{r}}={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \phi $

Complete answer:
In the question we have a double slit experiment where the width of one slit is greater than the other.
It is said that the amplitude of the wave from one slit is twice the amplitude of the other wave.
If the amplitude of one wave is ‘${{A}_{1}}$’ and the other wave is ‘${{A}_{2}}$’ then we have,
${{A}_{1}}=2{{A}_{2}}$
We know that the intensity is directly proportional to the square of the amplitude, i.e.
$I\propto {{A}^{2}}$
Therefore we can write,
${{I}_{1}}={{2}^{2}}{{I}_{2}}$
$\Rightarrow {{I}_{1}}=4{{I}_{2}}$
We know that the maximum intensity can be given as,
${{I}_{m}}={{I}_{1}}+{{I}_{2}}$
$\Rightarrow {{I}_{m}}={{A}_{1}}^{2}+{{A}_{2}}^{2}$
$\Rightarrow {{I}_{m}}=2{{A}_{2}}^{2}+{{A}_{2}}^{2}$
$\Rightarrow {{I}_{m}}={{\left( 3{{A}_{2}} \right)}^{2}}$
$\Rightarrow {{I}_{m}}=9{{A}_{2}}^{2}$
Here we can replace ${{A}_{2}}^{2}$ with ${{I}_{2}}$. Therefore the maximum intensity will become,
$\Rightarrow {{I}_{m}}=9{{I}_{2}}$
We know that when two waves interfere, the resultant intensity after interference is given by the equation,
${{I}_{r}}={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \phi $, were ‘${{I}_{r}}$’ is the resultant intensity and ‘$\phi $’ is the phase difference.
Since ${{I}_{1}}=4{{I}_{2}}$, we can write the above equation as,
$\Rightarrow {{I}_{r}}=4{{I}_{2}}+{{I}_{2}}+2\sqrt{4{{I}_{2}}{{I}_{2}}}\cos \phi $
$\Rightarrow {{I}_{r}}=5{{I}_{2}}+2\sqrt{4{{I}_{2}}^{2}}\cos \phi $
$\Rightarrow {{I}_{r}}=5{{I}_{2}}+2\times 2{{I}_{2}}\cos \phi $
$\Rightarrow {{I}_{r}}=5{{I}_{2}}+4{{I}_{2}}\cos \phi $
From earlier calculations we know that,
${{I}_{m}}=9{{I}_{2}}$
From this equation we get,
$\Rightarrow {{I}_{2}}=\dfrac{{{I}_{m}}}{9}$
By substituting this in the equation for resultant intensity, we get
$\Rightarrow {{I}_{r}}=\dfrac{5{{I}_{m}}}{9}+\dfrac{4{{I}_{m}}}{9}\cos \phi $
$\Rightarrow {{I}_{r}}=\dfrac{{{I}_{m}}}{9}\left( 5+4\cos \phi \right)$
We know that $\cos 2\theta =2{{\cos }^{2}}\theta -1$
Therefore we can write,
$\cos \phi =2{{\cos }^{2}}\left( \dfrac{\phi }{2} \right)-1$
Now let us substitute this in the equation for resultant intensity, thus we get
$\Rightarrow {{I}_{r}}=\dfrac{{{I}_{m}}}{9}\left( 5+4\left( 2{{\cos }^{2}}\left( \dfrac{\phi }{2} \right)-1 \right) \right)$
$\Rightarrow {{I}_{r}}=\dfrac{{{I}_{m}}}{9}\left( 5+8{{\cos }^{2}}\left( \dfrac{\phi }{2} \right)-4 \right)$
$\Rightarrow {{I}_{r}}=\dfrac{{{I}_{m}}}{9}\left( 1+8{{\cos }^{2}}\left( \dfrac{\phi }{2} \right) \right)$
Therefore the resultant intensity when the waves interfere at a phase difference $\phi $ is $\dfrac{{{I}_{m}}}{9}\left( 1+8{{\cos }^{2}}\left( \dfrac{\phi }{2} \right) \right)$

So, the correct answer is “Option D”.

Note:
Young’s double slit experiment demonstrates the constructive and destructive interference of light and also proves the wave nature of light.
In the experiment we place two coherent sources at a distance and pass the light from the source through two slits. The lights coming through the slit are made to fall and the screen and depending on the superposition they form constructive and destructive interferences.