Question

In Young’s double slit experiment, if the phase difference between the two waves interfering at a point is \[\phi \], then the intensity at this point can be expressed as,
$\begin{align}
  & \text{A}\text{. }I=\sqrt{{{A}^{2}}+{{B}^{2}}+{{\cos }^{2}}\phi } \\
 & \text{B}\text{. }I=\dfrac{A}{B}\cos \phi \\
 & \text{C}\text{. }I=A+B\cos \left( \dfrac{\phi }{2} \right) \\
 & \text{D }I=A+B\cos \phi \\
\end{align}$

Answer 1

Hint: We are given the phase difference between two waves at an interfering point. We need to find the total intensity at that point. We have an equation to find the net intensity at a point after interference and we also know that intensity is directly proportional to square of amplitude. By using these concepts we will get the required solution.
Formula used:
$I={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}\cos \phi }$
$I\propto {{A}^{2}}$

Complete answer:
In the question Young’s double slit experiment is performed.
We are given the phase difference between two waves interfering at a point as ‘$\phi $’.
Let ‘${{I}_{1}}$’ be the intensity of one of the waves and ‘${{I}_{2}}$’ be the intensity of the other wave.
Then the net intensity after interference is given by the equation,
$I={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}\cos \phi }$
We know that intensity of a wave is directly proportional to the amplitude of the wave, i.e.
$I\propto {{A}^{2}}$, where ‘A’ is the amplitude and ‘I’ is the intensity of the wave.
Therefore we can write,
${{I}_{1}}\propto {{A}_{1}}^{2}$
${{I}_{2}}\propto {{A}_{2}}^{2}$
By substituting this in the equation for net intensity, we will get
 \[\Rightarrow I={{A}_{1}}^{2}+{{A}_{2}}^{2}+2{{A}_{1}}{{A}_{2}}\cos \phi \]
Now let us consider that,
\[{{A}_{1}}^{2}+{{A}_{2}}^{2}=A\]
And, \[2{{A}_{1}}{{A}_{2}}=B\]
Now we get the net intensity at the point as,
$\Rightarrow I=A+B\cos \phi $
This is the equation for the intensity at the interfering point.

So, the correct answer is “Option D”.

Note:
Young’s double slit experiment demonstrates the constructive and destructive interference of light.
In the experiment two coherent light sources are placed at a distance and are made to pass through two separate slits. The lights thus passed through the slits are observed on a screen.
Constructive interference occurs only when the path difference of the light from the two slits is an integral multiple of the wavelength. And the destructive interference occurs when the path difference of the lights is half the integral number of the wavelength.