Two coherent monochromatic light beams of intensities I and 4I are superimposed. The maximum and minimum possible intensities in the resulting beam are
A. 5I and I
B. 5I and 3I
C. 9I and I
D. 9I and 3I

Hint: Intensity of a light beam depends on the amplitude of the light. When two coherent light beams merge, it can have different types of superpositions. The maximum intensity occurs when they are in-phase and minimum superposition occurs when they are completely out of phase.

Complete step-by-step answer:
Superposition of two light beams occurs when two light beams converge at a position. The resultant light beam has an amplitude which depends on the amplitude of the constituent waves and the phase of those waves.
The intensity of light waves depends on the amplitude of the wave in this manner.
$I\propto {{A}^{2}}$
When two light waves converge, we cannot add the intensity of the light waves.
We need to find the amplitude of the waves and then calculate the resultant amplitude of the new superimposed wave.

Let’s assume the amplitude of the wave that has intensity I, is A.
As intensity depends on the square of the amplitude, we can write
$I=k{{A}^{2}}$
Hence, for the other light wave amplitude can be given by,
$4I=k{{(2A)}^{2}}$
If the two light waves converge in-phase, the amplitudes can be added.
Hence, the resultant amplitude will be = (A+2A) = 3A.
So, the intensity will be,
$I'=k{{(3A)}^{2}}=9k{{A}^{2}}=9I$
When the two light waves converge out-of-phase, the amplitude will be,
(2A-A) =A
Hence, the intensity of the resultant light will be,
$I''=k{{(A)}^{2}}=k{{A}^{2}}=I$
So, the maximum intensity will be 9I and the minimum intensity will be I.
The correct answer is (C).

Note: There is another formula to calculate the maximum and minimum intensity of the light wave.
The maximum intensity can be given by,
$I={{(\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}})}^{2}}$
The minimum intensity can be given by,
$I={{(\sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}})}^{2}}$
The intensity due to a superposition can be given by,
$I={{({{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \theta )}^{2}}$

Where.
${{I}_{1}}$ is the intensity of one light beam
${{I}_{2}}$ is the intensity of the second light beam
$\theta $ phase difference between the two light beams