Question

Two coherent monochromatic light beams of intensities I and 4I are superimposed. The difference between maximum and minimum possible intensities in the resulting beam is x I. The value of x is______.

Answer 1

Correct Answer: 8

The maximum intensity \( I_{\text{max}} \) in the superimposed beam is given by:
\[I_{\text{max}} = \left( \sqrt{I} + \sqrt{4I} \right)^2 = \left( \sqrt{I} + 2\sqrt{I} \right)^2 = (3\sqrt{I})^2 = 9I\]
The minimum intensity \( I_{\text{min}} \) is given by:
\[I_{\text{min}} = \left( \sqrt{4I} - \sqrt{I} \right)^2 = \left( 2\sqrt{I} - \sqrt{I} \right)^2 = (\sqrt{I})^2 = I\]
Therefore, the difference \( I_{\text{max}} - I_{\text{min}} \) is:
\[x = I_{\text{max}} - I_{\text{min}} = 9I - I = 8I\]