Question

If the intensities of two interfering light waves in a medium are I and 4I, then the maximum and minimum intensities are respectively:

• A. 3I, I
• B. 4I, I
• C. 9I, I
• D. 16I, 0

Hint:

For interference, use $I_{{max}} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{{min}} = (\sqrt{I_1} - \sqrt{I_2})^2$ to find the resultant intensities.

Answer 1

The Correct Option is C

For two interfering waves with intensities $I_1$ and $I_2$, 
the maximum intensity is $I_{{max}} = (\sqrt{I_1} + \sqrt{I_2})^2$ 
and the minimum intensity is $I_{{min}} = (\sqrt{I_1} - \sqrt{I_2})^2$. 
Given: $I_1 = I$, $I_2 = 4I$. 
So, $\sqrt{I_1} = \sqrt{I}$, $\sqrt{I_2} = \sqrt{4I} = 2\sqrt{I}$. 
Maximum intensity: $I_{{max}} = (\sqrt{I} + 2\sqrt{I})^2 = (3\sqrt{I})^2 = 9I$. 
Minimum intensity: $I_{{min}} = (\sqrt{I} - 2\sqrt{I})^2 = (-\sqrt{I})^2 = I$. 
Thus, $I_{{max}}, I_{{min}} = 9I, I$.