Question

Two coherent monochromatic light beams of intensities \(I\) and \(4I\) are superposed. What are the maximum and minimum possible intensities in the resulting beam?

• A. \(I_{\max}=2I,\ I_{\min}=I\)
• B. \(I_{\max}=4I,\ I_{\min}=I\)
• C. \(I_{\max}=9I,\ I_{\min}=3I\)
• D. \(I_{\max}=9I,\ I_{\min}=I\)

Hint:

Remember: \[ I_{\max} = (\sqrt{I_1}+\sqrt{I_2})^2 \] \[ I_{\min} = (\sqrt{I_1}-\sqrt{I_2})^2 \] for two coherent interfering sources.

Answer 1

The Correct Option is D

Step 1: Think in terms of amplitudes, not intensities.
Intensity is proportional to the square of amplitude, so $a \propto \sqrt{I}$. This lets us add the two waves as amplitudes first and square only at the end.
Step 2: Write the two amplitudes.
For $I_1 = I$ we get $a_1 = \sqrt{I}$, and for $I_2 = 4I$ we get $a_2 = \sqrt{4I} = 2\sqrt{I}$.
Step 3: Maximum case (waves in step).
When the crests line up, amplitudes add: $a_{\max} = a_1 + a_2 = \sqrt{I} + 2\sqrt{I} = 3\sqrt{I}$.
Step 4: Square it for maximum intensity.
\[ I_{\max} = (3\sqrt{I})^2 = 9I \]
Step 5: Minimum case (waves opposite).
When one crest meets the other trough, amplitudes subtract: $a_{\min} = a_2 - a_1 = 2\sqrt{I} - \sqrt{I} = \sqrt{I}$.
Step 6: Square it for minimum intensity.
\[ I_{\min} = (\sqrt{I})^2 = I \]
So the brightest fringe carries $9I$ and the darkest carries $I$.
\[ \boxed{I_{\max}=9I,\quad I_{\min}=I} \]