Question

If two waves of intensities \(I\) and \(4I\) superpose, the ratio between maximum and minimum intensities is

• A. \(9:1\)
• B. \(5:2\)
• C. \(4:3\)
• D. \(3:1\)
• E. \(6:1\)

Hint:

For interference: \[ I_{\max}=(\sqrt{I_1}+\sqrt{I_2})^2,\qquad I_{\min}=(\sqrt{I_1}-\sqrt{I_2})^2 \]

Answer 1

The Correct Option is A

If intensities are \(I_1=I\) and \(I_2=4I\), then amplitudes are proportional to square roots: \[ a_1\propto \sqrt{I},\qquad a_2\propto \sqrt{4I}=2\sqrt{I} \] Maximum intensity: \[ I_{\max}=(a_1+a_2)^2=(\sqrt{I}+2\sqrt{I})^2=(3\sqrt{I})^2=9I \] Minimum intensity: \[ I_{\min}=(a_2-a_1)^2=(2\sqrt{I}-\sqrt{I})^2=(\sqrt{I})^2=I \] Therefore, \[ I_{\max}:I_{\min}=9I:I=9:1 \] Hence, \[ \boxed{(A)\ 9:1} \]