Question:
In Young’s double slit experiment, the ratio of intensities of maxima and minima is \(25:9\). The ratio of intensities of two slits is:
In Young’s double slit experiment, the ratio of intensities of maxima and minima is \(25:9\). The ratio of intensities of two slits is:
Show Hint
Remember:
\[
\frac{I_{\max}}{I_{\min}}
=
\left(
\frac{\sqrt{I_1}+\sqrt{I_2}}
{\sqrt{I_1}-\sqrt{I_2}}
\right)^2
\]
Take square root first to simplify calculations quickly.
Updated On: Jun 17, 2026
- \(18:3\)
- \(4:1\)
- \(8:1\)
- \(16:1\)
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The Correct Option is D
Solution and Explanation
Concept:
In YDSE:
\[
I_{\max}=(\sqrt{I_1}+\sqrt{I_2})^2
\]
\[
I_{\min}=(\sqrt{I_1}-\sqrt{I_2})^2
\]
Given:
\[
\frac{I_{\max}}{I_{\min}}=\frac{25}{9}
\]
Step 1: Take square root. \[ \frac{\sqrt{I_1}+\sqrt{I_2}} {\sqrt{I_1}-\sqrt{I_2}} =\frac{5}{3} \] Let: \[ \sqrt{I_1}=a,\qquad \sqrt{I_2}=b \] Then: \[ \frac{a+b}{a-b}=\frac{5}{3} \] Cross multiplying: \[ 3a+3b=5a-5b \] \[ 8b=2a \] \[ a=4b \] Squaring: \[ I_1:I_2=16:1 \] \[ \boxed{16:1} \]
Step 1: Take square root. \[ \frac{\sqrt{I_1}+\sqrt{I_2}} {\sqrt{I_1}-\sqrt{I_2}} =\frac{5}{3} \] Let: \[ \sqrt{I_1}=a,\qquad \sqrt{I_2}=b \] Then: \[ \frac{a+b}{a-b}=\frac{5}{3} \] Cross multiplying: \[ 3a+3b=5a-5b \] \[ 8b=2a \] \[ a=4b \] Squaring: \[ I_1:I_2=16:1 \] \[ \boxed{16:1} \]
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