Question:
In Young’s double slit experiment, the ratio of intensities of dark and bright fringes is 16:36. The ratio of amplitudes of the two light waves is:
In Young’s double slit experiment, the ratio of intensities of dark and bright fringes is 16:36. The ratio of amplitudes of the two light waves is:
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In YDSE, amplitude ratio is square root of intensity ratio.
Updated On: Jun 20, 2026
- \(3:2\)
- \(36:16\)
- \(5:1\)
- \(2:1\)
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The Correct Option is C
Solution and Explanation
Step 1: Intensity relation in YDSE.
Resultant intensity: \[ I = I_1 + I_2 \pm 2\sqrt{I_1 I_2} \]
Step 2: Bright and dark fringe intensities.
\[ I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2 \] \[ I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2 \] Given: \[ I_{\min} : I_{\max} = 16 : 36 \]
Step 3: Take square roots.
\[ \frac{(\sqrt{I_1} - \sqrt{I_2})}{(\sqrt{I_1} + \sqrt{I_2})} = \frac{4}{6} = \frac{2}{3} \]
Step 4: Solve ratio.
Let: \[ \sqrt{I_1} = a, \quad \sqrt{I_2} = b \] \[ \frac{a - b}{a + b} = \frac{2}{3} \]
Step 5: Cross multiply.
\[ 3(a - b) = 2(a + b) \] \[ 3a - 3b = 2a + 2b \] \[ a = 5b \]
Step 6: Final amplitude ratio.
Since amplitude \(\propto \sqrt{I}\): \[ \boxed{5:1} \]
Resultant intensity: \[ I = I_1 + I_2 \pm 2\sqrt{I_1 I_2} \]
Step 2: Bright and dark fringe intensities.
\[ I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2 \] \[ I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2 \] Given: \[ I_{\min} : I_{\max} = 16 : 36 \]
Step 3: Take square roots.
\[ \frac{(\sqrt{I_1} - \sqrt{I_2})}{(\sqrt{I_1} + \sqrt{I_2})} = \frac{4}{6} = \frac{2}{3} \]
Step 4: Solve ratio.
Let: \[ \sqrt{I_1} = a, \quad \sqrt{I_2} = b \] \[ \frac{a - b}{a + b} = \frac{2}{3} \]
Step 5: Cross multiply.
\[ 3(a - b) = 2(a + b) \] \[ 3a - 3b = 2a + 2b \] \[ a = 5b \]
Step 6: Final amplitude ratio.
Since amplitude \(\propto \sqrt{I}\): \[ \boxed{5:1} \]
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