Question

In Young's double slit experiment with monochromatic light, the maximum intensity is 4 times the minimum intensity in the interference pattern. What is the ratio of the intensities of the two interfering waves?

• A. 1/9
• B. 1/3
• C. 1/16
• D. 1/2

Answer 1

The Correct Option is A

Given:
In Young's double slit experiment, the maximum intensity \( I_{\text{max}} = 4 \times I_{\text{min}} \)

Let the individual intensities of the two waves be:
\( I_1 \) and \( I_2 \) 

Then, we use the formulas:
\[ I_{\text{max}} = (\sqrt{I_1} + \sqrt{I_2})^2 \] \[ I_{\text{min}} = (\sqrt{I_1} - \sqrt{I_2})^2 \] 

Given: \[ \frac{I_{\text{max}}}{I_{\text{min}}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2} = 4 \] 

Taking square root on both sides: \[ \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} = 2 \] 

Cross-multiplying: \[ \sqrt{I_1} + \sqrt{I_2} = 2(\sqrt{I_1} - \sqrt{I_2}) \] \[ \sqrt{I_1} + \sqrt{I_2} = 2\sqrt{I_1} - 2\sqrt{I_2} \] \[ 3\sqrt{I_2} = \sqrt{I_1} \] 

Squaring both sides: \[ I_1 = 9 I_2 \] 

So, the ratio of intensities: \[ \frac{I_2}{I_1} = \frac{1}{9} \] Answer: 1/9