The ratio of intensities at two points P and Q on the screen in a Young’s double slit experiment where phase difference between two wave of same amplitude are \(\frac{π}{3}\) and \(\frac{π}{2}\), respectively are
- 3:2
- 2:3
- 1:3
- 3:1
The Correct Option is A
Solution and Explanation
Step 1: Using the formula for intensity in a Young's double slit experiment.
The intensity at any point in Young's double slit experiment is given by: \[ I_{\text{res}} = 4I_0 \cos^2 \left( \frac{\theta}{2} \right) \] Where \( I_0 \) is the maximum intensity, and \( \theta \) is the phase difference. For the first point \( P \), where \( \theta = \frac{\pi}{3} \): \[ I_1 = 4I_0 \cos^2 \left( \frac{\pi}{6} \right) = 4I_0 \left( \frac{\sqrt{3}}{2} \right)^2 = 4I_0 \times \frac{3}{4} = 3I_0 \] For the second point \( Q \), where \( \theta = \frac{\pi}{2} \): \[ I_2 = 4I_0 \cos^2 \left( \frac{\pi}{4} \right) = 4I_0 \left( \frac{1}{\sqrt{2}} \right)^2 = 4I_0 \times \frac{1}{2} = 2I_0 \] Thus, the ratio of intensities is: \[ \frac{I_1}{I_2} = \frac{3I_0}{2I_0} = \frac{3}{2} \]
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