Question

In a Young's double slits experiment, the ratio of amplitude of light coming from slits is 2:1. The ratio of the maximum to minimum intensity in the interference pattern is

• A. 2:1
• B. 9:4
• C. 9:1
• D. 25:9

Answer 1

The Correct Option is C

The given ratio of amplitudes is: \[ \frac{A_1}{A_2} = \frac{2}{1}. \] The ratio of maximum to minimum intensity is given by the formula: \[ \frac{I_{\text{max}}}{I_{\text{min}}} = \left( \frac{A_1 + A_2}{A_1 - A_2} \right)^2. \] Substituting the values of \( A_1 \) and \( A_2 \): \[ \frac{I_{\text{max}}}{I_{\text{min}}} = \left( \frac{2 + 1}{2 - 1} \right)^2 = \left( \frac{3}{1} \right)^2 = 9:1. \] Thus, the ratio of maximum to minimum intensity in the interference pattern is \( \boxed{9:1} \).