Question

In Young's double slit experiment, the ratio of maximum and minimum intensities in the fringe system is 9:1. The ratio of amplitudes of coherent sources is

• A. 9:1
• B. 3:1
• C. 2:1
• D. 1:1

Hint:

In Young's double slit experiment, the intensity ratio between maximum and minimum fringes is related to the square of the amplitude ratio of the two sources. Use the formula \( \frac{I_{\text{max}}}{I_{\text{min}}} = \left( \frac{A_1 + A_2}{A_1 - A_2} \right)^2 \) to solve for the amplitude ratio.

Answer 1

The Correct Option is C

In Young's double slit experiment, the intensity of the interference fringes depends on the amplitudes of the coherent sources. 
The maximum intensity \( I_{\text{max}} \) and the minimum intensity \( I_{\text{min}} \) are related to the amplitudes \( A_1 \) and \( A_2 \) of the two waves as follows: \[ I_{\text{max}} = (A_1 + A_2)^2 \] \[ I_{\text{min}} = (A_1 - A_2)^2 \] The ratio of maximum and minimum intensities is given by: \[ \frac{I_{\text{max}}}{I_{\text{min}}} = \frac{(A_1 + A_2)^2}{(A_1 - A_2)^2} \] Given that the ratio of maximum and minimum intensities is 9:1, we have: \[ \frac{(A_1 + A_2)^2}{(A_1 - A_2)^2} = 9 \] Taking the square root of both sides: \[ \frac{A_1 + A_2}{A_1 - A_2} = 3 \] Solving for the ratio of the amplitudes: \[ A_1 + A_2 = 3(A_1 - A_2) \] Expanding and simplifying: \[ A_1 + A_2 = 3A_1 - 3A_2 \] \[ A_1 + A_2 - 3A_1 + 3A_2 = 0 \] \[ -2A_1 + 4A_2 = 0 \] \[ A_1 = 2A_2 \] 
Thus, the ratio of the amplitudes is \( A_1 : A_2 = 2:1 \). Therefore, the correct answer is: \[ \text{(C) } 2:1 \]