Question

Two monochromatic light beams have intensities in the ratio 1:9. An interference pattern is obtained by these beams. The ratio of the intensities of maximum to minimum is

• A. 8 : 1
• B. 9 : 1
• C. 3 : 1
• D. 4 : 1

Hint:

In an interference pattern formed by two sources of intensities \( I_1 \) and \( I_2 \), the maximum intensity is \( (\sqrt{I_1} + \sqrt{I_2})^2 \) and the minimum intensity is \( (\sqrt{I_1} - \sqrt{I_2})^2 \). Use the given ratio of intensities to express one intensity in terms of the other and then calculate the ratio of maximum to minimum intensity.

Answer 1

The Correct Option is D

Let the intensities of the two monochromatic light beams be \( I_1 \) and \( I_2 \). 
Given that the ratio of their intensities is 1:9, we can write \( \frac{I_1}{I_2} = \frac{1}{9} \). 
Let \( I_1 = I \) and \( I_2 = 9I \). The maximum intensity \( I_{max} \) in the interference pattern is given by: \[ I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2 \] Substituting the values of \( I_1 \) and \( I_2 \): \[ I_{max} = (\sqrt{I} + \sqrt{9I})^2 = (\sqrt{I} + 3\sqrt{I})^2 = (4\sqrt{I})^2 = 16I \] The minimum intensity \( I_{min} \) in the interference pattern is given by: \[ I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2 \] Substituting the values of \( I_1 \) and \( I_2 \): \[ I_{min} = (\sqrt{I} - \sqrt{9I})^2 = (\sqrt{I} - 3\sqrt{I})^2 = (-2\sqrt{I})^2 = 4I \] The ratio of the maximum to the minimum intensity is: \[ \frac{I_{max}}{I_{min}} = \frac{16I}{4I} = 4 \] So, the ratio of the intensities of maximum to minimum is 4:1.