Question

Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is :

• A. \( (2\sqrt{2} + 1) : (2\sqrt{2} - 1) \)
• B. \( (3 + 2\sqrt{2}) : (3 - 2\sqrt{2}) \)
• C. \( 9 : 1 \)
• D. \( 3 : 1 \)

Hint:

In Young's double slit experiment, the intensity of light is proportional to the width of the slit. The maximum intensity is \( (\sqrt{I_1} + \sqrt{I_2})^2 \) and the minimum intensity is \( (\sqrt{I_1} - \sqrt{I_2})^2 \), where \( I_1 \) and \( I_2 \) are the intensities from the two slits. Use the given relationship between the widths to find the ratio of intensities and then calculate the ratio of maximum to minimum intensity.

Answer 1

The Correct Option is B

In Young's double-slit experiment, light intensity through a slit is directly proportional to its width. Let the slit widths be \( w_1 \) and \( w_2 \). If one slit's width is half the other's, let \( w_1 = w \) and \( w_2 = 2w \). The intensities, \( I_1 \) and \( I_2 \), are proportional to these widths: \( I_1 \propto w_1 = w \implies I_1 = I_0 \) and \( I_2 \propto w_2 = 2w \implies I_2 = 2I_0 \). The maximum intensity \( I_{max} \) in the interference pattern, resulting from constructive interference, is \( I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2 \). Substituting the intensity values yields: \( I_{max} = (\sqrt{I_0} + \sqrt{2I_0})^2 = (\sqrt{I_0} (1 + \sqrt{2}))^2 = I_0 (1 + \sqrt{2})^2 = I_0 (1 + 2 + 2\sqrt{2}) = I_0 (3 + 2\sqrt{2}) \). The minimum intensity \( I_{min} \), from destructive interference, is \( I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2 \). Substituting the intensity values yields: \( I_{min} = (\sqrt{I_0} - \sqrt{2I_0})^2 = (\sqrt{I_0} (1 - \sqrt{2}))^2 = I_0 (1 - \sqrt{2})^2 = I_0 (1 + 2 - 2\sqrt{2}) = I_0 (3 - 2\sqrt{2}) \). The ratio of maximum to minimum intensity is \( \frac{I_{max}}{I_{min}} = \frac{I_0 (3 + 2\sqrt{2})}{I_0 (3 - 2\sqrt{2})} = \frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}} \). Therefore, the ratio \( I_{max} : I_{min} \) is \( (3 + 2\sqrt{2}) : (3 - 2\sqrt{2}) \).