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(D) Given the intensity ratio $\frac{I_1}{I_2} = 4$,we can write $I_1 = 4I_2$. The ratio of maximum to minimum intensity in an interference pattern is

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) Given the intensity ratio $\frac{I_1}{I_2} = 4$,we can write $I_1 = 4I_2$. The ratio of maximum to minimum intensity in an interference pattern is given by: $\frac{I_{\max}}{I_{\min}} = \left[ \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} \right]^2$ Substituting $I_1 = 4I_2$: $\frac{I_{\max}}{I_{\min}} = \left[ \frac{\sqrt{4I_2} + \sqrt{I_2}}{\sqrt{4I_2} - \sqrt{I_2}} \right]^2 = \left[ \frac{2\sqrt{I_2} + \sqrt{I_2}}{2\sqrt{I_2} - \sqrt{I_2}} \right]^2 = \left( \frac{3}{1} \right)^2 = 9$. Now,we need to calculate $\frac{I_{\max} + I_{\min}}{I_{\max} - I_{\min}}$: $\frac{I_{\max} + I_{\min}}{I_{\max} - I_{\min}} = \frac{9 + 1}{9 - 1} = \frac{10}{8} = \frac{5}{4}$. Comparing this with the given expression $\frac{5}{x}$,we get: $\frac{5}{x} = \frac{5}{4} \implies x = 4$.

The interference pattern is obtained with two coherent light sources of intensity ratio $4:1$. If the ratio $\frac{I_{\max} + I_{\min}}{I_{\max} - I_{\min}}$ is $\frac{5}{x}$,then the value of $x$ will be equal to:

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