In Young's double-slit experiment,if the rati | vedclass

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(C) The intensity of light $I$ is directly proportional to the width of the slit $W$,so $\frac{I_1}{I_2} = \frac{W_1}{W_2} = \frac{4}{9}$.Let $I_1 = 4

Vedclass · Accepted Answer

$25:1$

Vedclass · Answer

(C) The intensity of light $I$ is directly proportional to the width of the slit $W$,so $\frac{I_1}{I_2} = \frac{W_1}{W_2} = \frac{4}{9}$.Let $I_1 = 4k$ and $I_2 = 9k$. The amplitudes are proportional to the square root of intensity,so $A_1 = \sqrt{4k} = 2\sqrt{k}$ and $A_2 = \sqrt{9k} = 3\sqrt{k}$.The ratio of maximum intensity to minimum intensity is given by the formula $\frac{I_{max}}{I_{min}} = \frac{(A_1 + A_2)^2}{(A_1 - A_2)^2}$.Substituting the values: $\frac{I_{max}}{I_{min}} = \frac{(2\sqrt{k} + 3\sqrt{k})^2}{(2\sqrt{k} - 3\sqrt{k})^2} = \frac{(5\sqrt{k})^2}{(-\sqrt{k})^2} = \frac{25k}{k} = \frac{25}{1}$.Thus,the ratio is $25:1$.

In Young's double-slit experiment,if the ratio of the widths of the two slits is $4:9$,then the ratio of the maximum to minimum intensity will be:

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