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(B) The intensity $I$ of light is directly proportional to the width $W$ of the slit,and also to the square of the amplitude $A$ of the wave.$	herefo

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$\frac{9}{4}$

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(B) The intensity $I$ of light is directly proportional to the width $W$ of the slit,and also to the square of the amplitude $A$ of the wave.$	herefore \frac{I_1}{I_2} = \frac{W_1}{W_2} = \frac{A_1^2}{A_2^2}$Given the ratio of widths $\frac{W_1}{W_2} = \frac{1}{25}$,we have:$\frac{A_1^2}{A_2^2} = \frac{1}{25} \implies \frac{A_1}{A_2} = \sqrt{\frac{1}{25}} = \frac{1}{5}$The ratio of maximum intensity to minimum intensity is given by:$\frac{I_{max}}{I_{min}} = \frac{(A_1 + A_2)^2}{(A_1 - A_2)^2} = \left( \frac{\frac{A_1}{A_2} + 1}{\frac{A_1}{A_2} - 1} ight)^2$Substituting $\frac{A_1}{A_2} = \frac{1}{5}$:$\frac{I_{max}}{I_{min}} = \left( \frac{\frac{1}{5} + 1}{\frac{1}{5} - 1} ight)^2 = \left( \frac{\frac{6}{5}}{-\frac{4}{5}} ight)^2 = \left( -\frac{6}{4} ight)^2 = \left( -\frac{3}{2} ight)^2 = \frac{9}{4}$

Two slits in Young's experiment have widths in the ratio $1 : 25$. The ratio of intensity at the maxima and minima in the interference pattern,$\frac{I_{max}}{I_{min}}$ is

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