The width of one of the two slits in a Young' | vedclass

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(A) The intensity $I$ of light passing through a slit is directly proportional to the width $w$ of the slit,i.e.,$I \propto w$.Given that the width of

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$9$

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(A) The intensity $I$ of light passing through a slit is directly proportional to the width $w$ of the slit,i.e.,$I \propto w$.Given that the width of one slit is $4$ times the other,let the intensities be $I_1 = I$ and $I_2 = 4I$.The maximum intensity is given by $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$.Substituting the values: $I_{\max} = (\sqrt{I} + \sqrt{4I})^2 = (\sqrt{I} + 2\sqrt{I})^2 = (3\sqrt{I})^2 = 9I$.The minimum intensity is given by $I_{\min} = (\sqrt{I_2} - \sqrt{I_1})^2$.Substituting the values: $I_{\min} = (\sqrt{4I} - \sqrt{I})^2 = (2\sqrt{I} - \sqrt{I})^2 = (\sqrt{I})^2 = I$.The ratio of maximum to minimum intensity is $\frac{I_{\max}}{I_{\min}} = \frac{9I}{I} = \frac{9}{1}$.

The width of one of the two slits in a Young's double slit experiment is $4$ times that of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is: (in $: 1$)

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