The width of one of two slits in a young's double slit experiment is four times that of the other. The ratio of maximum to minimum intensity in the interference pattern will be

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answer is D.

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Detailed Solution

If ‘b’ be the width of slit then intensity of light emerging from the slit $I \propto b$ .

$∴$ If $I_{1} = I, then I_{2} = 4 I$

$∴ \frac{I_{\max}}{I_{\min}} = {(\frac{\sqrt{I_{2}} + \sqrt{I_{1}}}{\sqrt{I_{2}} - \sqrt{I}})}^{2} = {(\frac{\sqrt{4} + 1}{\sqrt{4} - 1})}^{2} = {(\frac{3}{1})}^{2} = \frac{9}{1}$

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