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

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Solution

Q.
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 [2024]

1 1 : 1

2 16 : 1

3 9 : 1

4 4 : 1

Ans.
(3)

$Since, Intensity \propto width of slit (W)$

So, $I_{1} = I_{0}, I_{2} = 4 I_{0}$

$\frac{I_{\max}}{I_{\min}} = \frac{{(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}}{{(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}} = \frac{9 I_{0}}{I_{0}} = 9$

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