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 minimum intensity in the interference pattern is:

Question

Smart Achievers · Accepted Answer

(1)

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

$I_{\min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$

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

$I_{\max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} = 9 I$

$\frac{I_{\max}}{I_{\min}} = \frac{9 I}{I} = \frac{9}{1}$

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 minimum intensity in the interference pattern is:

1 9 : 1

2 16 : 1

3 1 : 1

4 4 : 1

Ans.
(1)

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

$I_{\min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$

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

$I_{\max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} = 9 I$

$\frac{I_{\max}}{I_{\min}} = \frac{9 I}{I} = \frac{9}{1}$

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