Width of one of the two slits in Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is: [2025]

Question

Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is: [2025]

Smart Achievers · Accepted Answer

(2)

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

$∴ I_{1} = I_{0}, I_{2} = 2 I_{0}$ $I_{\min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$

$\frac{I_{\max}}{I_{\min}} = \frac{{(\sqrt{2} + 1)}^{2}}{{(\sqrt{2} - 1)}^{2}} \Rightarrow \frac{3 + 2 \sqrt{2}}{3 - 2 \sqrt{2}}$

Solution

Q.
Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is: [2025]

1 $(2 \sqrt{2} + 1) : (2 \sqrt{2} - 1)$

2 $(3 + 2 \sqrt{2}) : (3 - 2 \sqrt{2})$

3 9 : 1

4 3 : 1

Ans.
(2)

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

$∴ I_{1} = I_{0}, I_{2} = 2 I_{0}$ $I_{\min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$

$\frac{I_{\max}}{I_{\min}} = \frac{{(\sqrt{2} + 1)}^{2}}{{(\sqrt{2} - 1)}^{2}} \Rightarrow \frac{3 + 2 \sqrt{2}}{3 - 2 \sqrt{2}}$

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Solution

Q. Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is: [2025]

1 (22+1) : (22–1)

2 (3+22) : (3–22)