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Q.

In a Young’s double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit – width. Find the ratio of the maximum to the minimum intensity in the interference pattern

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a

1 : 4

b

3 : 1

c

4 : 1

d

2 : 1

answer is C.

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Detailed Solution

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Amplitude $\propto$ Width of slit
$\Rightarrow A_{2} = 3 A_{1}$
$\frac{I_{max}}{I_{min}} = {(\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{| \sqrt{I_{1}} - \sqrt{I_{2}} |})}^{2}$
$∴$ Intensity $I \propto A^{2}$
$\Rightarrow \frac{I_{max}}{I_{min}} = {(\frac{A_{1} + A_{2}}{| A_{1} - A_{2} |})}^{2} = (\frac{A_{1} + 3 A_{1}}{| A_{1} - 3 A_{1} |}) = {(\frac{{4A}_{1}}{{2A}_{1}})}^{2} = 4 : 1$

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