Q. The width of one slit is double the another slit in Young's double slit experiment, then the ratio of maximum intensities to minimum intensities in the interference pattern is

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A

$9:1$

B

$1:9$

C

$3:1$

D

$1:3$

Solution:

Given, $a_{1}=2a\Rightarrow a_{2}=a$
$a_{max}=a_{1}+a_{2}=3a$
$a_{min}=a_{1}-a_{2}=a$
$\therefore $ $\frac{I_{max}}{I_{min}}=\frac{a_{max}^{2}}{a_{min}^{2}}=\frac{\left(\right. 3 a \left(\left.\right)^{2}}{\left(\right. a \left(\left.\right)^{2}}=\frac{9}{1}$
$I_{max}:I_{min}=9:1$

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