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In an interference pattern the ratio of the intensity of light at the bright fringe to that at the dark fringe is $9: 1$. Then, the ratio of the amplitudes of the two interfering waves
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$2: 1$
Given, $\frac{I_{\max }}{I_{\min }}=9$
Since, intensity $\propto(\text { amplitude })^{2}$
$\therefore \quad \frac{A_{\max }}{A_{\min }}=\frac{a_{1}+a_{2}}{a_{1}-a_{2}}=\sqrt{\frac{9}{1}}=\frac{3}{1}$
$\Rightarrow \quad a_{1}+a_{2}=3 a_{1}-3 a_{2}$
$\Rightarrow \quad 2 a_{1}=4 a_{2}$
$\therefore \quad a_{1}: a_{2}=2: 1$
Since, intensity $\propto(\text { amplitude })^{2}$
$\therefore \quad \frac{A_{\max }}{A_{\min }}=\frac{a_{1}+a_{2}}{a_{1}-a_{2}}=\sqrt{\frac{9}{1}}=\frac{3}{1}$
$\Rightarrow \quad a_{1}+a_{2}=3 a_{1}-3 a_{2}$
$\Rightarrow \quad 2 a_{1}=4 a_{2}$
$\therefore \quad a_{1}: a_{2}=2: 1$
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