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In an interference experiment, intensity ratio at the bright to dark fringe is $9: 1$. Amplitudes of interfering waves are in the ratio
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Verified Answer
The correct answer is:
$2: 1$
Given, $\frac{I_{\max }}{I_{\min }}=\frac{9}{1}$
As we know,
$\begin{array}{ll}
\Rightarrow & \frac{I_{\mathrm{max}}}{I_{\min }}=\frac{\left(a_{1}+a_{2}\right)^{2}}{\left(a_{1}-a_{2}\right)^{2}}=\frac{9}{1} \\
\Rightarrow & \frac{a_{1}+a_{2}}{a_{1}-a_{2}}=\frac{3}{1} \\
\Rightarrow & a_{1}+a_{2}=3 a_{1}-3 a_{2} \\
\Rightarrow & \frac{a_{1}}{}=2 a_{2} \\
\text { or } & \frac{a_{1}}{a_{2}}=2: 1
\end{array}$
As we know,
$\begin{array}{ll}
\Rightarrow & \frac{I_{\mathrm{max}}}{I_{\min }}=\frac{\left(a_{1}+a_{2}\right)^{2}}{\left(a_{1}-a_{2}\right)^{2}}=\frac{9}{1} \\
\Rightarrow & \frac{a_{1}+a_{2}}{a_{1}-a_{2}}=\frac{3}{1} \\
\Rightarrow & a_{1}+a_{2}=3 a_{1}-3 a_{2} \\
\Rightarrow & \frac{a_{1}}{}=2 a_{2} \\
\text { or } & \frac{a_{1}}{a_{2}}=2: 1
\end{array}$
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