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Two monochromatic light beams of intensity 16 and 9 units are interfering. The ratio of intensities of bright and dark parts of the resultant pattern is:
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1148 Upvotes
Verified Answer
The correct answer is:
$\frac{49}{1}$
$\frac{49}{1}$
Intensity $\propto(\text { amplitude })^2$
$$
\begin{aligned}
&\frac{I_1}{I_2}=\frac{16}{9}=\frac{a_1^2}{a_2^2} \\
&\Rightarrow \mathrm{a}_1=4 ; \mathrm{a}_2=3
\end{aligned}
$$
Therefore the ratio of intensities of bright and dark parts
$$
\frac{I_{\text {Bright }}}{I_{\text {Dark }}}=\frac{\left(a_1+a_2\right)^2}{\left(a_1-a_2\right)^2}=\frac{(4+3)^2}{(4-3)^2}=\frac{49}{1}
$$
$$
\begin{aligned}
&\frac{I_1}{I_2}=\frac{16}{9}=\frac{a_1^2}{a_2^2} \\
&\Rightarrow \mathrm{a}_1=4 ; \mathrm{a}_2=3
\end{aligned}
$$
Therefore the ratio of intensities of bright and dark parts
$$
\frac{I_{\text {Bright }}}{I_{\text {Dark }}}=\frac{\left(a_1+a_2\right)^2}{\left(a_1-a_2\right)^2}=\frac{(4+3)^2}{(4-3)^2}=\frac{49}{1}
$$
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