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Two coherent monochromatic beams of intensities $I$ and $4 I$ respectively are superposed. The maximum and minimum intensitien in the reaulting pattern are
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Verified Answer
The correct answer is:
9I and I
We know that The maximum intensities
$$
I_{\max }=\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}
$$
The minimum intensities
$$
I_{\min }=\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2}
$$
So, the ratio of the maximum and minimum of intensities is
$$
\begin{aligned}
\frac{I_{\max }}{I_{\min }} &=\frac{\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}}{\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2}} \\
\frac{I_{\max }}{I_{\min }} &=\frac{(\sqrt{4 I}+\sqrt{I})^{2}}{(\sqrt{4 I}-\sqrt{I})^{2}} \\
&=\left(\frac{3 \sqrt{I}}{\sqrt{I}}\right)^{2} \\
=& \frac{9}{1}=9: 1
\end{aligned}
$$
$$
I_{\max }=\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}
$$
The minimum intensities
$$
I_{\min }=\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2}
$$
So, the ratio of the maximum and minimum of intensities is
$$
\begin{aligned}
\frac{I_{\max }}{I_{\min }} &=\frac{\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}}{\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2}} \\
\frac{I_{\max }}{I_{\min }} &=\frac{(\sqrt{4 I}+\sqrt{I})^{2}}{(\sqrt{4 I}-\sqrt{I})^{2}} \\
&=\left(\frac{3 \sqrt{I}}{\sqrt{I}}\right)^{2} \\
=& \frac{9}{1}=9: 1
\end{aligned}
$$
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