MHT CET · Physics · Wave Optics
The intensity ratio of the maxima and minima in an interference pattern produced by two coherent sources of light is \(9: 1\). The intensities of the light sources used are in the ratio
- A \(3: 1\)
- B \(4: 1\)
- C \(9: 1\)
- D \(10: 1\)
Answer & Solution
Correct Answer
(B) \(4: 1\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll}
& \frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2}{\left(\sqrt{I_1}-\sqrt{I_2}\right)^2} \\
\therefore & \frac{9}{1}=\frac{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2}{\left(\sqrt{I_1}-\sqrt{I_2}\right)^2} \\
\therefore & \frac{3}{1}=\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}
\end{array}\)
...(given)
\(\begin{array}{ll}\therefore & 3\left(\sqrt{I_1}-\sqrt{I_2}\right)=1\left(\sqrt{I_1}+\sqrt{I_2}\right) \\ \therefore & 3 \sqrt{I_1}-3 \sqrt{I_2}=\sqrt{I_1}+\sqrt{I_2} \\ \therefore & 2 \sqrt{I_1}=4 \sqrt{I_2} \\ \therefore & \sqrt{\frac{I_1}{I_2}}=\frac{2}{1} \\ & \frac{I_1}{I_2}=\frac{4}{1}\end{array}\)
& \frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2}{\left(\sqrt{I_1}-\sqrt{I_2}\right)^2} \\
\therefore & \frac{9}{1}=\frac{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2}{\left(\sqrt{I_1}-\sqrt{I_2}\right)^2} \\
\therefore & \frac{3}{1}=\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}
\end{array}\)
...(given)
\(\begin{array}{ll}\therefore & 3\left(\sqrt{I_1}-\sqrt{I_2}\right)=1\left(\sqrt{I_1}+\sqrt{I_2}\right) \\ \therefore & 3 \sqrt{I_1}-3 \sqrt{I_2}=\sqrt{I_1}+\sqrt{I_2} \\ \therefore & 2 \sqrt{I_1}=4 \sqrt{I_2} \\ \therefore & \sqrt{\frac{I_1}{I_2}}=\frac{2}{1} \\ & \frac{I_1}{I_2}=\frac{4}{1}\end{array}\)
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