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Wave Optics — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsWave Optics1 Mark
MCQ
Two coherent sources of intensity ratio $\alpha$ interfere.In interference pattern $\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}=$
  • A
    $\frac{2 \alpha}{1+\alpha}$
  • ✓
    $\frac{2 \sqrt{\alpha}}{1+\alpha}$
  • C
    $\frac{2 \alpha}{1+\sqrt{\alpha}}$
  • D
    $\frac{1+\alpha}{2 \alpha}$
✓

Answer

Correct option: B.
$\frac{2 \sqrt{\alpha}}{1+\alpha}$
(b) : Let two coherent sources of intensity $I_1$ and $I_2$ interfere. Then
$
\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{\left(\sqrt{\frac{I_1}{I_2}}+1\right)^2}{\left(\sqrt{\frac{I_1}{I_2}}-1\right)^2}
$
As $\frac{I_1}{I_2}=\alpha$ (Given)
$
\therefore \frac{I_{\max }}{I_{\min }}=\frac{(\sqrt{\alpha}+1)^2}{(\sqrt{\alpha}-1)^2}=\left(\frac{\sqrt{\alpha}+1}{\sqrt{\alpha}-1}\right)^2
$
$
\begin{aligned}
& \text { Then, } \frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}=\frac{\frac{I_{\max }}{I_{\min }}-1}{\frac{I_{\max }}{I_{\min }}+1} \\
& =\frac{\left(\frac{\sqrt{\alpha}+1}{\sqrt{\alpha}-1}\right)^2-1}{\left(\frac{\sqrt{\alpha}+1}{\sqrt{\alpha}-1}\right)^2+1}=\frac{(\sqrt{\alpha}+1)^2-(\sqrt{\alpha}-1)^2}{(\sqrt{\alpha}+1)^2+(\sqrt{\alpha}-1)^2} \\
& =\frac{\alpha+1+2 \sqrt{\alpha}-\alpha-1+2 \sqrt{\alpha}}{\alpha+1+2 \sqrt{\alpha}+\alpha+1-2 \sqrt{\alpha}} \Rightarrow=\frac{4 \sqrt{\alpha}}{2+2 \alpha}=\frac{4 \sqrt{\alpha}}{2(1+\alpha)}=\frac{2 \sqrt{\alpha}}{1+\alpha}
\end{aligned}
$

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