Question:
Two coherent light sources having intensity in the ratio $2 x$ produce an interference pattern.
The ratio $\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}$ will be :
Correct Option: , 4
Solution:
Given that, $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=2 \mathrm{x}$
We know,
$I_{\max }=\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2} \& I_{\min }=\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2}$
$\therefore \frac{\mathrm{I}_{\max }-\mathrm{I}_{\min }}{\mathrm{I}_{\max }+\mathrm{I}_{\min }}=\frac{2 \sqrt{\mathrm{I}_{1} \mathrm{I}_{2}}}{\mathrm{I}_{1}+\mathrm{I}_{2}}=\frac{2 \sqrt{\mathrm{I}_{1} / \mathrm{I}_{2}}}{\frac{\mathrm{I}_{\mathrm{J}}}{\mathrm{I}_{2}}+1}=\frac{2 \sqrt{2 \mathrm{x}}}{2 \mathrm{x}+1}$