Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Statement I: In Young's experiment, for two coherent sources, the resultant intensity is given by $I=4 I_{0} \cos ^{2} \frac{\phi}{2}$.
Statement II: Ratio of maximum to minimum intensity
is $\frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}}{\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2}}$

Wave Optics

Solution:

For two coherent sources,
$I=I_{1}+I_{2}+\sqrt{I_{1} I_{2}} \cos \theta$
Putting, $I_{1}=I_{2}=I_{0}$
We have $I=I_{0}+I_{0}+2 \sqrt{I_{0} \times I_{0}} \cos \phi$
Simplifying the above expression
$I=2 I_{0}(1+\cos \phi)$
$=2 I_{0}\left(1+2 \cos ^{2} \frac{\phi}{2}-1\right)$
$=2 I_{0} \times 2 \cos ^{2} \frac{\phi}{2}=4 I_{0} \cos ^{2} \frac{\phi}{2}$
Also, $I_{\max }=\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}$
$I_{\min }=\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)$
$\therefore \frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}}{\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2}}$