Q. In Young’s double slit experiment, one of the slit is wider than other, so that amplitude of the light from one slit is double of that from other slit. If $I_{m}$ be the maximum intensity, the resultant intensity I when they interfere at phase difference $\phi$ is given by
A) $\frac{I_{m}}{9}(4 + 5\cos{\phi)}$
B) $\frac{I_{m}}{3}\left( 1 + 2\cos^{2}\frac{\phi}{2} \right)$
C) $\frac{I_{m}}{5}\left( 1 + 4\cos^{2}\frac{\phi}{2} \right)$
D) $\frac{I_{m}}{9}\left( 1 + 8\cos^{2}\frac{\phi}{2} \right)$
Explanation
(d) Let $A_{1} = A_{0}$, Then $A_{2} = 2A_{0}$
Intensity $I \propto A^{2}$
Hence $I_{1} = I_{0},I_{2} = 4I_{0}$
We have $I = I_{0} + 4I_{0} + 2\sqrt{I_{0} \times 4I_{0}}\cos\phi$
For $I_{\max},\cos\phi = 1$
Hence $I_{m} = 9I_{0}$ or $I_{0} = \frac{I_{m}}{9}$
When phase difference is $\phi$ then
$I = I_{0} + 4I_{0} + 2\sqrt{4I_{0}^{2}}\cos\phi$
$= I_{0} + 4I_{0}(1 + \cos\phi)$
$= I_{0}\left( 1 + 8\cos^{2}\frac{\phi}{2} \right)\ \ \left\lbrack \because 1 + \cos{\phi = 2\cos^{2}\frac{\phi}{2}} \right\rbrack$
$= \frac{I_{m}}{9}(1 + 8\cos^{2}{\phi/2})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.