Question

In Young's double slit experiment, one of the slit is wider than another, 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. $\frac{I_{m}}{9}\left(4 + 5 cos \phi\right)$
• 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. $\frac{I_m}{9}\left(1+8 \cos ^2 \frac{\Phi}{2}\right)$

Answer 1

Answer: (D)

Given, $a_{1}=2a_{2}$
$I_{1}=4I^{′}, \, I_{2}=I′$
We know that, $I_{m}=\left(\sqrt{I_{1}} + \sqrt{I_{2}}\right)^{2}$
$=\left(\sqrt{4 I^{′}} + \sqrt{I^{′}}\right)^{2}=9I′$
$\therefore I=I_{1}+I_{2}+2\sqrt{I_{1} I_{2}}cos \phi$
$=4I^{′}+I^{′}+2\sqrt{4 I^{′} \cdot I ′}cos \phi$
$=4I^{′}+I^{′}+4I^{′}cos \phi$
$=5I^{′}+4I^{′}cos \phi$
$=I^{′}\left(5 + 4 cos \phi\right)$
$=\frac{I_{m}}{9}\left[5 + 4 \left(2 \left(cos\right)^{2} \frac{\phi}{2} - 1\right)\right]$
$=\frac{I_{m}}{9}\left[5 + 8 cos^{2} \frac{\phi}{2} - 4\right]$
$=\frac{I_{m}}{9}\left[1 + 8 cos^{2} \frac{\phi}{2}\right]$