Q. In Young's double-slit experiment, the intensity of light at a point on the screen where the path difference is $\lambda $ is $I,\lambda $ being the wavelength of light used. The intensity at a point where the path difference is $\frac{\lambda }{4}$ will be
NTA AbhyasNTA Abhyas 2022
Solution:
For path difference $x$ , phase difference $\phi$ is given by,
$\phi=\frac{2 \pi }{\lambda }x$ .
For, $x=\lambda $ , $\phi=\frac{2 \pi }{\lambda }\times \lambda =2\pi $ .
The net intensity is $I$ ,
$I_{1}+I_{2}+2\sqrt{I_{1} I_{2}}cos\phi$
If $I_{1}=I_{2}=I_{0}$ then
$I=I_{0}+I_{0}+2I_{0}cos2\pi $
$\Rightarrow I=4I_{0} \left(\right. \because cos 2 \pi = 1 \left.\right)$ .
For $x = \frac{\lambda }{4}$ , phase difference $\phi=\frac{\pi }{2}$ .
The intensity is $I'$ ,
$\therefore I^{'}=I_{1}+I_{2}+2\sqrt{I_{1} I_{2}}cos\frac{\pi }{2}$
If $I_{1}=I_{2}=I_{0}\text{then}I^{'}=2I_{0}=2.\frac{I}{4}=\frac{I}{2}$ .
$\phi=\frac{2 \pi }{\lambda }x$ .
For, $x=\lambda $ , $\phi=\frac{2 \pi }{\lambda }\times \lambda =2\pi $ .
The net intensity is $I$ ,
$I_{1}+I_{2}+2\sqrt{I_{1} I_{2}}cos\phi$
If $I_{1}=I_{2}=I_{0}$ then
$I=I_{0}+I_{0}+2I_{0}cos2\pi $
$\Rightarrow I=4I_{0} \left(\right. \because cos 2 \pi = 1 \left.\right)$ .
The intensity is $I'$ ,
$\therefore I^{'}=I_{1}+I_{2}+2\sqrt{I_{1} I_{2}}cos\frac{\pi }{2}$
If $I_{1}=I_{2}=I_{0}\text{then}I^{'}=2I_{0}=2.\frac{I}{4}=\frac{I}{2}$ .