Question

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

• A. $\frac{I}{4}$
• B. $\frac{I}{2}$
• C. $I$
• D. Zero

Answer 1

Answer: (B)

For path difference $x$ , phase difference $\varphi$ is given by,
$\varphi =\frac{2\pi }{\lambda }x$ .
For, $x=\lambda$ , $\varphi =\frac{2\pi }{\lambda }\times \lambda =2\pi$ .
The net intensity is $I$ ,
$I_1+I_2+2\sqrt{I_1{I}_2}cos\varphi$
If $I_1=I_2=I_0$ then
$I=I_0+I_0+2I_{0}cos2\pi$
$\Rightarrow I=4I_0(\because cos2\pi =1)$ .

For $x=\frac{\lambda }{4}$ , phase difference $\varphi =\frac{\pi }{2}$ .
The intensity is $I^{\prime }$ ,
$\therefore I^{{}^{\prime }}=I_1+I_2+2\sqrt{I_1{I}_2}cos\frac{\pi }{2}$
If $I_1=I_2=I_0\text{then}{I}^{{}^{\prime }}=2I_0=2.\frac{I}{4}=\frac{I}{2}$ .