Question

In Young's double slit experiment, the intensity on the screen at a point where path differences is $\lambda$ is $K$. What will be the intensity at the point where path difference is $\frac{\lambda }{4}$ ?

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

Answer 1

The correct option is B $\frac{K}{2}$

The relation between the phase difference$\left(\varphi \right)$ and the path difference$\left(\Delta \text{x}\right)$ is given by,

$\varphi =\left(\frac{2\pi }{\lambda }\right)\Delta \text{x}$

For path difference $\lambda$,

${\varphi }_1=\left(\frac{2\pi }{\lambda }\right)\lambda =2\pi$

For path difference $\frac{\lambda }{4}$,

${\varphi }_2=\left(\frac{2\pi }{\lambda }\right)\frac{\lambda }{4}=\frac{\pi }{2}$

In YDSE, Intensity on screen is given by,

$I=4I_0{\cos }^2(\frac{\varphi }{2})$

For ${\varphi }_1=2\pi$,

$I_1=K=4I_0{\cos }^2(\frac{2\pi }{2})=4I_0$

$\therefore I_0=\frac{K}{4}$

For ${\varphi }_2=\frac{\pi }{2}$,

$I_2=4\times \frac{K}{4}{\cos }^2(\frac{\pi }{2}\times \frac{1}{2})$

$\therefore I_2=\frac{K}{2}$

Hence, option $\left(\text{B}\right)$ is correct.