Question

In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is $\lambda$ is K unit; $\lambda$ being the wavelength of light used. The intensity at a point where the path difference is $\frac{\lambda }{4}$ will be:

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

Answer 1

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

$K_1=I+I+2Icos\delta$ or $K_1=2I,K_1=\frac{K}{2}$
For path difference $\frac{\lambda }{4}$, phase difference, $\delta =\frac{2\pi }{\lambda }.\frac{\lambda }{4}=\frac{\pi }{2}$