• # question_answer In the Young's double slit experiment, the intensity on the screen at a point where path difference is$\lambda$ is K. What will be the intensity at the point where path difference is$(\lambda /4)?$ A) $\frac{K}{4}$           B)   $\frac{K}{2}$                       C)    K                    D)    0

$I=2{{I}_{0}}(1+cos\delta ).$When path difference $=\lambda ,$then phase difference $\delta =\frac{2\pi }{\lambda }\times$path difference $=2\pi {{I}_{1}}$ $=2{{I}_{0}}(1+\cos 2\pi )=4{{I}_{0}}=K$ When the difference $=\lambda /4.$Then phase difference $\delta =\frac{2\pi }{\lambda }=\frac{\lambda }{4}=\frac{\pi }{2}$ $\therefore$${{I}_{2}}=2{{I}_{0}}\left\{ 1+\cos \frac{\pi }{2} \right\}=2{{I}_{0}}=\frac{k}{2}$ Hence, the correction option is (b).