Question

In the Young's double slit experiment the intensity of light at a point on the screen where the path difference is $\lambda$ is K, ($\lambda$ being the wave length of light used). The intensity at a point where the path difference is $\lambda /4$, will be

• A. K
• B. K/4
• C. K/2
• D. zero

Answer 1

Answer: (C)

K/2

$I_{max}4I_0=k$
at the other point, path difference $=\frac{\lambda }{4}$
so phase difference $\theta =k\Delta x=\frac{2\pi }{\lambda }\times \frac{\lambda }{4}=\frac{\pi }{2}$
$I^1=I_0+I_0+2{\sqrt{I}}_0{\sqrt{I}}_{0}cos\frac{\pi }{2}$
$I^1=2I_0=\frac{K}{2}$