Question

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

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

Answer 1

Answer: (C)

Intensity at any point on the screen is
$I=4I_0{\cos }^2\frac{\varphi }{2}$
Where $I_0$ is the intensity of either wave
and $\varphi$ is the phase difference between two
waves. Phase difference, $\varphi =\frac{2\pi }{\lambda }\times$ path difference When path difference is $\lambda$, then
$\varphi =\frac{2\pi }{\lambda }\times \lambda =2\pi$
$\therefore I=4I_0{\cos }^2\left(\frac{2\pi }{2}\right)$
$=4I_0{\cos }^2(\pi )=4I_0=K\dots .\to (1)$
When path difference is $\frac{\lambda }{4}$ then
$\varphi =\frac{2\pi }{\lambda }\times \frac{\lambda }{4}=\frac{\pi }{2}$
$\therefore I=4I_0{\cos }^2\left(\frac{\pi }{4}\right)$
$=2I_0=\frac{K}{2}$ [ Using (1)]