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 $\frac{\lambda }{4}$ will be

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

Answer 1

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

For path difference $\lambda$, phase difference $=\frac{2}{\pi }\lambda \times \Delta x=\frac{2\pi }{\lambda }\times \lambda =2\pi$
Intensity is
$I=4I_0{\cos }^2\left(\frac{\theta }{2}\right)=4I_0{\cos }^2\left(\frac{2\pi }{4}\right)=4I_0=K$


For path difference $\frac{\lambda }{4},$ phase difference $=\frac{2\pi }{\lambda }\times \Delta x=\frac{2\pi }{\lambda }\times \frac{\lambda }{4}=\frac{\pi }{2}$
Intensity is
$I=4I_0{\cos }^2\left(\frac{\pi }{4}\right)=2I_0=\frac{K}{2}$