Question

The wavelength of the light used in Young’s double slit experiment is $\lambda$. The intensity at a point on the screen is I, where the path difference is $\frac{\lambda }{6}$. If $I_0$ denotes the maximum intensity, then the ratio of I and $I_0$ is:

• A. 0.866
• B. 0.5
• C. 0.707
• D. 0.75

Answer 1

The correct option is D

0.75

Phase difference $\varphi =\frac{2\pi }{\lambda }\Delta x$
$\varphi =\frac{2\pi }{\lambda }\frac{\lambda }{6}\varphi =\frac{\pi }{3}I=I_{0}co{s}^2\left[\frac{\varphi }{2}\right]\frac{I}{I_0}=co{s}^2\left[\frac{\pi }{3\times 2}\right]\frac{I}{I_0}=co{s}^2\frac{\pi }{6}\frac{I}{I_0}=\frac{3}{4}=0.75\therefore cos\frac{\pi }{6}=\frac{\sqrt{3}}{2}$