Q. 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
Explanation
(d) Phase difference, $\phi = \frac{2\pi}{\lambda} \times path\ difference\ $
$\ \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3} = 60{^\circ}$
Intensity, $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ I = I_{0}\cos^{2}\left( \frac{\phi}{2} \right)$
$\frac{I}{I_{0}} = \cos^{2}{(30{^\circ})} = \left( \frac{\sqrt{3}}{2} \right)^{2} = 0.75$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.