Q. In a Young’s double slit experiment the intensity at a point where the path difference is $\frac{\lambda}{6}$(λ being the wavelength of the light used)$\ is\ I.$ If $I_{0}$ denotes the maximum intensity, $I/I_{0}$ is equal to
A) $\frac{1}{\sqrt{2}}$
B) $\frac{\sqrt{3}}{2}$
C) $\frac{1}{2}$
D) $\frac{3}{4}$
Explanation
(d) Phase difference $= \frac{2\pi}{\lambda} \times$ path difference
ie, $\ \phi = \frac{2\pi}{\lambda}\ \times \frac{\lambda}{6} = \frac{\pi}{3}$
As, $\ I = I_{\max}\cos^{2}\left( \frac{\phi}{2} \right)$
Or $\frac{I}{I_{\max}} = \cos^{2}\left( \frac{\phi}{2} \right)$
Or $\frac{I}{I_{0}} = \cos^{2}{\left( \frac{\pi}{6} \right) = \frac{3}{4}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.