Q. In a Young’ s double slit experiment, the intensity at a point where the path difference $is\ \frac{\lambda\ }{6\ }$ where(λ is wavelength of the light)$\ is\ I.\ \ I\ f\ I_{0}$denotes the maximum intensity, then $\frac{I}{I_{0}}\ $is equal to
A) $\frac{1}{2}$
B) $\frac{\sqrt{3}}{2}$
C) $\frac{1}{\sqrt{2}}$
D) $\frac{3}{4}$
Explanation
(d) $\phi = \frac{\lambda}{6} = \frac{360{^\circ}}{6} = 60{^\circ}\ $
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ I = I_{0}\cos^{2}\theta$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = I_{0}\cos^{2}{60{^\circ}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{3}{4} \times I_{0}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{I}{I_{0}} = \frac{3}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.