Q. In Young’s double slit experiment, let $S_{1}\ and\ S_{2}$ be the two slits and C be the centre of the screen. If $\angle S_{1}CS_{2} = \theta$ and $\lambda$ is the wavelength, the fringe width will be
A) $\frac{\lambda}{\theta}$
B) $\lambda\theta$
C) $2\lambda/\theta$
D) $\frac{\lambda}{2\theta}$
Explanation
(a) $\beta = \frac{\lambda D}{d} = \frac{\lambda}{d/D} = \frac{\lambda}{\theta}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.