Q. In Young’s double slit experiment, the distance between sources is $1\ mm$ and distance between the screen and source is $1\ m$. If the fringe width on the screen is $0.06\ cm$, then $\lambda =$
A) $6000\ \mathring{\mathrm{A}}$
B) $4000\ \mathring{\mathrm{A}}$
C) $1200\ \mathring{\mathrm{A}}$
D) $2400\ \mathring{\mathrm{A}}$
Explanation
(a) $\beta = \frac{\lambda D}{d} \Rightarrow \left( 0.06 \times 10^{- 2} \right) = \frac{\lambda \times 1}{1 \times 10^{- 3}} \Rightarrow \lambda = 6000\mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.