Q. In a Young’s double slit experiment, distance between sources is 1 mm and distance between the screen and sources is 1 m. If the fringe width on the screen is 0.06 cm, then $\lambda$ is
A) $6000\ \mathring{\mathrm{A}}$
B) $4000\ \mathring{\mathrm{A}}$
C) $1200\ \mathring{\mathrm{A}}$
D) $2400\ \mathring{\mathrm{A}}$
Explanation
(a) From $\beta = \frac{\lambda D}{d} \Rightarrow \lambda = \frac{\beta d}{D} = \frac{0.06 \times 10^{- 2} \times 10^{- 3}}{1}$
$= 6 \times 10^{- 7}m = 6000\mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.