Q. In Young’s double slit experiment, the distance between the two slits is $0.1\ mm$ and the wavelength of light used is $4 \times 10^{- 7}m$. If the width of the fringe on the screen is $4\ mm$, the distance between screen and slit is
A) $0.1\ mm$
B) $1\ cm$
C) $0.1\ cm$
D) $1\ m$
Explanation
(d) $\beta = \frac{\lambda D}{d} \Rightarrow \left( 4 \times 10^{- 3} \right) = \frac{4 \times 10^{- 7} \times D}{0.1 \times 10^{- 3}} \Rightarrow D = 1m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.