Q. In the Young’s double slit experiment, the spacing between two slits is $0.1\ mm$. If the screen is kept at a distance of $1.0\ m$ from the slits and wavelength of light is $5000\ \mathring{\mathrm{A}}$, then the fringe width is
A) $1.0cm$
B) $1.5cm$
C) $0.5cm$
D) $2.0cm$
Explanation
(c) $\beta = \frac{\lambda D}{d} = \frac{500 \times 10^{- 10} \times 1}{0.1 \times 10^{- 3}}m = 5 \times 10^{- 3}m = 0.5\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.