Q. A single slit of width $0.20\ mm$ is illuminated with light of wavelength $500\ nm$. The observing screen is placed $80\ cm$ from the slit. The width of the central bright fringe will be
A) $1\ mm$
B) $2\ mm$
C) $4\ mm$
D) $5\ mm$
Explanation
(c) Width of central bright fringe
$= \frac{2\lambda D}{d} = \frac{2 \times 500 \times 10^{- 9} \times 80 \times 10^{- 2}}{0.20 \times 10^{- 3}} = 4 \times 10^{- 3}m = 4mm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.