Q. Two slits, 4 mm apart are illuminated by light of wavelength$600\ \mathring{\mathrm{A}}$. What will be the fringe width on a screen placed 2 m from the slits?
A) $0.12\ mm$
B) $0.3\ mm$
C) $3.0\ mm$
D) $4.0\ mm$
Explanation
(b) $\beta = \frac{\lambda D}{d} = \frac{6000 \times 10^{- 10} \times 2}{4 \times 10^{- 3}}$
$= 0.3 \times 10^{- 3}m = 0.3mm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.