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