Q. In double slit experiment, the distance between two slits is 0.6 mm and these are illuminated with light of wavelength 4800$\ \mathring{\mathrm{A}}$. The angular width of first dark fringe on the screen distant 120 cm from slits will be
A) $8 \times 10^{- 4}\ rad$
B) $6 \times 10^{- 4}\ rad$
C) $4 \times 10^{- 4}\ rad$
D) $16 \times 10^{- 4}\ rad$
Explanation
(a) Destructive interference occurs when the path difference is an odd multiple of $\lambda/2.$
Ie. $\frac{xd}{D} = \frac{(2n - 1)\lambda}{2}$
Angular width of first dark fringe is
$\frac{2x}{D} = \frac{2(2n - 1)\lambda}{2d}$
Given, $n = 1,\ \lambda = 4800\ \mathring{\mathrm{A}} = 4800 \times 10^{- 10}\ m,$
$d^{'} = 0.6\ mm = 0.6 \times 10^{- 3}\ m$
$\therefore\frac{2x}{D} = \frac{2(2 \times 1 - 1) \times 4800 \times 10^{- 10}}{2 \times 0.6 \times 10^{- 3}}$
$= 8 \times 10^{- 4}\ rad$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.