Q. In Young’s double slit experiment, distance between two sources is 0.1 mm. The distance of screen from the source is 20 cm. Wavelength of light used is 5460 $\mathring{\mathrm{A}}$hen angular position of first dark fringe is
A) $0.08{^\circ}$
B) $0.16{^\circ}$
C) $0.20{^\circ}$
D) $0.32{^\circ}$
Explanation
(b) $d = 0.1\ mm = 10^{- 4},\ D = 20\ cm = \frac{1}{5}m$
$\lambda = 5460\mathring{\mathrm{A}} = 5.46 \times 10^{- 7}m$
Angular position of first dark fringe is
$\theta = \frac{x}{D} = \frac{\lambda}{2d} = \frac{5.46 \times 10^{- 7}}{2 \times 10^{- 4}}$
$= 2.73 \times 10^{- 3}\ rad$
$= 2.73 \times 10^{- 3} \times \frac{180{^\circ}}{\pi} = 0.156{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.