Q. In a double slit interference experiment, the distance between the slits is 0.05 cm and screen is 2 m away from the slits. The wavelength of light is 6000$\mathring{\mathrm{A}}$. The distance between the fringe is
A) 0.24 cm
B) 0.12 cm
C) 1.24 cm
D) 2.28 cm
Explanation
(a) Distance between the fringe=fringe width
$x = \beta = \frac{\lambda D}{d} = \frac{6 \times 10^{- 7} \times 2}{0.05 \times 10^{- 2}}$
$= 24 \times 10^{- 4}\ m = 0.24\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.