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 $8.0 \times 10^{- 5}$ cm. The distance between successive fringes is
A) 0.24 cm
B) 3.2 cm
C) 1.28 cm
D) 0.32 cm
Explanation
(d) Distance between successive fringes-fringe width
$= \beta = \frac{\lambda D}{d} = \frac{8 \times 10^{- 5} \times 2}{0.05} = 0.32$ cm
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.