Q. In a Young’s double slit experiment, the slit separation is 1 mm and the screen is 1 m from the slit. For a monochromatic light of wavelength 500 nm, the distance of 3$^{rd}$ minima from the central maxima is
A) 0.50 mm
B) 1.25 mm
C) 1.50 mm
D) 1.75 mm
Explanation
(b) $d = 1\ mm = 1 \times 10^{- 3}\ m,\ D = 1\ m$
$\lambda = 500\ nm = 500 \times 10^{- 9}\ m$
Distance of $n$th minima from central maxima
$x_{n} = \frac{(2n - 1)\lambda}{2}\frac{D}{d}$
$= \frac{(2 \times 3 - 1) \times 500 \times 10^{- 9}}{2} \times \frac{1}{1 \times 10^{- 3}}$
$= {2.5 \times 500 \times 10}^{- 6}$
$= {12.5 \times 10}^{- 4}\ m$
$= 1.25\ mm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.