Q. In a Young’s double slit experiment, the slit separation is $1mm$ and the screen is $1m$ from the slit. For a monochromatic light of wavelength $500\ nm$, the distance of $3rd$ minima from the central maxima is
A) $0.50\ mm$
B) $1.25\ mm$
C) $1.50\ mm$
D) $1.75\ mm$
Explanation
(b) Distance of $n^{th}$ minima from central bright fringe
$x_{n} = \frac{(2n - 1)\lambda D}{2d}$
For $n = 3\ i.e.\ 3^{rd}$ minima
$x_{3} = \frac{(2 \times 3 - 1) \times 500 \times 10^{- 9} \times 1}{2 \times 1 \times 10^{- 3}}$
$= \frac{5 \times 500 \times 10^{- 6}}{2} = 1.25 \times 10^{- 3}m = 1.25\ mm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.