Q. In Young’s double slit experiment, carried out with light of wavelength $\lambda = 5000\ \mathring{\mathrm{A}}$, the distance between the slits is $0.2\ mm$ and the screen is at $200\ cm$ from the slits. The central maximum is at $x = 0$. The third maximum (taking the central maximum as zeroth maximum) will be at $x$ equal to
A) $1.67\ cm$
B) $1.5\ cm$
C) $0.5\ cm$
D) $5.0\ cm$
Explanation
(b) Distance of third maxima from central maxima is
$x = \frac{3\lambda D}{d} = \frac{3 \times 5000 \times 10^{- 10} \times (200 \times 10^{- 2})}{0.2 \times 10^{- 3}} = 1.5cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.