Q. Light of wavelength $\lambda = 5000\ \mathring{\mathrm{A}}$ falls normally on a narrow slit. A screen placed at a distance of $1m$ from the slit and perpendicular to the direction of light. The first minima of the diffraction pattern is situated at $5\ mm$ from the centre of central maximum. The width of the slit is
A) $0.1\ mm$
B) $1.0\ mm$
C) $0.5\ mm$
D) $0.2\ mm$
Explanation
(a) Position of $n^{th}$ minima $x_{n} = \frac{n\lambda D}{d}$
$\Rightarrow 5 \times 10^{- 3} = \frac{1 \times 5000 \times 10^{- 10} \times 1}{d}$
$\Rightarrow d = 10^{- 4}m = 0.1\ mm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.