Q. In Young’s double slit experiment, the spacing between the slits is $d$ and wavelength of light used is 6000$\mathring{\mathrm{A}}$. If the angular width of a fringe formed on a distance screen is $1{^\circ}$, then value of $d$ is
A) 1 mm
B) 0.05 mm
C) 0.03 mm
D) 0.01 mm
Explanation
(c) Here, $\sin{= \theta = \left( \frac{y}{D} \right)}$
So, $\mathrm{\Delta}\theta = \frac{\mathrm{\Delta}y}{D}$
Angular fringe width $\theta_{0} = \mathrm{\Delta}\theta($width $\mathrm{\Delta}y = \beta)$
$\theta_{0} = \frac{\beta}{D} = \frac{D\lambda}{d} \times \frac{1}{D} = \frac{\lambda}{d}$
$\theta_{0} = 1{^\circ} = \pi/180\ $rad
And $\lambda = 6 \times 10^{- 7}$m
$d = \frac{\lambda}{\theta_{0}} = \frac{180}{\pi} \times 6 \times 10^{- 7}$
$= 3.44 \times 10^{- 5}$m
$= 0.03\ mm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.