Q. In a two-slit experiment, with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by $5 \times 10^{- 2}m$ towards slits, the change in fringe width is $10^{- 3}m$ Then the wavelength of light used is (given that distance between the slits is $0.03mm$)
A) $4000\ \mathring{\mathrm{A}}$
B) $4500\ \mathring{\mathrm{A}}$
C) $5000\ \mathring{\mathrm{A}}$
D) $6000\ \mathring{\mathrm{A}}$
Explanation
(d) Fringe width $\beta = \frac{\lambda D}{d}$
Where $D$ is the distance between slit and screen, $d$ is the distance between two slits, $\lambda$ is the wavelength of light
$\therefore\mathrm{\Delta}\beta = \frac{\lambda\Delta D}{d}$
$\Rightarrow \lambda = \frac{\Delta\beta d}{\Delta D} = \frac{10^{- 3} \times 0.03 \times 10^{- 3}}{5 \times 10^{- 2}} = \frac{10^{- 3} \times 3 \times 10^{- 5}}{5 \times 10^{- 2}}$
$= 6 \times 10^{- 7}m = 6000\mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.