Q. In Young’s double slit experiment, distance between two slits is 0.28 mm and distance between slits and screen is 1.4 m. Distance between central bright fringe and third bright fringe is 0.9 cm, what is the wavelength of light used?
A) $4000\ \mathring{\mathrm{A}}$
B) $6000\ \mathring{\mathrm{A}}$
C) $3000\ \mathring{\mathrm{A}}$
D) $5000\ \mathring{\mathrm{A}}$
Explanation
(b) Given, $d = 0.28\ mm = 0.28 \times 10^{- 3}m,$
$\ D = 1.4\ m,$
$\ \ \beta = 0.9\ cm = 0.9 \times 10^{- 2}\ m$
$\ \ \beta = 3\frac{D\lambda}{d}$
Or $\ \ \lambda = \frac{\beta d}{3D}$
Or $\lambda = \frac{0.9\ \times \ 10^{- 2}\ \times \ 0.28\ \times {\ 10}^{- 3}}{3 \times 1.4}$
Or$\ \ \lambda = 6 \times 10^{- 7}\ m = 6000\ \mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.