Q. In Young’s double slit experiment distance between source is 1 mm and distance between the screen and source is 1m. If the fringe width on the screen is0.06 cm, then $\lambda$ is
A) $6000\ \mathring{\mathrm{A}}$
B) $4000\ \mathring{\mathrm{A}}$
C) $1200\ \mathring{\mathrm{A}}$
D) $2400\ \mathring{\mathrm{A}}$
Explanation
(a) When distance between screen and source is D, and d the distance between coherent sources, then fringe width (W) is given by
$W = \frac{D\lambda}{d}$
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Where $\lambda$ is wavelength of monochromatic light.
$\lambda = \frac{Wd}{D}$
Given, $D = 1\ m,\ d = 1\ mm = 10^{- 3}\ m,$
$W = 0.06\ cm = 0.06 \times 10^{- 2}m$
$\therefore\ \lambda = \frac{0.06 \times 10^{- 2} \times 10^{- 3}}{1}$
$= 6 \times 10^{- 7}m = 6000\ \mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.