Q. In young’s two slit experiment the distance between the two coherent sources is 2 mm and the screen is at a distance of 1 m. If the fringe width is found to be 0.03 cm, then the wavelength of the light used is
A) $4000\mathring{\mathrm{A}}$
B) $5000\mathring{\mathrm{A}}$
C) $5890\mathring{\mathrm{A}}$
D) $6000\mathring{\mathrm{A}}$
Explanation
(d) From $\beta = \frac{\lambda D}{d},$
$\lambda = \frac{\beta.d}{D} = \frac{0.3 \times 10^{- 2} \times 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.