Q. In Young’s double slit experiment, an interference pattern is obtained on a screen by a light of wavelength $6000\mathring{\mathrm{A}}$ coming from the coherent sources $S_{1}$ and $S_{2}.$ At certain point P on the screen third dark fringe is formed. Then the path difference $S_{1}P - S_{2}P\ $ in microns is
A) 0.75
B) 1.5
C) 3.0
D) 4.5
Explanation
(b) $\lambda = 6000\ \mathring{\mathrm{A}} = 6 \times 10^{- 7}\ m$
Path difference for dark fringe $\mathrm{\Delta}x = (2n + 1)\frac{\lambda}{2}$
For third dark fringe $n = 2$
$\therefore\ \mathrm{\Delta}x = (2 \times 2 + 1) \times \frac{6 \times 10^{- 7}}{2}$
$= \frac{5 \times 6 \times 10^{- 7}}{2}$
$= 15 \times 10^{- 7}$
$= 1.5 \times 10^{- 6}\ m = 1.5\mu$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.