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) For dark fringe at $P$
$S_{1}P - S_{2}P = \Delta = (2n - 1)\lambda/2$
Here $n = 3$ and $\lambda = 6000\mathring{\mathrm{A}}$
So, $\Delta = \frac{5\lambda}{2} = 5 \times \frac{6000\mathring{\mathrm{A}}}{2} = 15000\mathring{\mathrm{A}} = 1.5\ micron$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.