Q. The distance between the first dark and bright band formed in Young’s double slit experiment with band width B is
A) $\frac{B}{4}$
B) B
C) $\frac{B}{2}$
D) $\frac{3B}{2}$
Explanation
(c) Position of nth bright fringe $x_{1} = \frac{n\lambda D}{d}$
For first bright fringe$\ n = 1$
$\therefore\ \ x_{1} = \frac{\lambda D}{d}$
Position of nth dark fringe $x_{2} = \frac{(2n - 1)\lambda D}{2d}$
For first dark fringe $n = 1$
$\therefore x_{2} = \frac{\lambda D}{2d}$
Now, $x_{1} - x_{2} = \frac{\lambda D}{2d}$
If B is the band width, then
$x_{1} - x_{2} = \frac{B}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.