Question 264: Wave Optics

Question

In Young's double slit experiment, for the $n$th dark fringe ( $\mathrm{n}=1,2,3 \ldots$ ) the phase difference of the interfering waves in radian will be

Accepted Answer

For the $n$th dark fringe in Young's double slit experiment, the condition for darkness is:
$ 	ext{Path difference} = \left(n - \dfrac{1}{2}ight)\lambda,\quad 	ext{where } n = 1, 2, 3, \ldots $
The corresponding phase difference $(\Delta \phi)$ is given by:
$ \Delta \phi = \frac{2\pi}{\lambda} 	imes 	ext{path difference} $
Substituting the value of path difference:
$ \Delta \phi = \frac{2\pi}{\lambda} \left(n - \frac{1}{2}ight)\lambda $
$ = 2\pi\left(n - \frac{1}{2}ight) $
$ = 2n\pi - \pi $
$ = (2n - 1)\pi $
Correct answer:
$ \boxed{(2n-1)\pi} $
So, the correct option is
Option C
.

Answer

$\mathrm{n} \frac{\pi}{2}$

Answer

$\quad(2 n+1) \pi$

Answer

$\quad(2 n-1) \pi$

Answer

$(2 n-1) \frac{\pi}{2}$

Wave Optics

In Young's double slit experiment, for the nth dark fringe ( \mathrm{n}=1,2,3 \ldots ) the phase difference of the interfering waves in radian will be

Choose the correct answer: