Question

A double slit experiment is performed with sodium light of wavelength $600\text{nm}$ and interference pattern is observed on a screen $100\text{cm}$ away from the slits. The $4^{\text{th}}$ dark fringe is at a distance of $7\text{mm}$ from the central maximum. Find the separation between the slits.

• A. $0.15\text{mm}$
• B. $0.30\text{mm}$
• C. $0.60\text{mm}$
• D. $1.2\text{mm}$

Answer 1

The correct option is B $0.30\text{mm}$

Given:
$\lambda =600\text{nm}=6\times {10}^{-7}\text{m}$
$D=100\text{cm}=1\text{m}$
$4^{\text{th}}$ dark fringe is formed at $y=7\text{mm}=7\times {10}^{-3}\text{m}$

Path difference, $\Delta x=\frac{yd}{D}$
For the $4^{th}$ dark fringe,
$\Delta x=\frac{7\lambda }{2}$
$\because \Delta x=(n-\frac{1}{2})\lambda$, for a dark fringe.

So, $\frac{yd}{D}=\frac{7\lambda }{2}$

$\Rightarrow d=\frac{7\lambda D}{2y}=\frac{7\times 6\times {10}^{-7}\times 1}{2\times 7\times {10}^{-3}}$

$d=3\times {10}^{-4}\text{m}=0.3\text{mm}$

Hence, $\left(B\right)$ is the correct answer.

$$\begin{array}{c}\text{Why this question :}\\ \text{Concept : For dark fringes to form, path difference :}\\ \Delta x=(2n-1)\frac{\lambda }{2}\\ \text{Where},n=1,2,3,.....\end{array}$$