Question

In Young's double slit experiment, the separation between the slits is $0.1\text{mm}$, the wavelength of light used is $600\text{nm}$ and the interference pattern is observed on a screen $1\text{m}$ away from the slits. Find the separation between the successive bright fringes.

• A. $3\text{mm}$
• B. $2\text{mm}$
• C. $4\text{mm}$
• D. $6\text{mm}$

Answer 1

The correct option is D $6\text{mm}$

Given:
$d=0.1\text{mm}={10}^{-4}\text{m}$
$\lambda =600\text{nm}=6\times {10}^{-7}\text{m}$
$D=1\text{m}$

Now, the separation between the successive bright fringes is,
$\beta =\frac{\lambda D}{d}=\frac{6\times {10}^{-7}\times 1}{{10}^{-4}}=6\text{mm}$

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

$\begin{array}{c}\text{Why this question?}\\ \text{Tip : Separation between successive bright fringes}\\ \text{and dark fringes is equal to the fringe width.}\end{array}$