Question

In a Young's double slit experiment, the slits are $2\text{mm}$ apart and the screen is positioned $140\text{cm}$ away from the plane of the slits. The slits are illuminated with light of wavelength $600\text{nm}$. Find the distance of the third bright fringe from the central maximum.

• A. $1.26\text{mm}$
• B. $0.15\text{mm}$
• C. $3.76\text{mm}$
• D. $2.97\text{mm}$

Answer 1

The correct option is A $1.26\text{mm}$

Given:
$d=2\text{mm}=2\times {10}^{-3}\text{m}$
$D=140\text{cm}=1.4\text{m}$
$\lambda =600\text{nm}=6\times {10}^{-7}\text{m}$

Distance of $n^{\text{th}}$ bright fringe, from the central bright fringe, is
${\beta }_{n,b}=n\beta =n\frac{\lambda D}{d}$
Here, $n=3$
${\beta }_{3,b}=3\times \frac{6\times {10}^{-7}\times 1.4}{2\times {10}^{-3}}$

$\therefore {\beta }_{3,b}=12.6\times {10}^{-4}\text{m}=1.26\text{mm}$

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

$$\begin{array}{c}\text{Why this question :}\\ \text{Concept : For bright fringe, the path difference between interfering waves should be integral multiple of wavelength.}\end{array}$$