Question

In a Young's double slit experiment, the distance between the $3^{\text{rd}}$ bright fringe and $6^{\text{th}}$ dark fringe both lying on the same side of the central maximum is $3\text{mm}$. What is the distance of $2^{\text{nd}}$ dark fringe from the central maximum?

• A. $0.6\text{mm}$
• B. $1.2\text{mm}$
• C. $1.8\text{mm}$
• D. $2.4\text{mm}$

Answer 1

The correct option is C $1.8\text{mm}$

Distance of $n^{\text{th}}$ bright fringe from central bright fringe, is
$y=\frac{n\lambda D}{d}$
For third bright fringe $n=3$,
$\therefore y_{3,b}=\frac{3\lambda D}{d}$

Distance of $n^{\text{th}}$ dark fringe from central bright fringe, is
$y=\left(\frac{2n-1}{2}\right)\frac{\lambda D}{d}$
For sixth dark fringe $n=6$
$\therefore y_{6,d}=\frac{11\lambda D}{2d}$

Distance between, third bright fringe and sixth dark fringe, both lying on the same side of central bright fringe, is,
$y=y_{6,d}-y_{3,b}$

$\therefore \frac{11\lambda D}{2d}-\frac{3\lambda D}{d}=3\text{mm}$

$\Rightarrow \frac{5\lambda D}{2d}=3\text{mm}$

$\therefore \frac{\lambda D}{d}=\frac{6}{5}\text{mm}$ .......(i)

Now, the distance of $2$nd dark fringe from the central maximum is :
$y_{2,d}=\left(\frac{2\times 2-1}{2}\right)\frac{\lambda D}{d}$

$y=\frac{3\lambda D}{2d}$

substituting value from (i), we get,
$y_{2,d}=\frac{3}{2}\times \frac{6}{5}=1.8\text{mm}$

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

Why this question? To understand layout of interference pattern and find distance between different maxima and/or minima