Question

In a Young's double-slit experiment, the slits are $0.5\text{mm}$ apart and the interference pattern is observed on a screen at a distance of $100\text{cm}$ from the slits. It is found that the ninth bright fringe on one side of the central bright fringe, is at a distance of $10.5\text{mm}$ from the second dark fringe lying on the opposite side of central bright fringe. The wavelength of the light used is,

• A. $5000\mathring{\text{A}}$
• B. $\frac{5000}{7}\mathring{\text{A}}$
• C. $2500\mathring{\text{A}}$
• D. $\frac{2500}{7}\mathring{\text{A}}$

Answer 1

The correct option is A $5000\mathring{\text{A}}$

Given:
$d=0.5\text{mm}D=100\text{cm}$

Distance of ninth bright fringe, from central bright fringe is,
${\beta }_{9,b}=9\left(\frac{\lambda D}{d}\right)$

Distance of second dark fringe, from central bright fringe, is,
${\beta }_{2,d}=(2-\frac{1}{2})\frac{\lambda D}{d}=\frac{3\lambda D}{2d}$

As both the above fringes are on the opposite sides of central bright fringe, the distance between them will be equal to, ${\beta }_{9,b}+{\beta }_{2,d}=10.5\text{mm}$

$\Rightarrow \frac{\lambda D}{d}(9+\frac{3}{2})=10.5\times {10}^{-3}$

$\therefore \lambda =(10.5\times {10}^{-3})\times \left(\frac{2}{21}\right)\times \left(\frac{0.5\times {10}^{-3}}{100\times {10}^{-2}}\right)$

$=5\times {10}^{-7}\text{m}=5000\mathring{\text{A}}$

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

Why this question? Caution: while finding the distance between any two fringes (dark or bright), it is also important to know, on which side (same or opposite) of the central maximum, do they lie.