In Young's double slit experiment, two slits are 2 mm apart and fringes are observed on a screen 2 m away from the slits. A light source having two wavelengths, 4000 Å and 6000 Å, is used. The minimum distance from the centre of the screen where the bright fringes produced by both wavelengths coincide is:

$1.2 × 10^{-3} m$

The correct answer is Option (2) → $1.2 × 10^{-3} m$

Given:

Slit separation: $d = 2 \times 10^{-3}\ \text{m}$

Screen distance: $D = 2\ \text{m}$

Wavelengths: $\lambda_1 = 4000\ \text{Ã…} = 4 \times 10^{-7}\ \text{m}$, $\lambda_2 = 6000\ \text{Ã…} = 6 \times 10^{-7}\ \text{m}$

Position of $n^{th}$ bright fringe: $y = \frac{n\lambda D}{d}$

For coincidence of fringes, $\frac{n_1\lambda_1 D}{d} = \frac{n_2\lambda_2 D}{d}$

$\Rightarrow n_1\lambda_1 = n_2\lambda_2$

Substitute values:

$n_1 \cdot 4000 = n_2 \cdot 6000$

$\Rightarrow \frac{n_1}{n_2} = \frac{6000}{4000} = \frac{3}{2}$

Smallest integers: $n_1 = 3$, $n_2 = 2$

Position of coincidence fringe:

$y = \frac{n_1\lambda_1 D}{d} = \frac{3 \cdot 4 \times 10^{-7} \cdot 2}{2 \times 10^{-3}}$

$y = \frac{24 \times 10^{-7}}{2 \times 10^{-3}}$

$y = \frac{24 \times 10^{-7}}{2 \times 10^{-3}} = 12 \times 10^{-4} = 1.2 \times 10^{-3}\ \text{m}$

$y = 1.2\ \text{mm}$

Answer: $1.2\ \text{mm}$ from the central fringe