In Young's double slit experiment, the slits are 2\ mm apart and...

Given:
$d = 2\ \text{mm} = 2 \times 10^{-3}\ \text{m}$
$D = 2\ \text{m}$
$\lambda_1 = 6000\ \text{\AA} = 6 \times 10^{-7}\ \text{m}$
$\lambda_2 = 5000\ \text{\AA} = 5 \times 10^{-7}\ \text{m}$

Let the $n_1$-th bright fringe of $\lambda_1$ coincide with the $n_2$-th bright fringe of $\lambda_2$. The position of the $n$-th bright fringe from the central maximum is given by $y = \dfrac{n \lambda D}{d}$.

For coincidence:
$\dfrac{n_1 \lambda_1 D}{d} = \dfrac{n_2 \lambda_2 D}{d}$

$n_1 \lambda_1 = n_2 \lambda_2$

$\dfrac{n_1}{n_2} = \dfrac{\lambda_2}{\lambda_1} = \dfrac{5000}{6000} = \dfrac{5}{6}$

For the minimum distance from the central maximum, we take the smallest integer values for $n_1$ and $n_2$, which are $n_1 = 5$ and $n_2 = 6$.

Substituting $n_1 = 5$ into the position formula:
$y = \dfrac{5 \times 6 \times 10^{-7} \times 2}{2 \times 10^{-3}}$

$y = 30 \times 10^{-4}\ \text{m}$

$y = 3 \times 10^{-3}\ \text{m} = 3\ \text{mm}$

Answer: $3\ \text{mm}$