A Young's double slit apparatus is illuminated simultaneously by a...

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 the fringes to coincide, $y_1 = y_2$:

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

$n_1 \lambda_1 = n_2 \lambda_2$

Substituting the given wavelengths:

$n_1 (420) = n_2 (560)$

$\dfrac{n_1}{n_2} = \dfrac{560}{420} = \dfrac{4}{3}$

For the minimum non-zero distance, $n_1$ and $n_2$ must be the smallest possible integers. Thus, $n_1 = 4$ and $n_2 = 3$.

The minimum distance $y$ is:

$y = \dfrac{n_1 \lambda_1 D}{d}$

$y = \dfrac{4 \times 420 \times 10^{-9} \times 1.5}{1.0 \times 10^{-3}}$

$y = 2520 \times 10^{-6} \text{ m}$

$y = 2.52 \text{ mm}$

Answer: $2.52 \text{ mm}$