In a Young's double slit experiment employing white light,...

Let $n_v$ be the order of the violet fringe and $n_r$ be the order of the red fringe.

The position of the $n$-th bright fringe in Young's double slit experiment is given by $y = \dfrac{n \lambda D}{d}$.

For the fringes to overlap, their positions on the screen must be the same:

$\dfrac{n_v \lambda_v D}{d} = \dfrac{n_r \lambda_r D}{d}$

$n_v \lambda_v = n_r \lambda_r$

Substituting the given wavelengths $\lambda_v = 400 \text{ nm}$ and $\lambda_r = 700 \text{ nm}$:

$n_v (400) = n_r (700)$

$4 n_v = 7 n_r$

To find the minimum order, $n_v$ and $n_r$ must be the smallest possible positive integers. Since $4$ and $7$ are coprime, we get:

$n_v = 7$ and $n_r = 4$

The minimum order of the violet fringe that overlaps with a red fringe is $7$.