In a Young's double slit setup, the slits are illuminated by a...

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

For the first-order bright fringe, $n = 1$, which gives $y_1 = \dfrac{\lambda D}{d}$.

The distance $y_1$ is directly proportional to the wavelength $\lambda$ of the light.

Among the colours of visible light, violet has the shortest wavelength ($\lambda_{violet} < \lambda_{green} < \lambda_{yellow} < \lambda_{red}$).

Since violet light has the minimum wavelength, its first-order bright fringe will be formed at the minimum distance from the central maximum. Therefore, violet will be closest to the central white maximum.

Answer: Violet