In Young's double slit experiment, in an interference pattern, a minimum is observed exactly in front of one slit. The distance between the two coherent sources is 'd' and 'D' is the distance between source and screen. The possible wavelengths used are inversely proportional to

The position of the point on the screen exactly in front of one slit is at a distance $y = \dfrac{d}{2}$ from the central maximum.

The path difference at this point is given by $\Delta x = \dfrac{y d}{D}$.

Substituting $y = \dfrac{d}{2}$, we get:

$\Delta x = \dfrac{\left(\dfrac{d}{2}\right) d}{D} = \dfrac{d^2}{2D}$

For a minimum (destructive interference) to be observed, the path difference must be an odd multiple of $\dfrac{\lambda}{2}$:

$\Delta x = (2n - 1) \dfrac{\lambda}{2}$, where $n = 1, 2, 3, \ldots$

Equating the two expressions for path difference:

$\dfrac{d^2}{2D} = (2n - 1) \dfrac{\lambda}{2}$

Solving for $\lambda$, we get:

$\lambda = \dfrac{d^2}{(2n - 1) D}$

Substituting $n = 1, 2, 3, \ldots$, the possible wavelengths are $\dfrac{d^2}{D}, \dfrac{d^2}{3D}, \dfrac{d^2}{5D}, \ldots$

Therefore, the possible wavelengths used are inversely proportional to $D, 3D, 5D, \ldots$

Answer: $D, 3D, 5D, \ldots$