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 proportional to

The distance between the coherent sources is $d$ and the distance to the screen is $D$. For a point at position $y = d/2$ directly in front of one slit, the path difference becomes $\Delta x = \frac{(d/2)d}{D} = \frac{d^2}{2D}$.

Destructive interference requires $\Delta x = (2n - 1)\frac{\lambda}{2}$ for $n = 1, 2, 3, \dots$. Equating the expressions gives $\frac{d^2}{2D} = (2n - 1)\frac{\lambda}{2}$.

Simplifying yields $\lambda = \frac{d^2}{(2n - 1)D}$. The possible wavelengths are therefore proportional to $\frac{1}{D}$, $\frac{1}{3D}$, $\frac{1}{5D}$, … matching the sequence in option B.

Final answer: $\boxed{B}$