Step 1: Set up the geometry.
In Young's experiment the two slits are separated by $d$ and the screen sits a distance $D$ away. We want the path setup for a point that lies straight ahead of one slit.
Step 2: Locate the point.
The centre of the pattern lies on the perpendicular bisector of the two slits. A point directly in front of one slit is shifted from that centre by half the slit spacing, so $y = \dfrac{d}{2}$.
Step 3: Apply the dark fringe condition.
A minimum forms when the path difference is an odd multiple of half a wavelength, which on the screen reads \[ y = (2n-1)\frac{\lambda D}{2d}, \quad n = 1, 2, 3, \dots \]
Step 4: Insert the position.
Replace $y$ with $\dfrac{d}{2}$: \[ \frac{d}{2} = (2n-1)\frac{\lambda D}{2d}. \]
Step 5: Solve for the wavelength.
Cancel the common factor of $2$ and rearrange: \[ d^2 = (2n-1)\lambda D \;\Rightarrow\; \lambda = \frac{d^2}{(2n-1)D}. \]
Step 6: Read off the sequence.
Putting $n = 1, 2, 3, \dots$ gives the odd numbers $1, 3, 5, \dots$ multiplying $D$, so $\lambda \propto \dfrac{1}{(2n-1)D}$. The wavelengths are inversely proportional to $D, 3D, 5D, \dots$ Well done, that matches option (B). \[ \boxed{\lambda \propto \frac{1}{D,\,3D,\,5D,\dots}} \]