Physics - Wave Optics Question with Solution | TestHub

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

The path difference at this point is given by $\Delta x=\frac{yd}{D}$.

Substituting $y=\frac{d}{2}$, we get $\Delta x=\frac{d^2}{2D}$.

For a minimum (dark fringe) in the interference pattern, the path difference must be an odd multiple of $\frac{\lambda }{2}$:

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

Equating the two expressions for path difference:

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

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

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

Thus, the possible wavelengths are proportional to $\frac{1}{D},\frac{1}{3D},\frac{1}{5D},\dots$

Answer: $\frac{1}{D},\frac{1}{3D},\frac{1}{5D}$,...........