Physics - Wave Optics Question with Solution | TestHub

The position of the point on the screen 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:

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

For a minimum (destructive interference) to be observed, 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:

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

Solving for $\lambda$, we get:

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

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

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

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