Question

In Young's double slit experiment the two slits are d distance apart. Interference pattern is observed on a screen at a distance D from the slits. A dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light is

• A. $\frac{D^2}{2d}$
• B. $\frac{d^2}{2D}$
• C. $\frac{D^2}{d}$
• D. $\frac{d^2}{D}$

Answer 1

Answer: (D)

We have the Young's double slit experiment given by

From the question, we see that the distance between the slits is equal to $d$ and the distance between the slit and screen is equal to $D$.
Hence for the nth dark fringe, we have $ (2n -1) \frac{D \lambda}{2d} = \frac{d}{2}$
Hence, we get $\lambda = \frac{d^2}{(2n - 1)D} = \frac{d^2}{D} \, for \, n = 1$