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MHT CET · Physics · Wave Optics

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

  1. A \(\frac{\mathrm{D}^2}{2 \mathrm{~d}}\)
  2. B \(\frac{\mathrm{D}^2}{\mathrm{~d}}\)
  3. C \(\frac{\mathrm{d}^2}{2 \mathrm{D}}\)
  4. D \(\frac{\mathrm{d}^2}{\mathrm{D}}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{\mathrm{d}^2}{\mathrm{D}}\)

Step-by-step Solution

Detailed explanation

The dark fringe is produced at a point just opposite to the slit, i.e., \(y=\frac{d}{2}\)
For dark fringe: \(\mathrm{y}=(2 \mathrm{n}-1) \frac{\lambda \mathrm{D}}{2 \mathrm{~d}}\) where, \(\mathrm{n}=1,2,3, \ldots \ldots .\). is the order of the fringe.
\(\begin{aligned} & \Rightarrow \frac{d}{2}=(2 n-1) \frac{\lambda D}{2 d} \\ & \Rightarrow \lambda=\frac{d^2}{(2 n-1) D}\end{aligned}\)
For \(\mathrm{n}=1, \lambda=\frac{\mathrm{d}^2}{\mathrm{D}}\)
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