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WBJEE · Physics · Wave Optics

Light of wavelength \(6000 Ã…\) is incident on a thin glass plate of \(\mu = 1.5\) such that the angle of refraction into the plate is \(60^{\circ}\). Calculate the smallest thickness of the plate which will make dark fringe by reflected beam interference.

  1. A \(1.5 \times 10^{-7} \mathrm{~m}\)
  2. B \(2 \times 10^{-7} \mathrm{~m}\)
  3. C \(3.5 \times 10^{-7} \mathrm{~m}\)
  4. D \(4 \times 10^{-7} \mathrm{~m}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(4 \times 10^{-7} \mathrm{~m}\)

Step-by-step Solution

Detailed explanation

For reflected light, condition of destructive interference, \(2 \mu t \cos r-\frac{\lambda}{2}=(2 n+1) \frac{\lambda}{2}\) \(\Rightarrow 2 \mu \mathrm{t} \cos \mathrm{r}=(2 \mathrm{n}+1) \frac{\lambda}{2}+\frac{\lambda}{2}\) For minimum thickness of the plate, \(\mathrm{n}=0\),…