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.
- A \(1.5 \times 10^{-7} \mathrm{~m}\)
- B \(2 \times 10^{-7} \mathrm{~m}\)
- C \(3.5 \times 10^{-7} \mathrm{~m}\)
- D \(4 \times 10^{-7} \mathrm{~m}\)
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, \(\mathrm{n}=0\),…
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