Optics Question 1147

Question: A light of wavelength $ 5890{AA} $ falls normally on a thin air film. The minimum thickness of the film such that the film appears dark in reflected light

[Pb. PMT 2003]

Options:

A) $ 2.945\times {{10}^{-7}}m $

B) $ 3.945\times {{10}^{-7}}m $

C) $ 4.95\times {{10}^{-7}}m $

D) $ 1.945\times {{10}^{-7}}m $

Show Answer

Answer:

Correct Answer: A

Solution:

If thin film appears dark 2mt cos r = nl for normal incidence $ r=0^{o} $

Therefore 2 m t = nl

Therefore $ t=\frac{n\lambda }{2\mu } $

Therefore $ {t _{\min }}=\frac{\lambda }{2\mu } $ = $ \frac{5890\times {{10}^{-10}}}{2\times 1} $

$ =2.945\times {{10}^{-7}}m $ .



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