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 $ .
