A light of wavelength 5890 Å falls normally on a thin air film. The minimum thickness of the film such that the film appears dark in reflected light is
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a
2.945×10−7m
b
3.945×10−7m
c
4.95×10−7m
d
1.945×10−7m
answer is A.
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Detailed Solution
If thin film appears dark2μt cos r=nλ for normal incidence r=0∘⇒2μt=nλ ⇒ t=nλ2μ⇒tmin=λ2μ=5890×10−102×1=2.945×10−7m.