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A light of wavelength 5890Å falls normal...

A light of wavelength `5890Å` falls normally on a thin air film. The minimum thickness of the film such that the film appears dark in reflected light is

A

(a) `2.945xx10^-7m`

B

(b) `3.945xx10^-7m`

C

(c) `4.95xx10^-7m`

D

(d) `1.945xx10^-7m`

Text Solution

Verified by Experts

The correct Answer is:
A

If thin film appears dark
`2mutcosr=nlambda` for normal incidence `r=0^@`
`implies2mut=nlambdaimpliest=(nlambda)/(2mu)`
`impliest_(min)=(lambda)/(2mu)=(5890xx10^(-10))/(2xx1)`
`=2.945xx10^-7m`.
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