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A light beam of wavelength 600nm in air ...

A light beam of wavelength `600nm` in air passes through film `1(n_1=1.2)` of thickness `1.0 mu m`,then through film `2`(air) of thickness `1.5mu m`, and finally through film `3(n_3=1.8)` of `thickness `1.0mu m`.
(a) Which film does the light cross in the least time, and what is that least time?
(b) What are the total number of wavelength (at any instant) across all three films together?

Text Solution

Verified by Experts

The correct Answer is:
A, B, D

(a) `t_1=l/v_1=l/(c//n_1)=(n_1l)/c="minimum"`
`=(1.2xx10^-6)/(3xx10^8)`
`=4xx10^-15s`
(b) `lambda=(lambda_0)/n`
Number of wavelength across any film,
`N=l/lambda=(ln)/(lambda_0)`
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