Consider a tank made of glass (refractive index 1.5) with a thick bottom. It is filled with a liquid of refractive index A student finds that, irrespective of what the incident angle i (see figure) is for a beam of light entering the liquid, the light reflected from the liquid glass interface is never completely polarized. For this to happen, the minimum value of `mu` is `x//sqrt(5)`. Find x.

Answer

3

Explanation

At air -liquid interface
`( sin i)/(sin theta) =mu`…….i
At liquid glass interface, for reflected ray to be completely polarised
`tan theta =1.5/(mu)`…….ii
For `mu` to min,` theta` max
So from (i)
`(sin theta)_("max")=1/(mu)`
`implies(tan theta)_("max")=1/(sqrt(mu^(2)-1))implies1/(sqrt(mu^(2)-1))=3/(2mu)`
`9mu^(2)-9=4mu^(2),5mu^(2)=9`
`mu=3/(sqrt(5))`