Optics 7 Question 5

5. Consider a tank made of glass (refractive index is 1.5 ) with a thick bottom. It is filled with a liquid of refractive index μ. 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 polarised.

For this to happen, the minimum value of μ is

(2019 Main, 9 Jan I)

(a) 35

(b) 53

(c) 43

(d) 53

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Solution:

Key Idea When a beam of unpolarised light is reflected from a transparent medium of refractive index μ, then the reflected light is completely plane polarised at a certain angle of incidence iB, which is known as Brewster’s angle.

In the given condition, the light reflected irrespective of an angle of incidence is never completely polarised. So,

iC>iB

where, iC is the critical angle.

siniC<siniB

From Brewster’s law, we know that

taniB=wμg=μglass μwater =1.5μ

From Eqs. (i) and (ii), we get

1μ<1.5(1.5)2+(μ)2(1.5)2+μ2<1.5μμ2+(1.5)2<(1.5μ)2 or μ<35 The minimum value of μ should be 35.



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