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Consider a tank made of glass (refractive index 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 polarized. For this to happen, the minimum value of μ is
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(A) 35
(B) 53
(C) 53
(D) 43

Answer
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Hint: For finding out the minimum value of the refractive index, we need to use the relation between Brewster angle and the critical angle which is given as sin c<sin ib . Brewster angle is an angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection.

Formula used:
 sin c<sin ib
  sin90=μsincsinc=1μ
Where, the critical angle be  c , the Brewster angle be  ib ,  μ is the refractive index.

Complete step by step solution:
Let us consider the critical angle be c, the Brewster angle be ib.
The relation comes between the critical angle and the Brewster angle is,
 sin c<sin ib
For the ray travelling from air to liquid,
 sin90=μsinc
Now the value of sin90 is 1 so we get
 sinc=1μ
Since, we know that,
 tanib=μ0rel
And,
 sin c<sin ib
Then substituting the values in the equation we get,
 1μ<1.5μ2+(1.5)2
Thus, after simplification, we get,
 μ<35
Hence, the correct answer is option A.

Note:
There are numerous applications of Brewster angle in real life. It includes,
-Polarized sunglasses use the principle of Brewster's angle to reduce glare from the sun reflecting off horizontal surfaces such as water or road.
-Photographers use the same principle to remove reflections from water so that they can photograph objects beneath the surface.
-Brewster angle prisms are used in laser physics. The polarized laser light enters the prism at Brewster's angle without any reflective losses.
-In surface science, Brewster angle microscopes are used in imaging layers of particles or molecules at air-liquid interfaces.
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