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

Correct answer: A. 3 5 For reflected light not be completely polarized, tan r ≠ μ _ g μ _ w = (1.5)/(μ) For minimum is maximum. r is maximum when i is maximum i.e. i = 90 ^ o ⇒ sin r = (1)/(μ) ⇒ tan r = 1 μ ^ 2 - 1 ∴ 1 μ ^ 2 - 1 = (1.5)/(μ) ⇒ μ ^ 2 = 2.25( μ ^ 2 - 1) ⇒ 1.25 μ ^ 2 = 2.25 ⇒ μ = (2.25)/(1.25) = (9)/(5) = 3 5

JEE Main 20199 JanuaryMorning Shift

Exam: JEE Main
Year: 2019
Date: 9 January
Shift: Morning Shift
Subject: Physics
Chapter: Wave Optics

JEE Main 2019 · 9 January · Morning Shift · Physics · Wave Optics

Consider a tank made of glass (refractive index 1.5 ) with a thick...

Consider a tank made of glass (refractive index

1.5

) with a thick bottom. It is filled with a liquid of refractive index

\mu .

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:

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

Options

A $\dfrac{3}{\sqrt{5}}$
B $\dfrac{4}{3}$
C $\dfrac{5}{\sqrt{3}}$
D $\sqrt{\dfrac{5}{3}}$

Correct answer

A. $\dfrac{3}{\sqrt{5}}$

Step-by-step solution

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

For reflected light not be completely polarized,

tan r \ne \dfrac{{\mu}_{g}}{{\mu}_{w}} = \dfrac{1.5}{\mu}

For minimum

\mur

is maximum.

r

is maximum when

i

is maximum i.e.

i = {90}^{o}

⇒ sin r = \dfrac{1}{\mu}

⇒ tan r = \dfrac{1}{\sqrt{{\mu}^{2} - 1}}

∴ \dfrac{1}{\sqrt{{\mu}^{2} - 1}} = \dfrac{1.5}{\mu}

⇒ {\mu}^{2} = 2.25\left({\mu}^{2} - 1\right)

⇒ 1.25{\mu}^{2} = 2.25

⇒ \mu = \sqrt{\dfrac{2.25}{1.25}}

= \sqrt{\dfrac{9}{5}} = \dfrac{3}{\sqrt{5}}

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