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QJEEM19Phfb4a: JEE Main 2019 Physics - Class 12 JEEM19Phfb4a Solution: Consider a tank made of glass(refractive index 1.5) with a…

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

JEE Main 2019 (Online) 9th January Morning Slot Physics - Wave Optics Question 114 English

Multiple Choice Options

  • A 53\sqrt {{5 \over 3}}
  • B 35{3 \over {\sqrt 5 }}
  • C 53{5 \over {\sqrt 3 }}
  • D 43{4 \over 3}

Step-by-Step Verified Solution

JEE Main 2019 (Online) 9th January Morning Slot Physics - Wave Optics Question 114 English Explanation
When light is incident on the liquid at 90o, then from ssnells law,

1.sin90o = μ\mu sin θ\theta

⇒\Rightarrow    sin θ\theta = 1μ{1 \over \mu } . . . . . . (1)

Between liquid and glass,

μ\mu sin θ\theta = 1.5 sin r

⇒\Rightarrow     μ\mu sin θ\theta = 1.5 sin (90o −- θ\theta )

⇒\Rightarrow     μ\mu tan θ\theta = 1.5

⇒\Rightarrow    tan θ\theta = 1.5μ{{1.5} \over \mu }

∴\therefore    sin θ\theta = 1.5μ2+(1.5)2{{1.5} \over {\sqrt {{\mu ^2} + {{\left( {1.5} \right)}^2}} }} . . . . . . (2)

∴\therefore     From (1) and (2) we get,

1μ{1 \over \mu } = 1.5μ2+(1.5)2{{1.5} \over {\sqrt {{\mu ^2} + {{\left( {1.5} \right)}^2}} }}

⇒\Rightarrow     μ\mu 2 + (1.5)2 = μ\mu 2 ×\times (1.5)2

⇒\Rightarrow     μ\mu = 35{3 \over {\sqrt 5 }}

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