JEE MAIN - Physics (2019 - 9th January Morning Slot - No. 16)
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 :
_9th_January_Morning_Slot_en_16_1.png)
$$\sqrt {{5 \over 3}} $$
$${3 \over {\sqrt 5 }}$$
$${5 \over {\sqrt 3 }}$$
$${4 \over 3}$$
Explanation
_9th_January_Morning_Slot_en_16_2.png)
When light is incident on the liquid at 90o, then from ssnells law,
1.sin90o = $$\mu $$sin$$\theta $$
$$ \Rightarrow $$ sin$$\theta $$ = $${1 \over \mu }$$ . . . . . . (1)
Between liquid and glass,
$$\mu $$sin$$\theta $$ = 1.5 sin r
$$ \Rightarrow $$ $$\mu $$sin$$\theta $$ = 1.5 sin (90o $$-$$ $$\theta $$)
$$ \Rightarrow $$ $$\mu $$ tan$$\theta $$ = 1.5
$$ \Rightarrow $$ tan$$\theta $$ = $${{1.5} \over \mu }$$
$$ \therefore $$ sin$$\theta $$ = $${{1.5} \over {\sqrt {{\mu ^2} + {{\left( {1.5} \right)}^2}} }}$$ . . . . . . (2)
$$ \therefore $$ From (1) and (2) we get,
$${1 \over \mu }$$ = $${{1.5} \over {\sqrt {{\mu ^2} + {{\left( {1.5} \right)}^2}} }}$$
$$ \Rightarrow $$ $$\mu $$2 + (1.5)2 = $$\mu $$2 $$ \times $$ (1.5)2
$$ \Rightarrow $$ $$\mu $$ = $${3 \over {\sqrt 5 }}$$
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