JEE Main 2014 · 19 Apr · Shift 1 · Physics Q23

← JEE Main 2014 · 19 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 23 of 90

A ray of light is incident from a denser to a rarer medium. The critical angle for total internal reflection is $\theta_{iC}$ and the Brewster's angle of incidence is $\theta_{iB}$ , such that $\sin\theta_{iC}/\sin\theta_{iB}=\eta=1.28$ . The relative refractive index of the two media is :

0.2

0.4

0.8

0.9

Answer and worked solution

Official answer: C

Let $n$ be the index of the denser medium relative to the rarer. Then $\sin\theta_{iC}=\frac1n$ and $\tan\theta_{iB}=\frac1n$ , so $\sin\theta_{iB}=\frac1{\sqrt{1+n^2}}$ .
$\eta=\frac{\sqrt{1+n^2}}n=1.28\implies n^2=\frac1{1.28^2-1}=1.566,\ n=1.25.$
The relative index of the rarer medium with respect to the denser one is $\frac1n=0.8$ .

Topics:Polarisation and Brewster's law Total internal reflection

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JEE Main 2014 · 19 Apr · Shift 1, Physics question 23