A ray of light is incident from a denser to a rarer medium. The critical angle for total internal reflection is θ iC and the Brewster's angle of incidence is θ iB, such that sin θ iC/sin θ iB = η = 1.28. The relative refractive index of the two media is: (a) 0.2 (b) 0.4 (c) 0.8 (d) 0.9

Answer: (c). Let n = nrarerndenser < 1. Then sin θC = n and tan θB = n, so sin θB = (n)/(√(1+n²)). η = (sin θC)/(sin θB) = √(1+n²) = 1.28 n² = 1.6384 - 1 = 0.6384 ⇒ n ≈ 0.8 So the answer is (c) 0.8.

A ray of light is incident from a denser to a rarer medium. The… | JEE Main 2014 (19 Apr, Shift 1)

Q59JEE Main 2014 · 19 APR · SHIFT 1Difficult

Polarisation

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:

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Answer: (c)

Let $n = \frac{n_{rarer}}{n_{denser}} < 1$ . Then $\sin\theta_C = n$ and $\tan\theta_B = n$ , so $\sin\theta_B = \frac{n}{\sqrt{1+n^2}}$ .
$\eta = \frac{\sin\theta_C}{\sin\theta_B} = \sqrt{1+n^2} = 1.28$
$n^2 = 1.6384 - 1 = 0.6384 \Rightarrow n \approx 0.8$
So the answer is (c) $0.8$ .

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Polarisation question from JEE Main 2014 (19 Apr, Shift 1)

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