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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 :

Option: 1
0.2

Option: 2
0.4

Option: 3
0.8

Option: 4
0.9

Answers (1)

best_answer

$\sin\theta _{ic}=\frac{\mu_ r}{\mu _d}$

$\mu_r$ = refractive index of the rarer medium.

$\mu_d$ = refractive index of the rarer medium.

In the case of Brewster’s angle

$r = 90-\theta _{iB}$

From Brew's law: $\mu _{d}.\sin\theta _{iB}=\mu _{r}.\sin r$

$\frac{\sin\theta _{iB}}{\cos\theta _{iB}}=\frac{\mu _{r}}{\mu _{d}}$ or $\tan\theta _{iB}=\frac{\mu _{r}}{\mu _{d}}$

$\sin\theta _{iB}= \frac{\mu _{r}}{\sqrt{\mu _{r}^{2}+\mu _{d}^{2}}}....(2)$

$\because\frac{ \sin\theta _{ic}}{ \sin\theta _{iB}}=1.28$

$\mu _{1}^{2}+\mu _{d}^{2}=1.638\mu _d^{2}$

$or \ \ 0.638\mu d^{2}=\mu _{r}^{2}$

$\frac{\mu _{r}}{\mu _{d}}=\sqrt{0.638}=0.8$

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