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Question

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 $$sin\theta_{iC}/sin\theta_{iB}=n=1.28$$. The relative refractive index of the two media is :


A
0.2
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B
0.4
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C
0.8
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D
0.9
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Solution

The correct option is C 0.8
Let $$'\theta_c'$$ be the critical angle.
$$n_1sin\theta_c=n_2sin90^o$$
$$\displaystyle sin\theta_c=\frac{n_2}{n_1}$$
Brewster's angle 
$$\displaystyle tan\theta_b= \frac{n_2}{n_1}$$
$$\displaystyle sin\theta_b= \frac{n_2}{\sqrt{n_1^2+n_2^2}}$$
Given, $$\displaystyle \frac{sin\theta_c}{sin\theta_b}=1.28$$
$$\displaystyle \frac{\frac{n_2}{n_1}}{\frac{n_2}{\sqrt{n_1^2+n_2^2}}}=1.28$$
Solving the above equation, we get,
$$\displaystyle \frac{n_2}{n_1}=0.8$$

Physics

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