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
B
0.4
C
0.8
D
0.9

Solution

## The correct option is C 0.8Let $$'\theta_c'$$ 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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