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

Answer 1

The correct option is C 0.8

Let ${}^{\prime }{\theta }_c^{\prime }$ be the critical angle.
$n_{1}sin{\theta }_c=n_{2}sin{90}^o$
$sin{\theta }_c=\frac{n_2}{n_1}$
Brewster's angle
$tan{\theta }_b=\frac{n_2}{n_1}$
$sin{\theta }_b=\frac{n_2}{\sqrt{n_1^2+n_2^2}}$
Given, $\frac{sin{\theta }_c}{sin{\theta }_b}=1.28$
$\frac{\frac{n_2}{n_1}}{\frac{n_2}{\sqrt{n_1^2+n_2^2}}}=1.28$
Solving the above equation, we get,
$\frac{n_2}{n_1}=0.8$