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 Brewster's angle of incidence is $\theta_{ iB }$, such that $\frac{\sin \theta_{ iC }}{\sin \theta_{ iB }}=\eta=1.28$. The relative refractive index of the two media is

• A. $\text{0.4}$
• B. $\text{0.2}$
• C. $\text{0.9}$
• D. $\text{0.8}$

Answer 1

Answer: (D)

$\mu_{1}:$ refractive index of rarer medium
$\mu_{2}:$ retractive index of denser medium
$\sin \theta_{C}=\frac{\mu_{1}}{\mu_{2}}$
$\tan \theta_{B}=\frac{\mu_{1}}{\mu_{2}}$
$\sin \theta_{C}=\tan \theta_{B}$
$\frac{\sin \theta_{C}}{\sin \theta_{B}}=\frac{\tan \theta_{B}}{\sin \theta_{B}}=\frac{1}{\cos \theta_{B}}=1.28$
$\tan (\theta)_{B}=\frac{(\mu)_{1}}{(\mu)_{2}}=\frac{\sqrt{1-\left(\frac{1}{1.28}\right)^{2}}}{\left(\frac{1}{1.28}\right)}$
$=0.8$