Question

A ray of light is incident from a denser to a rarer medium. The critical angle for total internal reflection is $\theta _{\text{iC}}$ and Brewster's angle of incidence is $\theta _{\text{iB}}$ , such that $\frac{\text{sin} \theta _{\text{iC}}}{\text{sin} \theta _{\text{iB}}}=\eta=1\text{.}28$ . 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
$sin\theta _{C}=\frac{\mu _{1}}{\mu _{2}}$
$tan\theta _{B}=\frac{\mu _{1}}{\mu _{2}}$
$sin \theta _{C} = tan \theta _{B}$
$\frac{sin \theta _{C}}{sin \theta _{B}} = \frac{tan \theta _{B}}{sin \theta _{B}} = \frac{1}{cos \theta _{B}} = 1.28$
$\tan \theta_B=\frac{\mu_1}{\mu_2}=\frac{\sqrt{1-\left(\frac{1}{1.28}\right)^2}}{\left(\frac{1}{1.28}\right)}$
$= 0.8$