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
$\sin {\theta }_C=\frac{{\mu }_1}{{\mu }_2}$
$\tan {\theta }_B=\frac{{\mu }_1}{{\mu }_2}$
$\sin {\theta }_C=\tan {\theta }_B$
$\frac{\sin {\theta }_C}{\sin {\theta }_B}=\frac{\tan {\theta }_B}{\sin {\theta }_B}=\frac{1}{\cos {\theta }_B}=1.28$
$\tan (\theta {)}_B=\frac{(\mu {)}_1}{(\mu {)}_2}=\frac{\sqrt{1-{\left(\frac{1}{1.28}\right)}^2}}{\left(\frac{1}{1.28}\right)}$
$=0.8$