Physics - Geometrical Optics Question with Solution | TestHub

${\mathrm{\mu }}_1:$refractive index of rarer medium

${\mathrm{\mu }}_2:$retractive index of denser medium

${\mathrm{sin\theta }}_{\mathrm{C}}=\frac{{\mathrm{\mu }}_1}{{\mathrm{\mu }}_2}$

${\mathrm{tan\theta }}_{\mathrm{B}}=\frac{{\mathrm{\mu }}_1}{{\mathrm{\mu }}_2}$

$\sin {\theta }_C=\tan {\theta }_B$

$\frac{\sin {\theta }_C}{\sin {\theta }_B}=\frac{\tan {\theta }_B}{\sin {\theta }_B}=\frac{1}{\cos {\theta }_B}=1.28$

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

$=0.8$