Physics - Geometrical Optics Question with Solution | TestHub

$\mathrm{A}=60°$and${\mathrm{\delta }}_{\mathrm{m}}=30°$

$\mathrm{\mu }=\frac{\sin \left(\frac{\mathrm{A}+{\mathrm{\delta }}_{\mathrm{m}}}{2}\right)}{\sin \frac{\mathrm{A}}{2}}$

$\mathrm{\mu }=\frac{\sin \left(\frac{60+30}{2}\right)}{\sin \frac{60}{2}}$

$\mathrm{\mu }=\frac{\sin 45°}{\sin 30°}$

$\mathrm{\mu }=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}$

$\mathrm{\mu }=\sqrt{2}$

Therefore,${\mathrm{\theta }}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}}={\sin }^{-1}\left(\frac{1}{\mathrm{\mu }}\right)$

${\mathrm{\theta }}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}}=45°$

Therefore after$1^{\mathrm{s}\mathrm{t}}$incidence light goes straight. At second incidence, total internal reflection takes place, and hence deviation is$\mathrm{\delta }=60°$.