Physics - Geometrical Optics Question with Solution | TestHub

Solid angle$\left(\varphi \right)=2\pi \left(1-\cos \theta \right)$

where$\theta =$plane angle

$\text{escape energy}=\frac{\text{total energy}}{4\text{π}}\times \text{solid angle}$

$\frac{\text{escape energy}}{\text{total energy}}=\frac{1}{4\text{π}}\times 2\text{π}\left(1-\text{cosC}\right)$

$=\frac{1}{2}\left(1-\text{cosC}\right)\mathrm{ \; \; \; \; \; ....(1)}$

from law of refraction

${\mu }_w\sin i={\mu }_a\sin r$

$\frac{4}{3}\times \text{sin C}=1\times {\text{sin 90}}^{\text{o}}$

$\text{sin C}=\frac{3}{4}\Rightarrow \text{cos C}=\sqrt{1-{\text{sin}}^2\text{C}}\Rightarrow \text{cos C}=\frac{\sqrt{7}}{4}\text{....}\left(2\right)$

from (1) and (2)

${\mu }_w\sin i={\mu }_a\sin r$

$\frac{\text{escaped energy}}{\text{total energy}}=\text{0.17}$

total$\%$energy enscaped$=17\%$