Question

The critical angle for glass-water surface is (refractive index of glass = 1.5 and refractive index of water $=\frac{4}{3}$)

• A. $si{n}^{-1}\left(\frac{8}{9}\right)$
• B. $si{n}^{-1}\left(\frac{9}{8}\right)$
• C. $si{n}^{-1}\left(\frac{2}{3}\right)$
• D. $si{n}^{-1}\left(\frac{4}{3}\right)$

Answer 1

The correct option is A $si{n}^{-1}\left(\frac{8}{9}\right)$

$\frac{{\mu }_{water}}{{\mu }_{glass}}=\frac{\sin i}{\sin r}=\frac{\sin {\theta }_c}{\sin {90}^{\circ }}$

$\Rightarrow \sin {\theta }_c=\frac{{\mu }_{water}}{{\mu }_{glass}}$

$\Rightarrow {\theta }_c={\sin }^{-1}\left(\frac{{\mu }_{water}}{{\mu }_{glass}}\right)$

$\Rightarrow {\theta }_c={\sin }^{-1}\left(\frac{4}{3\times 1.5}\right)({\mu }_{water}=\frac{4}{3}{\mu }_{glass}=\frac{1}{1.5})$
$={\sin }^{-1}\left(\frac{8}{9}\right)$
Hence, the correct answer is option (a).