Question

If refractive index of glass is $1.50$ and of water is $1.33$, then critical angle is

• A. ${{\sin }^{-1}}\left( \frac{8}{9} \right)$
• B. ${{\sin }^{-1}}\left( \frac{2}{3} \right)$
• C. ${{\cos }^{-1}}\left( \frac{8}{9} \right)$
• D. None of these

Answer 1

Answer: (A)

${{\sin }^{-1}}\left( \frac{8}{9} \right)$

Answer 2

When a ray of light passes from a denser medium to a rarer medium, it bends away from the normal at the interface of the two media. The angle of incidence is measured with respect to the normal at the refractive boundary. It is given by $C=\sin ^{-1}\left(\frac{n_{2}}{n_{1}}\right)$ where, $C$ is critical angle, $n_{2}$ is the refractive index of rarer medium and $n_{1}$ of the denser medium. Given, $n_{2} =1.33, n_{1}=1.50$ $C =\sin ^{-1}\left(\frac{1.33}{1.50}\right)$ $C =\sin ^{-1}\left(\frac{8}{9}\right)$