Question

The critical angle of incidence in a glass salb placed in air is 45°. What will be the critical angle when it is immersed in water of refractive index 1.33.

Answer 1

Dear Student,

From Snell's Law,

$$\begin{gathered} {\mu }_1\sin i_c={\mu }_2\sin 90 \\ whenglassisinair, \\ {\mu }_2=1 \\ or{\mu }_1=\frac{1}{\sin i_c}..\left(i\right) \\ whenglassisinwater, \\ {\mu }_1\sin i_c={\mu }_2\sin 90 \\ here, \\ {\mu }_2=1.33 \\ or,\sin i_c=\frac{{\mu }_2}{{\mu }_1}\Rightarrow {\mu }_2\sin i_c \\ {i}_c={\sin }^{-1}\left({\mu }_2\sin i_c\right)\Rightarrow {\sin }^{-1}\left(1.33\times \frac{1}{\sqrt{2}}\right)\approx 70° \end{gathered}$$
Regards.