Answer to Question #93218 in Optics for modrick

Question

calculate the critical angle of a material having a refractive index of 1.52 when light pass from this material to air

Answer 1

The critical angle

$$\theta=\sin^{-1}\left(\frac{n_r}{n_i}\right)=\sin^{-1}\left(\frac{1.00}{1.52}\right)=41.1^{\circ}$$