Question

1. The speed of light in a material is 0.52c. What is the critical angle of a light ray at the interface between the material and a vacuum?

Answer 1

${v\,of\,light\,in\,Material=0.52c}\\ {\theta_c=sin^{-1}(\frac{n_{light}}{n_{material}})}\\ {n_{material}=\frac{c}{v}}\\ {n_{material}=\frac{c}{0.52c}}\\ {n_{material}=\frac{1}{0.52}}\\ {n_{material}=1.92}\\ {\theta_c=sin^{-1}(\frac{1}{1.92})}\\ {\theta_c=31.39\degree}\\ {\theta_c=31\degree}\\$