Given,
The refractive index of carbon bisulphide for blue light is n₀=0.48
The critical angle for the red light, θ_r=49°
The critical angle for carbon bisulphide is given by,
[tex]\theta_c=\sin ^{-1}(\frac{n_0}{n})[/tex]Where n=1 is the refractive index of the air.
On substituting the known values,
[tex]\begin{gathered} \theta_c=\sin ^{-1}(\frac{0.48}{1}) \\ =28.7\degree \end{gathered}[/tex]Thus the critical angle for carbon bisulphide for blue light is 28.7°
