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,
![\theta_c=\sin ^(-1)((n_0)/(n))](https://img.qammunity.org/2023/formulas/physics/college/vr9p5nof5onw9hffm2lhamyhmf0y1hhmmh.png)
Where n=1 is the refractive index of the air.
On substituting the known values,
![\begin{gathered} \theta_c=\sin ^(-1)((0.48)/(1)) \\ =28.7\degree \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/p9679ny2fh0gz8cl4g9wsw8ptct96gnho0.png)
Thus the critical angle for carbon bisulphide for blue light is 28.7°