If the critical angle is 48°42’ when the media concerned are air and water and 36°47’ when they are in air and glass, what is it when they are in water and glass.

52°52'

We know that

$\sin C =\frac{1}{m_w}$

$m_w =\frac{1}{\sin C}=\frac{1}{\sin 48°42'}$

Similarly $m_g =\frac{1}{\sin 36°47'}$

 Now for glass and water media

sin C = refractive index of water w.r.t. glass

$\frac{m_w}{m_g}=\frac{1/\sin 48°42'}{\sin 36°47'}=\frac{\sin 36°72'}{\sin 48°42'}$ or $C=52°52'$