If the critical angle is `48^(@)42'` when the media concerned are air and water and `36^(@)47'` when they are air and glass, what is the critical angle when the media are water and glass.

Explanation

We know that
`SinC = (1)/(mu_(w))`
`mu_(w) = (1)/(Sin C) = (1)/(sin 48^(@)42')`
Similarly, `mu_(g) = (1)/(sin 36^(@)47')`
Now for glass and water media
sin C = refractive index of water w.r.t. glass
`= (mu_(w))/(mu_(g)) = (1//sin48^(@)42')/(1//sin 36^(@)47') = sin C`
`C = 52^(@)52'`