Figures (a) and (b) show refraction of a ray in air incident at `60^(@)` with the normal to a glass-air and water-air interface, respectively. Predict the angle of a refraction in glass when the angle of incidence in water is `45^(@)` with the normal to a water-glass interface .

Explanation

Fig (a)
`n_(ga) = (sin 60)/( sin 35) ` = 1.51
` n_(gw)= (n_(ga))/(n_(wa)) = (1.51)/(1.32) = 1.14`
Also `n_(gw) = (sin 45)/(sin r) " " therefore sin r = (sin 45)/(1.14) = (0.7071)/(1.14) = 0.6203`
`r = sin^(-1) (0.6203) = 38^(@)21`