Fig. (a) and (b) show refraction of an incident ray in air at `60^@` with the normal to a glass-air and water-air interface respectively. Predict the angle of refraction of an incident ray in water at `45^@` with the normal to a water glass interface. Take `.^a mu_g = 1.32`.
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Explanation

As per the given figure, for the glass-air interface:
Angle of incidence, I = `60^(@)`.
Angle of refraction, `r=p35^(@)`.
The relative index of glass with respect to air is given by Snell's law as:
`mu_(s)^(a) = (sini)/(sinr)`
`=(sin60^(@))/(sin35^(@)) = 0.8660/0.5736=1.51`..............(1)
As per the given figure, for the air-water interface.
Angle of incidence, i=`60^(@)`
Angle of refraction, r=`47^(@)`
the relative index of water with respect to air is given by Snell's law as:
`u_(w)^(s) = (sini)/(sinr)`
`=(sin60)/(sin47) = 0.8660/0.7314=1.184`.............(2)
Using 1 and 2, the relative refractive index of glass with respect to water can be obtained as:
`mu_(g)^(w)=mu_(g)^(s)/mu_(w)^(a)`
`=(sin60)/sin47)= 0.8660/0.7314=1.184`.............(2)
using 1 and 2, the relative index of glass with respect to water can be obtained as
`u_(g)^(w)= mu_(g)^(a)/mu_(w)^(a) = (1.51)/(1.184) = 1.275`
The following figure shows the situations involving the glass-water interface.

Angle of incidence, i=`45^(@)`.
Angle of refraction =r
From snell's law, r can be calcualed as: `(sini)/(sin3)=mu_(g)^(w)`
`(sin45^(@))/(sinr)= 1.275`
`sinr = (1/sqrt(2))/(1.275) = 0.5546`
`therefore r=sin^(-1)(0.5546) = 36.68^(@)`
Hence,the angle of refraction at the water - glass interface is `38.68^(@)`.