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

Case (a), Angle of incidence, `i=60^(@)`
Angle of refraction, `r=35^(@)`
Refractive index of glass with respect to air is
`""_(a)mu_(g)=(sini)/(sinr)`
`rArr" "_(a)mu_(g)=(sin 60^(@))/(sin35^(@))=1.51`
Case (b), Angle of incidence, `i=60^(@)`
Angle of refraction, `r=47^(@)`
Refractive index of water with respect to air is
`""_(a)mu_(w)=(sin 60^(@))/(sin 47^(@))=1.18`
Case (c),
Refractive index of glass with respect to water is
`""_(w)mu_(g)=""_(a)mu_(g)xx""_(w)mu_(a)`
`=(""_(a)mu_(g))/(""_(a)mu_(w))=(1.51)/(1.18)=1.28`
Angle of incidence `i=45^(@)`
Angle of refraction `r=?`
`therefore" "(w)mu_(g)=(sini)/(sinr)`
`=(sini)/(sinr)=1.28`
`sinr=(sin45^(@))/(1.28)=0.5524`
`sinr=sin33^(@)36.`
`r=33^(@)36.`