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

From Fig. (a)
i=angle of incidence=`60^(@)`
r= angle of refraction=`35^(@)`
`therefore` Refractive index of glass with repect to water,
`""_(a)mu_(g)=sini/sinr=sin 60^(@)/sin35^(@)=0.8660/0.5736=1.51`
From Fig. (b), here, i=`60^(@)` and r=`47^(@)`
`therefore` Refractive index of water with respect to air,
`""_(a)mu_(w)=sini/sinr=sin60^(@)/sin47^(@)=0.8660/0.7314=1.18`
`therefore` we have `""_(a)mu_(g)=""_(a)mu_(w)xx""_(w)mu_(g)`
or `""_(w)mu_(g)=(""_(a)mu_(g))/(""_(a)mu_(w))=1.51/1.81=1.28`
From Fig.(c ), i=angle of incidence=`45^(@)`
`therefore` Using the relation
`""_(w)mu_(g)=sini/sinror1.28=sin45^(@)/sinr`
or `sinr=(sin45^(@))/1.28=(1//sqrt2)/1.28`
`=0.707/1.28=0.5525=sin33.54'`
`therefore` Angle of refraction,r=`33.54^(@)`