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

In Fig. `I = 60^(@), r = 35^(@)`
`.^(a)mu_(g) = (sin i)/(sin r) = (sin 60^(@))/(sin 35^(@)) = (0.8660)/(0.5736) = 1.51`
In Fig. `I = 60^(@), r = 41^(@)`
`.^(a)mu_(w) = (sin i)/(sin r) = (sin 60^(@))/(sin 47^(@)) = (0.8660)/(0.7314) = 1.18`
In Fig. `i = 45^(@), r = ?`
`.^(w)mu_(g) = (.^(a)mu_(g))/(.^(a)mu_(w)) = (sin i)/(sin r)`
`(1.51)/(1.32) = (sin 45^(@))/(sin r) = (0.7071)/(sin r) :. sin r = (1.32 xx 0.7071)/(1.51) = 0.6181`
`r = 38.2^(@)`