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Fig. 9.02(a) and (b), show refraction of...

Fig. 9.02(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 refraction in glass when the angle of incidence in water is `45^(@)` with the normal to a water - glass interface [Fig. 9.02(c)].

Text Solution

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From fig (a) `n_(ga) =(sin 60^(@))/(sin 35^(@)) =0.8660/0.5736 = 1.51`
and from fig (b) : `n_(wa) = (sin 60^(@))/(sin 47^(@)) = 0.8660/0.6561 = 1.32`
`therefore n_(gw) = n_(ga)/n_(wa) = 1.51/1.32 = 1.144`
Now, in fig. ( c) if angle of refraction be r, then
`(sin 45^(@))/(sin r) = n_(gw) = 1.144 rArr sin r =(sin 45^(@))/(1.144) = 0.7071/1.144 = 0.6181` and `r = sin^(-1) (0.6181) = 38^(@)`
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