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Figures (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 a refraction in glass when the angle of incidence in water is 45^(@) with the normal to a water-glass interface . |
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Answer» Solution :FIG (a) `n_(ga) = (SIN 60)/( sin 35) ` = 1.51 ` n_(gw)= (n_(ga))/(n_(wa)) = (1.51)/(1.32) = 1.14` Also `n_(gw) = (sin 45)/(sin R)"" therefore sin r = (sin 45)/(1.14) = (0.7071)/(1.14) = 0.6203` `r = sin^(-1) (0.6203) = 38^(@)21` |
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