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Figures 9.31(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. 9.31(c)]. |
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Answer» Solution :For (a), refractive index of glass with respect to air, `""^(a)mu_g = (SIN i)/(sin r) = (sin 60^@)/(sin 35^@) = (0.8660)/(0.5736) = 1.51 ` (B) Refactive index of glass with respect to air, `""^(a) mu_(W) = (sin i)/(sin r) = (sin 60^@)/(sin 47^@) = (0.8660)/(0.6561) = 1.32 ` (c) Refractive index of glass w.r.t water. `""^(w)mu_(G) = (a_(mug))/(a_(mug))= 1.51/1.32= 1.144` `""^(w)mug = (sin i)/(sinr) = (sin 45^@)/(sinr)` `(""^(a)mu_g)/(""^(a)mu_g) = (sin 45^@)/(sinr )` `THEREFORE (1.51)/(1.32) = (0.7071)/(sinr)` `therefore sin r = (0.7071)/(sinr)` `therefore sin r = (0.7071 xx 1.32)/(1.51) = 0.6181` From log table` r = 38.2^@` |
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