Question

# Figures 9.34(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 incidencein water is 45° with the normal to a water-glass interface [Fig. 9.34(c)].

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Solution

## Given: The incident angle of the ray in the air is 60° and the angle of incidence in water is 45° with the normal to a water-glass interface. For air-glass interface: The refractive index of glass with respect to air is given as, μ g a = sini sinr where, the angle of incidence is i and the angle of refraction is r. By substituting the given values in the above expression, we get, μ g a = sin 60 0 sin 35 0 =1.51 (1) For air-water interface: The refractive index of glass with respect to water is given as, μ w a = sini sinr where, the angle of incidence is i and angle of refraction is r. By substituting the given values in the above expression, we get, μ w a = sin 60 0 sin 47 0 = 0.866 0.731 =1.184 (2) From (1) and (2), the refractive index of glass with respect to water is given as, μ g w = μ g a μ w a = 1.51 1.184 =1.275 For water-glass interface: The refractive index of glass with respect to water is given as, μ g w = sini sinr where, the angle of incidence is i and angle of refraction is r. By substituting the given values in the above expression, we get, μ w a = sin 45 0 sinr 1.275= 0.7071 sinr r= sin −1 ( 0.5546 ) = 38.68 0 Therefore, the angle of refraction at the water-glass interface is 38.68 0 .

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