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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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