A prism is made of glass of unknown refractive index. A parallelbeam of light is incident on a face of the prism. The angle of minimumdeviation is measured to be 40°. What is the refractive index of thematerial of the prism? The refracting angle of the prism is 60°. Ifthe prism is placed in water (refractive index 1.33), predict the newangle of minimum deviation of a parallel beam of light.
Given: The angle of minimum deviation is 40 0 , refractive angle of the prism is 60 0 , the angle of prism is 60 0 , and the refractive index of water is 1.33 .
The refractive index of prism is given as,
μ ' = sin ( A + δ 2 ) sin ( A 2 )
Where, the angle of minimum deviation is δ and the angle of prism is A .
By substituting the given values in the above expression, we get,
μ ' = sin ( 60 0 + 40 0 2 ) sin ( 60 0 2 ) = sin 50 0 sin 30 0 = 1.532
Thus, the refractive index of prism is 1.532 .
The refractive index of glass with respect to water is given as,
μ g w = μ ' μ
Substitute the given values in the above expression.
μ g w = 1.532 1.33 = 1.518
And,
μ g w = sin ( A + δ ' 2 ) sin ( A 2 )
By substituting the given values in the above expression, we get,
1.518 = sin ( 60 0 + δ ' 2 ) sin ( 60 0 2 ) sin ( 60 0 + δ ' 2 ) = 0.5759 ( 60 0 + δ ' 2 ) = sin − 1 0.5759 δ ' = 10.32 0
Thus, the new minimum angle of deviation is 10.32 0 .
Given: The angle of minimum deviation is
The refractive index of prism is given as,
Where, the angle of minimum deviation is
By substituting the given values in the above expression, we get,
Thus, the refractive index of prism is
The refractive index of glass with respect to water is given as,
Substitute the given values in the above expression.
And,
By substituting the given values in the above expression, we get,
Thus, the new minimum angle of deviation is
