The diagram shows an equilateral prism. The medium on one side of the prism is μ 1 . The refractive index of the prism is μ = 
. The diagram shows variation of magnitude of angle of deviation with respect to μ 1 . Consider the light ray to be normally incident on the first face.
(i) Value of k 2 is–
Text Solution
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Ans.
(i)
Sol.
r 1 = 0
r 1 + r 2 = A
r 2 = Α = 60°
μ sin r 2 = sin e × μ 1
μ 1 sin e = 
× sin 60°
sin e = 
= 
β = e – r 2 (Deviation)
when μ 1 = k 2
B = e – r 2 = 0
∴ e = r 2 = 60°
sin 60° = 2 × 
= 
μ 1 = 
∴ k 2 = 
(ii)
Sol. When μ 1 < k 1
light will not emerge, it can be seen from graph
∴ r 2 = θ c when μ 1 = k 1
sin r 2 = sin θ c = 
sin 60° = 
= 
× 
μ 1 = 
∴ k 1 = 
(iii)
Sol. β 1 is maximum deviation β 2 is minimum deviation
β 2 = 
– 60 ° = 30°
β 1 – β 2 = 60° – 30° = 30°
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