A ray of light travels through the material of an equilateral prism with speed 2.12×10⁸ m·s⁻¹. Taking the speed of light in vacuum as 3.0×10⁸ m·s⁻¹, what is the minimum angle of deviation produced by the prism?
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Question

A ray of light travels through the material of an equilateral prism with speed 2.12×10⁸ m·s⁻¹. Taking the speed of light in vacuum as 3.0×10⁸ m·s⁻¹, what is the minimum angle of deviation produced by the prism?

Options

G1 Physics · Accepted Answer

The prism is equilateral, so A = 60°. The refractive index is μ = c ÷ v = 3.0×10⁸ ÷ 2.12×10⁸ ≈ 1.415 ≈ √2. At minimum deviation, μ = sin[(A+δ(min)) ÷ 2] ÷ sin(A ÷ 2). Thus, √2 = sin[(60°+δ(min)) ÷ 2] ÷ sin30°. Since sin30° = 1 ÷ 2, sin[(60°+δ(min)) ÷ 2] = 1 ÷ √2. Therefore, (60°+δ(min)) ÷ 2 = 45°. Hence, δ(min) = 30°. Correct option: B.