A beam of polychromatic light passes through a thin prism of prism angle 6°. The refractive index of the material of the prism varies with wavelength λ as n(λ) = αλ + β/λ², where α = 3 μm⁻¹ and β = 0.096 μm². If λ(min) is the wavelength at which the angle of minimum deviation D(min) is smallest, then the correct value of D(min) at λ(min) is:
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Question

A beam of polychromatic light passes through a thin prism of prism angle 6°. The refractive index of the material of the prism varies with wavelength λ as n(λ) = αλ + β/λ², where α = 3 μm⁻¹ and β = 0.096 μm². If λ(min) is the wavelength at which the angle of minimum deviation D(min) is smallest, then the correct value of D(min) at λ(min) is:

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G1 Physics · Accepted Answer

For a thin prism, D(min) ≈ [n(λ) − 1]A. Therefore minimize n(λ) = αλ + β/λ². Setting dn/dλ = α − 2β/λ³ = 0 gives λ³ = 2β/α = 0.064 μm³, hence λ = 0.4 μm. Then n = 3(0.4) + 0.096/(0.4)² = 1.8. Thus D(min) = (1.8 − 1)×6° = 4.8°.