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Smallest angle of minimum deviation of a thin prism whose refractive index varies with wavelength

A thin 6 deg prism with n = alpha lambda + beta/lambda^2: rotate it to find the minimum deviation, scan the wavelength, and find where Dm is smallest. lambda_min = (2 beta/alpha)^(1/3) = 0.4 um, n = 1.8, Dm = 4.8 deg - option B.

Sub-topics covered

• Angle of minimum deviation: the symmetric path through a prism
• Thin prism: Dm = (n - 1)A; exact: n = sin((A + Dm)/2)/sin(A/2)
• Dispersion: refractive index as a function of wavelength
• Minimising n(lambda) = alpha lambda + beta/lambda^2 by calculus

What it teaches

• For a thin prism Dm = (n - 1)A, so the smallest Dm across wavelengths is at the smallest n(lambda)
• dn/dlambda = alpha - 2 beta/lambda^3 = 0 gives lambda_min = (2 beta/alpha)^(1/3) = 0.4 um; n'' > 0 so it is a minimum
• n(0.4) = 1.2 + 0.6 = 1.8, so Dm = 0.8 x 6 = 4.8 deg - option B; the exact prism formula gives 4.81 deg
• Options C (3.2) and D (2.4) need n = 1.53 and 1.4, below the material's minimum 1.8; option A (6.4) needs n = 2.07, reached only at 0.28 and 0.6 um
• Minimum deviation is found experimentally by rotating the prism: the deviation dips to its least value on the symmetric path

Simulation ID

• ADV-2026-P2-PHY-Q03 · revision 1

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