Find the wavelength where the prism gives its smallest deviation
For a thin prism, the angle of minimum deviation is controlled by the refractive index: Dm ≈ (n − 1)A. The material has competing wavelength terms, so the demonstration searches for the wavelength where n(λ), and therefore Dm, is minimum.
2026 Paper 2 Q3 · thin-prism dispersion
Drag the wavelength slider · the minimum appears at λ = 0.4 μm
Observation stage
Wavelength0.40 μm
Refractive-index curve
The prism material follows n(λ) = αλ + β/λ². One term rises with wavelength, while the other falls rapidly.
Expressionn = 3λ + 0.096/λ²
Current valuen = 1.800
Wavelength λshown in micrometres
Minimum pointwhere dn/dλ = 0
Deviation Dmthin-prism result (n − 1)A
Derivation path
01 / THIN PRISM
Deviation follows refractive index
For a thin prism with small prism angle A, the minimum deviation is approximately proportional to n − 1.
Dm = (n − 1)A
02 / MINIMISE n
Set the wavelength derivative to zero
Since A is fixed, the smallest Dm occurs where n(λ) is smallest.
dn/dλ = α − 2β/λ³ = 0
03 / SUBSTITUTE
The optimum wavelength is 0.4 μm
With α = 3 μm⁻¹ and β = 0.096 μm², λ³ = 2β/α = 0.064.