PRAYOGX
ADV-2026-P2-PHY-Q03

Prism Spectrum Lab — at which wavelength is the minimum deviation smallest?

JEE Advanced 2026 · Paper 2 · Physics · Q.3 · single correct option (+3 / 0 / −1) · Ray Optics and Optical Instruments

01The question

Q.3 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 + βλ2, where α = 3 μm−1 and β = 0.096 μm2. If λmin is the wavelength at which the angle of minimum deviation Dm is smallest, then the correct value of Dm at λmin is
  1. (A)6.4°
  2. (B)4.8°
  3. (C)3.2°
  4. (D)2.4°

02The experiment

Target:smallest minimum deviation Dm

Ready.

6.0°
3.0 μm⁻¹
0.096 μm²
0.600 μm
Speed

03Detailed solution

Answer:  Dm = 4.8° at λmin = 0.4 μm

  1. Thin prism. For a small prism angle A the minimum deviation is Dm = (n − 1)A. With A fixed, Dm is smallest exactly where the refractive index n(λ) is smallest.
  2. Minimise n(λ). dn/dλ = α − 2β/λ3 = 0 gives λ3 = 2β/α = 2 × 0.096/3 = 0.064 μm3, so λmin = 0.4 μm. It is a minimum: d2n/dλ2 = 6β/λ4 > 0.
  3. The smallest index. n = 3 × 0.4 + 0.096/0.16 = 1.2 + 0.6 = 1.8.
  4. The deviation. Dm = (1.8 − 1) × 6° = 4.8°: option (B).
  5. Check with the exact prism formula. n = sin[(A + Dm)/2]/sin(A/2) gives Dm = 2 sin−1(1.8 sin 3°) − 6° = 4.81°, within 0.3 % of the thin-prism value. The experiment measures this exact value by tracing rays.
  6. The other options. (C) 3.2° and (D) 2.4° would need n = 1.53 and 1.4, below the smallest index the material ever has (1.8), so no wavelength gives them. (A) 6.4° needs n = 2.07, which the material does reach, but at other wavelengths (0.28 μm and 0.6 μm), where Dm is not the smallest.

Understand the concept

A prism deviates light least when the ray passes through it symmetrically; that smallest deviation for a given colour is the angle of minimum deviation. For a thin prism it is simply (n − 1)A, so the colour that is bent least is the one for which the glass has the lowest refractive index. Here the index has two parts that pull in opposite directions: β/λ2 falls as the wavelength grows, while αλ rises. Their sum has a minimum at λ = (2β/α)1/3, and that wavelength is deviated less than any other. The law in the question describes a hypothetical material; ordinary glass has n falling steadily across the visible range.

Key takeaways

  • Thin prism: Dm = (n − 1)A; exact: n = sin[(A + Dm)/2]/sin(A/2).
  • Minimum deviation needs the symmetric path, with equal angles of incidence and emergence.
  • Smallest Dm across wavelengths ⇔ smallest n(λ): set dn/dλ = 0 and check that it is a minimum.
  • An option that needs n below the material's minimum can be ruled out at once.

Verification: an independent derivation; exact rational arithmetic for the thin-prism result; the exact prism formula; a brute-force ray trace over wavelength and angle of incidence that shares no code with the page; a check of every option against the range of n; and the page's own engine, run by tests/verify_p2phyq03.py. The MathonGo solution was read first only for the approach. The official final answer key (JEE Advanced 2026, Paper 2), read last, gives (B), in agreement. Status: script-verified by the PrayogX simulation factory; no review by a person has been recorded.

How to use this simulation

  1. Press ▶ START EXPERIMENT. White light passes through the prism and spreads into its colours on the screen.
  2. The monochromator then picks one wavelength and the turntable rotates the prism: the deviation dips to a minimum, the angle of minimum deviation.
  3. Next the wavelength is scanned and each minimum deviation is plotted; a search homes in on the wavelength where it is smallest, and the result is matched to the options.
  4. Then move the probe wavelength, change A, α or β, or try What if… rotate the prism at λmin.
  5. Replay runs it again. Reset clears it. Classroom fills the screen for a projector.