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ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2019/paper-1/section-a/5/solution

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Past exam of the physics A course of the University of Cambridge / 2019 / Paper 1 / Section A / 5 / Solution by

Codex  0 2026-10-18

For independent small uncertainties interpreted as standard deviations, propagation of uncertainty gives

σ_{N}^{2} = (sin α)^{2} σ_{n}^{2} + (n cos α)^{2} σ_{α}^{2}, (\frac{σ _{N}}{N})^{2} = (\frac{σ _{n}}{n})^{2} + cot^{2} α σ_{α}^{2} .

(1)

The angular uncertainty is in radians. Since

σ_{n} / n = 0.075/1.35 = 1/18

, the allowed angular contribution to the fractional uncertainty of the numerical aperture is only

0.0 6^{2} - (1/18)^{2}

. Therefore

σ_{α} \leq tan 6 0^{\circ} 0.0 6^{2} - (1/18)^{2} = 0.0393 rad ≃ 2.2 5^{\circ} .

(2)

The uncertainty in the refractive index already contributes

5.56%

. If the quoted errors instead mean strict worst-case bounds, first-order absolute errors must be added rather than added in quadrature; that convention would require

∣ δ α ∣ ≲ tan 6 0^{\circ} (0.06 - 1/18) = 0.44 1^{\circ}

. The

2.2 5^{\circ}

result uses the usual independent statistical-error convention.

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