Numerical Aperture of Optical Fibre: Formula, Angle and Common Slips

The numerical aperture of an optical fibre is n₀ sin θₐ, which equals √(n₁² − n₂²) for a step-index fibre, where n₀ is the index of the surrounding medium, n₁ is the core refractive index and n₂ is the cladding index. In air, n₀ = 1, so it is the sine of the largest entry angle measured from the fibre axis (Virtual Labs, RP Photonics).

A step-index fibre has a core of uniform index surrounded by a cladding of lower index. Light stays in the core by total internal reflection, which needs n₁ > n₂. The largest entry angle is the acceptance angle θₐ, the half-angle of the acceptance cone. The cone's full vertex angle is therefore 2θₐ.

Choose the row by reading what the question gives, before you touch the calculator.

The Δ row is the standard weak-guidance approximation. It follows from n₁² − n₂² = (n₁ − n₂)(n₁ + n₂) ≈ 2n₁²Δ when the indices are nearly equal. For core 1.48 and cladding 1.45, it gives 0.298 against the exact 0.2965.

NA stays fixed because √(n₁² − n₂²) contains only the fibre's own indices, while the angle comes from n₀ sin θₐ = NA. A denser surrounding medium gives a smaller θₐ, and NA stays unchanged (RP Photonics).

Take core n₁ = 1.48 and cladding n₂ = 1.45 (our worked example):

1. n₁² = 2.1904 and n₂² = 2.1025, so n₁² − n₂² = 0.0879.
2. NA = √0.0879 = 0.2965.
3. In air, θₐ = sin⁻¹(0.2965) = 17.25°, so the full cone is 34.5°.

Now keep the same fibre and change only the surroundings. We assumed n₀ = 1.33 for water and 1.47 for an oil, purely as illustration values.

As a cross-check on method, a core of 1.46 and cladding of 1.45 gives NA 0.1706 and θₐ 9.82°, matching the roughly 0.17 and 9.8° half-angle, about 20° cone, in the Fiber Optic Association reference guide.

In our view, four slips are easy to make: treating n₁ − n₂ as NA, stopping before the square root, mixing the half-angle with the cone, and ignoring the surrounding medium. Each has a quick test, using the 1.48 and 1.45 fibre above.

A value of 1 or more. If the formula returns a value of 1 or more (core 1.5 and cladding 1.0 give 1.118), sin⁻¹ cannot be applied. In air, our reading is that every entry angle up to 90° is then accepted, so the answer is "no limiting angle", not an error message.

Two Corning telecom fibres have lower NA than the textbook 1.48 and 1.45 fibre. Corning lists 0.14 for its SMF-28 ULL single-mode fibre and 0.200 ± 0.015 for its ClearCurve OM2, OM3 and OM4 multimode fibres (ClearCurve is Corning's product name). Our 1.48 and 1.45 exercise gives 0.2965. Other fibres, including some multimode types, can have higher NA, so these two are examples, not a rule.

The half-angles and cones are our calculation, sin⁻¹(NA) in air, using the nominal NA values. The links are to distributor-hosted copies of the sheets, the editions we checked as of November 2026.

The SMF-28 ULL sheet states that NA "is measured at the one percent power level of a one-dimensional far-field scan at 1310 nm". So the SMF-28 ULL NA is measured, not computed from n₁ and n₂.

Light in single-mode fibres cannot be described properly by simple ray geometry, so the ray-based formula is not the right tool there. For graded-index fibres (the index falls smoothly from the core outward) the NA idea is questionable too: some authors apply the step-index formula using the maximum index difference (RP Photonics). In an exam, use √(n₁² − n₂²) whenever n₁ and n₂ are given for a step-index fibre.

You measure NA by placing a detector at a fixed distance d from the fibre end, sliding it sideways, and finding the spot radius r where intensity falls to 37% of the maximum. Then sin θ = r ÷ √(r² + d²), and NA is that value (Virtual Labs procedure).

Steps from the procedure:

1. Fix d (the example uses 4 mm).
2. Move the detector sideways in 0.5 mm steps and record intensity at each position.
3. Plot intensity against displacement.
4. Read r at Imax ÷ 2.71, which is 37% of the maximum.
5. Calculate sin θ, then θ = sin⁻¹ of that value.

With assumed readings d = 4 mm and r = 1.2 mm (an illustration, not a real run): √(1.44 + 16) = 4.176, so sin θ = 0.2874 and θ = 16.7°.

Likely viva questions and short answers:

• Define NA. The sine of the acceptance half-angle in air.
• Why must n₁ be greater than n₂? Total internal reflection at the core and cladding boundary needs the core to have the higher index.
• What does a larger NA mean? The fibre collects light from a wider cone (Virtual Labs).
• Does NA change in water? No, but θₐ becomes smaller.
• What do you plot in the experiment? Intensity against sideways detector displacement.

In the SPPU First Year Engineering 2024 Pattern, ESC102 Engineering Physics, optical fibres sit in Unit I, Fundamentals of Photonics (8 hours). The syllabus lists critical angle, acceptance angle, acceptance cone, numerical aperture, total internal reflection, fibre types, attenuation and "numerical problems on parameters of optical fiber" (SPPU syllabus document, a college-hosted copy).

The course carries 30 marks of Comprehensive Continuous Evaluation (CCE), 70 marks in the end-semester theory exam and 25 marks of term work. Lab item 2 asks you to determine the numerical aperture, attenuation coefficient or critical angle. Newton's rings is another item on the same lab list, and our page on Newton's rings covers it.

For planning the rest of the course, see our guide to Engineering Physics 1. For a four-step routine you can apply to any numerical, see how to solve physics numericals.

At RG Lectures we teach Physics the Padhna Aasaan Hai way, with free lectures, live sessions and revision videos on our YouTube channel. For first-year SPPU engineering, our FE Engineering course covers Physics, Mechanics and Drawing: see it in the RG Lectures store.