---
title: "Glass-clad fiber numerical aperture"
subject: "Engineering Physics"
chapter: 6
key: "semester_1/engineering_physics:fiber-na-glass-clad-numerical"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/fiber-na-glass-clad-numerical
site: BCT Engineering
---

# Glass-clad fiber numerical aperture

Engineering Physics · Chapter 6: Laser and Fiber Optics

- Asked in: 2071 Shrawan, 5 marks
- Video: [Examples of Numerical Aperture, Acceptance Angle, Critical Angle & Refractive Index of Optical Fiber](https://www.youtube.com/watch?v=lGFCPaAI4FM) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/fiber-na-glass-clad-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-6

## Question

A glass-clad fiber is made with a core glass of refractive index 1.55 and the cladding is doped to give a fractional index difference of 5.5×10⁻⁴. Determine (i) Cladding index (ii) the critical internal reflection angle (iii) the external critical acceptance angle and (iv) numerical aperture (NA).

## Short answer

Cladding index n₂ = n₁(1−Δ) = 1.549; critical internal reflection angle φ_c = sin⁻¹(n₂/n₁) ≈ 88.1°; numerical aperture NA = n₁√(2Δ) ≈ 0.0514; external acceptance angle θ₀ = sin⁻¹(NA) ≈ 2.95°.

## Answer

The fractional index difference is Δ = (n₁ − n₂)/n₁.

**Given:** n₁ = 1.55, Δ = 5.5 × 10⁻⁴.

#### (i) Cladding index
```text
n₂ = n₁ (1 − Δ) = 1.55 × (1 − 5.5×10⁻⁴)
   = 1.55 × 0.99945
   = 1.5491
```

#### (ii) Critical internal reflection angle (core-cladding)
```text
sin φ_c = n₂/n₁ = 0.99945
φ_c = sin⁻¹(0.99945) = 88.10°
```

#### (iii) External critical acceptance angle (from air)
```text
sin θ₀ = NA = 0.0514
θ₀ = sin⁻¹(0.0514) = 2.95°
```

#### (iv) Numerical aperture
```text
NA = n₁ √(2Δ) = 1.55 × √(2 × 5.5×10⁻⁴)
   = 1.55 × √(1.1×10⁻³)
   = 1.55 × 0.03317
   = 0.0514
```
**Check:** √(n₁² − n₂²) = √(2.4025 − 2.39972) = 0.0514.

**Answer: (i) n₂ = 1.549; (ii) φ_c ≈ 88.1°; (iii) θ₀ ≈ 2.95°; (iv) NA ≈ 0.051.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The fractional index difference is Δ = (n1 - n2)/n1. How is the cladding index found?
   A. n2 = n1(1 + Δ) = 1.551
   B. n2 = n1(1 - Δ) = 1.549
   C. n2 = n1 Δ = 8.5 x 10^-4
   D. n2 = n1/Δ
   Answer: B. Rearranging Δ = (n1 - n2)/n1 gives n2 = n1(1 - Δ), which is slightly less than n1. (Key point: Cladding index from Δ)

2. Why is the internal critical angle for this fiber so close to 90° (88.1°)?
   A. n2/n1 = 0.99945 is almost 1, so sin φc is almost 1
   B. Because the NA is very large
   C. Because n2 is greater than n1
   D. Because Δ is large
   Answer: A. sin φc = n2/n1 and the indices differ only by 0.055%, so φc is near 90°. (Key point: Critical angle at core-cladding)

3. Which expression was used for the numerical aperture of this fiber?
   A. NA = n1 (2Δ) = 0.0017
   B. NA = n1 sqrt(Δ) = 0.0364
   C. NA = n1^2 sqrt(2Δ)
   D. NA = n1 sqrt(2Δ) = 0.0514
   Answer: D. For a small index difference, NA = sqrt(n1^2 - n2^2) ≈ n1 sqrt(2Δ). (Key point: NA formula)

4. How is the external acceptance angle θ0 (in air) related to the NA?
   A. tan θ0 = NA, giving 5.1°
   B. θ0 = NA x n1 = 0.08°
   C. sin θ0 = NA, giving θ0 ≈ 2.95°
   D. cos θ0 = NA, giving 87°
   Answer: C. For light entering from air (n = 1), n0 sin θ0 = NA reduces to sin θ0 = NA = 0.0514. (Key point: External acceptance angle)

## Related in Engineering Physics

- [Acceptance angle of fiber in a liquid](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Afiber-acceptance-angle-liquid-index-change) — Engineering Physics, chapter 6
- [Acceptance angle of fiber in water](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Afiber-acceptance-angle-water) — Engineering Physics, chapter 6
- [Derive numerical aperture of optical fiber](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Ana-derivation-optical-fiber) — Engineering Physics, chapter 6, asked 4×
- [Physical significance of numerical aperture](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Anumerical-aperture-significance-dependence) — Engineering Physics, chapter 6
- [Newton's rings, refractive index of liquid](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Anewton-rings-liquid-refractive-index) — Engineering Physics, chapter 4, asked 2×
- [Refractive index of liquid film by interference](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Acoherent-sources-film-refractive-index-method) — Engineering Physics, chapter 4

## Related in other subjects

- [Acceptance angle and NA of step-index fibre](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Aacceptance-angle-na-step-index) — Propagation and Antenna, chapter 6, asked 2×
- [Refractive index, critical frequency and MUF](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Arefractive-index-fc-muf) — Propagation and Antenna, chapter 4
- [Light propagation; NA and acceptance angle](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Alight-propagation-na-acceptance-relation) — Propagation and Antenna, chapter 6
- [Light propagation; define NA and acceptance angle](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Alight-propagation-define-na-acceptance) — Propagation and Antenna, chapter 6