Fibre Optics Notes FE Sem 1 Applied Physics | Mumbai University

Optical fibres carry most of the world's internet and telephone traffic as pulses of light. This module explains total internal reflection and the critical angle, derives the acceptance angle and numerical aperture, and classifies fibres into single mode and multimode, step index and graded index types. It then covers attenuation, the attenuation coefficient and the factors causing loss, and ends with the block diagram and advantages of a fibre optic communication system. It carries course outcome CO2, and numerical aperture and attenuation numericals are almost certain in the exam.

Summary

Critical angle and total internal reflection: When light travels from a denser medium of refractive index \(n_1\) to a rarer medium \(n_2\) (\(n_1 > n_2\)), it bends away from the normal. The angle of incidence for which the refracted ray grazes the boundary (refraction angle 90 degrees) is the critical angle, \(\sin\theta_c = n_2/n_1\). If the angle of incidence exceeds \(\theta_c\), no light is refracted and all of it is reflected back into the denser medium; this is total internal reflection (TIR), and it involves practically no loss.

Structure and propagation of light: An optical fibre is a thin, flexible strand of glass or plastic with a central core of refractive index \(n_1\) surrounded by a cladding of slightly lower index \(n_2\), and an outer protective buffer or jacket. Light launched into the core strikes the core-cladding boundary at angles greater than the critical angle and is therefore totally internally reflected again and again, travelling along the fibre in a zig-zag path even when the fibre is gently bent. The cladding keeps the light inside the core and protects the core surface.

Acceptance angle, acceptance cone and numerical aperture: Not every ray entering the fibre end is guided. The acceptance angle \(\theta_a\) is the maximum angle with the fibre axis at which a ray can enter the fibre and still undergo TIR at the core-cladding boundary. Applying Snell's law at the entry face and the critical angle condition gives \(n_0\sin\theta_a = \sqrt{n_1^2 - n_2^2}\). Rotating this angle about the axis gives the acceptance cone of half angle \(\theta_a\); all rays within this cone are guided. The numerical aperture is \(NA = n_0\sin\theta_a = \sqrt{n_1^2 - n_2^2} \approx n_1\sqrt{2\Delta}\), where \(\Delta = (n_1 - n_2)/n_1\) is the fractional refractive index change. NA measures the light gathering ability of the fibre and typically lies between 0.1 and 0.5.

Types of optical fibres: According to the number of modes, a single mode fibre has a very small core (about 8 to 10 μm diameter) and carries only one mode, so there is no intermodal dispersion and it is used for long distance, high bandwidth links with laser sources. A multimode fibre has a larger core (about 50 to 200 μm) and carries many modes; it is easy to couple and splice and works with LEDs, but intermodal dispersion limits it to short distances. The number of modes is governed by the V-number \(V = (2\pi a/\lambda)NA\); a fibre is single mode when \(V < 2.405\), and a step index multimode fibre carries about \(V^2/2\) modes.

Step index and graded index fibres: In a step index fibre the core has a uniform index \(n_1\) that drops abruptly to \(n_2\) at the cladding; rays travel in zig-zag straight paths, and rays at larger angles take longer, giving large intermodal dispersion. In a graded index fibre the core index is maximum on the axis and decreases gradually (nearly parabolically) towards the cladding; rays follow curved, sinusoidal paths and the outer rays travel faster in the lower index region, so all modes arrive at almost the same time. Graded index multimode fibre therefore has much less dispersion and greater bandwidth than step index multimode fibre.

Attenuation and attenuation coefficient: Attenuation is the decrease in optical power as light travels along the fibre. It is expressed by the attenuation coefficient \(\alpha = (10/L)\log_{10}(P_{in}/P_{out})\) in dB/km, where L is in km. Modern silica fibres have about 0.2 dB/km near 1550 nm. Factors affecting attenuation: (1) absorption, both intrinsic (by silica itself in the ultraviolet and infrared) and extrinsic (by impurities such as hydroxyl OH ions and transition metal ions); (2) scattering, mainly Rayleigh scattering from microscopic density fluctuations, which varies as \(1/\lambda^4\); (3) bending losses, macrobends of large curvature and microbends due to tiny irregularities; and (4) losses at splices and connectors and due to core-cladding imperfections. These losses explain why communication uses the low loss windows at 850, 1310 and 1550 nm.

Fibre optic communication system: The message from the information source is converted into electrical signals, encoded and used to modulate an optical source (LED or laser diode) in the transmitter. The light pulses are coupled into the optical fibre, which is the channel; for long links repeaters or optical amplifiers restore the weakened signal. At the receiving end a photodetector (PIN photodiode or avalanche photodiode) converts the light back into an electrical signal, which is amplified, demodulated and decoded to recover the original message at the destination.

Advantages of optical fibre communication: very large bandwidth and information carrying capacity; low loss, so repeaters can be far apart; immunity to electromagnetic interference and no crosstalk between fibres; security, since tapping is difficult; small size and light weight; electrical insulation and no sparks, so safe in hazardous areas; resistance to corrosion and temperature; and the raw material silica is cheap and plentiful. Limitations include costly joining and the need for careful handling.

Mind map

Derivation of numerical aperture

Important definitions

Formulas

Key points to remember

• TIR needs two conditions: light must go from denser to rarer medium, and the angle of incidence must exceed the critical angle.
• Core index is always slightly greater than cladding index.
• NA does not depend on the fibre dimensions, only on n1 and n2 (and n0).
• Larger NA means more light gathered but more modes and more dispersion.
• Acceptance angle is a half angle; the full cone angle is 2θa.
• Single mode: core about 8 to 10 μm, laser source, long distance; multimode: core 50 μm or more, LED source, short distance.
• Graded index fibre reduces intermodal dispersion because its refractive index profile equalises travel times.
• Attenuation in dB/km is 10 log10(Pin/Pout)/L; a 3 dB loss means half the power is lost.
• Rayleigh scattering is proportional to 1/λ^4, so longer wavelengths (1310, 1550 nm) have less loss.
• Fibre optic system: transmitter (source), channel (fibre), receiver (photodetector).

Solved examples

Example 1. An optical fibre has core index 1.48 and cladding index 1.46. Find the critical angle, numerical aperture, acceptance angle and fractional index change (outside medium air).

Critical angle: \(\sin\theta_c = 1.46/1.48 = 0.9865\), so \(\theta_c = 80.57^\circ\). NA: \(\sqrt{1.48^2 - 1.46^2} = \sqrt{2.1904 - 2.1316} = \sqrt{0.0588} = 0.2425\). Acceptance angle: \(\theta_a = \sin^{-1}(0.2425) = 14.03^\circ\) (acceptance cone full angle 28.06 degrees). \(\Delta = (1.48 - 1.46)/1.48 = 0.0135\).

Example 2. The numerical aperture of a fibre is 0.22 and its core index is 1.50. Find the cladding index.

\(NA^2 = n_1^2 - n_2^2\), so \(n_2 = \sqrt{n_1^2 - NA^2} = \sqrt{2.25 - 0.0484} = \sqrt{2.2016} = 1.4838\).

Example 3. A fibre has core index 1.50 and fractional index change 0.01. Find the cladding index, NA and acceptance angle.

\(n_2 = n_1(1 - \Delta) = 1.50\times0.99 = 1.485\). \(NA \approx n_1\sqrt{2\Delta} = 1.50\sqrt{0.02} = 0.2121\) (the exact formula \(\sqrt{n_1^2 - n_2^2}\) gives 0.2116; both are accepted). \(\theta_a = \sin^{-1}(0.2121) = 12.25^\circ\).

Example 4. The fibre of example 1 (NA = 0.2425 in air) is immersed in water of index 1.33. Find the new acceptance angle.

\(n_0\sin\theta_a = NA\) is fixed by n1 and n2, so \(\sin\theta_a = 0.2425/1.33 = 0.1823\) and \(\theta_a = 10.5^\circ\). The acceptance angle decreases in a denser outer medium.

Example 5. Optical power of 9 mW is launched into a 3 km fibre and 3 mW emerges. Find the attenuation coefficient.

\(\alpha = (10/L)\log_{10}(P_{in}/P_{out}) = (10/3)\log_{10}(9/3) = (10/3)(0.4771) = 1.59\) dB/km.

Example 6. A fibre has attenuation 0.5 dB/km. If 2 mW is launched, what power is received after 40 km?

Total loss \(= \alpha L = 0.5\times40 = 20\) dB. \(P_{out} = P_{in}\times10^{-20/10} = 2\times10^{-2}\) mW = 0.02 mW = 20 μW.

Example 7. A step index fibre has core diameter 50 μm, n1 = 1.53, n2 = 1.50. Find the V-number and approximate number of modes at λ = 1 μm.

\(NA = \sqrt{1.53^2 - 1.50^2} = \sqrt{0.0909} = 0.3015\). \(V = \pi d\,NA/\lambda = \pi\times50\times0.3015/1 = 47.36\). Number of modes \(\approx V^2/2 = 1121\). It is a multimode fibre.

Example 8. Find the maximum core radius for single mode operation at 1.3 μm if n1 = 1.480 and n2 = 1.478.

\(NA = \sqrt{1.480^2 - 1.478^2} = \sqrt{(0.002)(2.958)} = 0.0769\). Single mode needs \(V \le 2.405\): \(a_{max} = 2.405\lambda/(2\pi NA) = 2.405\times1.3/(2\pi\times0.0769) = 6.47\) μm (core diameter about 12.9 μm).

Important questions and answers

Q1. Define critical angle and total internal reflection. State the conditions for TIR. (3 marks)

Critical angle is the angle of incidence in the denser medium for which the refracted ray grazes the interface (angle of refraction 90 degrees); \(\sin\theta_c = n_2/n_1\). Total internal reflection is the complete reflection of light back into the denser medium when the angle of incidence exceeds the critical angle. Conditions: (1) light must travel from a denser to a rarer medium (n1 > n2); (2) the angle of incidence at the interface must be greater than the critical angle.

Q2. Define acceptance angle and numerical aperture. Derive the expression for numerical aperture of a step index fibre. (5 marks)

Acceptance angle: the maximum angle with the fibre axis at which light can be launched and still be guided by TIR. Numerical aperture: \(NA = n_0\sin\theta_a\), the light gathering ability. Derivation: a ray enters from medium n0 at angle θi; Snell's law gives \(n_0\sin\theta_i = n_1\sin\theta_r\). It meets the core-cladding boundary at \(\phi = 90^\circ - \theta_r\). For the limiting ray \(\phi = \theta_c\), so \(\sin\theta_r = \cos\theta_c = \sqrt{1 - (n_2/n_1)^2}\). Hence \(n_0\sin\theta_a = n_1\sqrt{1 - n_2^2/n_1^2} = \sqrt{n_1^2 - n_2^2}\). So \(NA = \sqrt{n_1^2 - n_2^2}\), and with \(\Delta = (n_1 - n_2)/n_1\), \(NA \approx n_1\sqrt{2\Delta}\). For air, \(\theta_a = \sin^{-1}(NA)\). Diagram: fibre end with core, cladding, incident ray at θa, refracted ray, TIR at boundary.

Q3. Compare single mode and multimode fibres. (4 marks)

Single mode: core diameter about 8 to 10 μm; one mode only; V < 2.405; no intermodal dispersion; very high bandwidth; small NA; needs laser source; difficult coupling and splicing; long distance telecom. Multimode: core 50 to 200 μm; many modes; V > 2.405; large intermodal dispersion; lower bandwidth; larger NA; LED source is enough; easy coupling; short distance links such as LANs.

Q4. Differentiate between step index and graded index fibres with refractive index profiles. (5 marks)

Step index: core index constant at n1 and falls abruptly to n2 at the cladding; rays travel in zig-zag straight lines by TIR; high intermodal dispersion; lower bandwidth; can be single or multimode; simpler and cheaper to make. Graded index: core index maximum at the axis and decreasing (nearly parabolically) to n2 at the core-cladding boundary; rays follow curved sinusoidal paths by continuous refraction; outer rays travel faster in lower index regions, so all modes arrive nearly together; low intermodal dispersion; higher bandwidth; usually multimode; costlier to make. Sketch: index versus radius as a rectangle (step) and as a parabola (graded), with ray paths.

Q5. What is attenuation? Explain the factors that contribute to attenuation in optical fibres. (5 marks)

Attenuation is the loss of optical power as light propagates along the fibre, \(\alpha = (10/L)\log_{10}(P_{in}/P_{out})\) dB/km. Factors: (1) Absorption: intrinsic absorption by silica in the UV (electronic) and IR (molecular vibrations) regions, and extrinsic absorption by impurities such as OH ions (peak near 1380 nm) and metal ions like Fe, Cu, Cr. (2) Scattering: Rayleigh scattering from microscopic density and composition fluctuations frozen in the glass, proportional to \(1/\lambda^4\). (3) Bending losses: macrobending when the fibre is bent with a small radius so some rays fall below the critical angle and escape; microbending due to small kinks during cabling. (4) Other losses: core-cladding interface irregularities, splices and connectors. Because of these, operation is done in low loss windows 850, 1310 and 1550 nm.

Q6. With a block diagram explain a fibre optic communication system. (5 marks)

Blocks: information source, transmitter, optical fibre channel, receiver, destination. Transmitter: the input (voice, data, video) is converted into an electrical signal, coded (usually digital) and fed to a driver circuit that modulates an optical source, an LED or laser diode, producing light pulses. Channel: the light is coupled into the fibre through a connector and travels by TIR; for long links repeaters or optical amplifiers regenerate or amplify the weak signal; splices join fibre lengths. Receiver: a photodetector (PIN or avalanche photodiode) converts light into electric current; it is amplified, filtered, demodulated and decoded. Destination: the output device such as a telephone, computer or TV. Draw the block diagram in this order with arrows.

Q7. State the advantages of optical fibre communication over conventional copper cables. (4 marks)

(1) Enormous bandwidth, so very high data rates. (2) Low attenuation, about 0.2 dB/km, so fewer repeaters. (3) Immune to electromagnetic and radio interference and lightning. (4) No crosstalk between neighbouring fibres. (5) Secure, since tapping is difficult to do without detection. (6) Small size and light weight. (7) Electrical isolation and no sparking, safe in explosive environments. (8) Silica is cheap and abundant and fibres resist corrosion.

Q8. Calculate NA, acceptance angle and critical angle of a fibre with n1 = 1.48 and n2 = 1.46. (3 marks)

\(NA = \sqrt{1.48^2 - 1.46^2} = 0.2425\); \(\theta_a = \sin^{-1}0.2425 = 14.03^\circ\); \(\theta_c = \sin^{-1}(1.46/1.48) = 80.57^\circ\).

Q9. Explain propagation of light through an optical fibre. Why must the core index be greater than the cladding index? (3 marks)

Light entering the core within the acceptance cone strikes the core-cladding boundary at an angle greater than the critical angle and is totally internally reflected. It keeps reflecting in a zig-zag path along the axis with negligible loss, even around gentle bends. TIR is possible only when light goes from a denser to a rarer medium, so the core index n1 must exceed the cladding index n2; otherwise light would refract into the cladding and be lost.

Q10. A signal of 100 μW is launched into a fibre and 40 μW is received after 8 km. Find the loss in dB and dB/km. (3 marks)

Total loss \(= 10\log_{10}(100/40) = 10\times0.398 = 3.98\) dB. Attenuation \(= 3.98/8 = 0.50\) dB/km.

Exam tips

• Always draw the ray diagram (core, cladding, acceptance cone) with the NA derivation; it is worth 1 to 2 marks.
• Remember the acceptance angle is the half angle; if asked for the acceptance cone angle, give 2θa.
• Put your calculator in degree mode for sin inverse.
• In attenuation problems use log to base 10 and write the unit dB/km.
• For comparison questions write a two-column table with at least five points.
• If Δ is given instead of n2, use NA = n1√(2Δ) or first find n2 = n1(1 − Δ).
• Label every block in the communication system diagram, including repeaters and photodetector.

FAQs

Does a larger numerical aperture always make a better fibre?

No. A larger NA collects more light and eases coupling, but it allows more modes and therefore more dispersion, which reduces bandwidth. Long distance fibres use small NA single mode designs.

Why is cladding needed if light is reflected at the glass-air surface anyway?

A bare core would lose light wherever dust, scratches or contact with other surfaces spoil the boundary. The cladding gives a clean, protected boundary with a controlled index difference and keeps fibres from leaking light into each other.

Why are 1310 nm and 1550 nm used for long distance links?

Rayleigh scattering falls as 1/λ^4 and silica's infrared absorption rises only beyond about 1600 nm, so attenuation is lowest near 1550 nm (about 0.2 dB/km), while 1310 nm offers very low dispersion in standard fibre.

What does a 3 dB loss mean?

Loss of 3 dB means \(P_{out}/P_{in} = 10^{-0.3} \approx 0.5\), so half the power is lost.

These notes follow the University of Mumbai first-year engineering syllabus (NEP 2020, Applied Physics) and were prepared with AI help and checked.