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Fiber Optics
Fiber optics is the technology of transmitting light through thin strands of high-purity glass. By utilizing Total Internal Reflection, these fibers guide light pulses across vast distances with minimal signal loss and massive Bandwidth.
Refractive Index
The Refractive Index of a material determines the speed of light within it and governs refraction at boundaries.
$$n = \frac{c}{v}$$
Ratio of light speed in vacuum to light speed in the medium
$n$=Refractive index of the material(dimensionless)
$c$=Speed of light in vacuum(m/s)
$v$=Speed of light in the material(m/s)
Vacuum
→$n = 1$ exactly (the reference point)
Air
→$n \approx 1.0003$, treated as 1.0 in most problems
Optically Denser: A material with a higher refractive index is called optically denser — light slows down more inside it.
Snell's Law: At any boundary, $n_1 \sin \theta_1 = n_2 \sin \theta_2$ governs the direction of the refracted ray.
The Foundation: Total Internal Reflection (TIR)
Light traveling from a dense medium (core) to a less dense medium (cladding) reflects entirely back if it strikes the boundary beyond the Critical Angle.
$$\sin \theta_c = \frac{n_{cladding}}{n_{core}}$$
Determines the threshold angle for light containment
$\theta_c$=Critical Angle(degrees)
$n_{cladding}$=Refractive index of outer layer(dimensionless)
$n_{core}$=Refractive index of inner core(dimensionless)
$n_{core} = n_{cladding}$
→$\sin \theta_c = 1$ → $\theta_c = 90°$; no TIR possible since no ray can exceed 90°
Glass-air boundary ($n_1 = 1.5$, $n_2 = 1.0$)
→$\theta_c = \arcsin(1.0/1.5) = 41.8°$ — the textbook standard case
Density Rule: TIR only occurs when light moves from a higher refractive index to a lower refractive index — never the reverse.
Boundary Behavior: At exactly the critical angle, the refracted ray travels along the interface at 90°. Beyond it, all light reflects back.
Proportionality Shortcut: $\theta_c \propto n_2/n_1$ — if the cladding index increases (getting closer to the core), the critical angle grows, making TIR harder to achieve.
Anatomy of an Optical Fiber
A standard fiber consists of a high-index Core surrounded by a lower-index Cladding, protected by a Buffer Coating.
Core: The central region where light is guided; made of high-purity silica.
Cladding: Surrounds the core to create the interface needed for TIR.
Buffer Coating: A plastic layer that provides mechanical strength and protection from moisture.
Numerical Aperture and Light Gathering
Numerical Aperture (NA) characterizes the light-gathering capability of the fiber and defines its Acceptance Cone.
$$NA = \sin \theta_{max} = \sqrt{n_{core}^2 - n_{cladding}^2}$$
Relates index difference to the maximum coupling angle
$\text{NA}$=Numerical Aperture(dimensionless)
$\theta_{max}$=Half-angle of the acceptance cone(degrees)
$n_{core} \gg n_{cladding}$
→Large NA; easier to couple light but higher dispersion
Acceptance Cone: The region at the fiber face where entering rays will be captured by TIR.
Launch Angle: Rays entering outside the Acceptance Angle leak into the cladding and are lost.
Proportionality Shortcut: $NA \propto \sqrt{n_1^2 - n_2^2}$ — a larger difference between core and cladding indices means a wider acceptance cone.
In External Medium: When the fiber is in a medium with index $n_0$ (e.g., water), the acceptance angle changes: $n_0 \sin \theta_a = NA$, so $\theta_a = \arcsin(NA / n_0)$.
Types of Optical Fibers
A Single-Mode Step-Index Fiber has a very thin core (~5 µm) that allows only one path (mode) for light, eliminating modal dispersion entirely.
Core Size: ~5 µm diameter — extremely thin, requiring a Laser source for coupling.
Light Source: Must use a strong monochromatic source (laser) since the tiny core cannot capture light from broad, divergent sources like LEDs.
Capacity: Can carry more than 14 TV channels or 14,000 phone calls simultaneously.
Best For: Long-distance telecommunications where signal purity matters most.
A Multimode Step-Index Fiber has a larger core (~50 µm) with a constant refractive index that abruptly drops at the cladding boundary.
Step Profile: The refractive index is constant at $n_1$ (e.g., 1.52) across the entire core, then drops sharply to $n_2$ (e.g., 1.48) in the cladding.
Multiple Modes: Many ray paths exist — axial rays travel the shortest path, while high-angle rays bounce more and travel longer distances.
Dispersion Problem: Different path lengths cause time delays of ~33 ns per km, limiting useful distance.
A Multimode Graded-Index Fiber has a core (50–1000 µm) where the refractive index gradually decreases from the center toward the periphery.
Gradual Profile: No sharp boundary between core and cladding — the index smoothly decreases outward.
Self-Correcting Paths: Rays traveling longer paths through the outer region move through lower-index material, where light speed is higher ($v = c/n$). This compensates for the longer distance.
Dispersion Reduction: Time difference reduced to ~1 ns per km (vs 33 ns/km for step-index) — a 33× improvement.
Key Insight: Speed is inversely proportional to refractive index, so outer rays travel faster, arriving at nearly the same time as axial rays.
Signal Transmission and Communication
A fiber optic communication system converts electrical signals to light, transmits them through the fiber, and converts them back at the receiving end.
Transmitter: A semiconductor Laser or LED converts electrical signals to infrared light (typical wavelength: 1.3 µm).
Digital Modulation: The light source is flashed on/off at extremely high rates — a pulse represents '1' and absence represents '0' (binary).
Receiver: A Photodiode at the far end converts light pulses back to electrical signals for amplification and decoding.
Repeaters: Despite ultra-pure glass (99.99%), signals dim over distance. Repeaters regenerate the signal every 30–100 km.
Advantages of Fiber Over Copper
•
Much wider bandwidth (thousands of calls through a single fiber)
•
Immune to electromagnetic interference
•
Thinner and lighter cables (6 mm fiber replaces 7.62 cm copper bundle)
•
Greater security — difficult to tap without detection
Signal Degradation: Attenuation and Dispersion
Signals traveling through fibers lose power (Attenuation) and spread out in time (Modal Dispersion).
Attenuation Sources: Absorption by impurities in the glass, and Rayleigh Scattering from microscopic density variations formed during manufacturing.
Scattering at Joints: Where fibers are spliced together, atomic groups scatter light, causing additional loss.
Modal Dispersion: Different ray paths (modes) travel different distances, causing pulses to spread and overlap at the receiver.
Step-Index Penalty: Time spread of ~33 ns per km — this limits how closely pulses can be packed.
Graded-Index Fix: Reduces time spread to ~1 ns per km by equalizing travel times across all modes.
Dimensional Check: Dispersion is measured in ns/km — multiplying by fiber length gives total pulse spread.
Application Strategy
Worked Examples
Problem Statement
An optical fiber has a core refractive index of 1.48 and a cladding refractive index of 1.45. Calculate the Numerical Aperture (NA).
Numerical Aperture formulaRefractive Index
Strategic Toolkit
$NA = \sqrt{n_{core}^2 - n_{cladding}^2}$
Step 1
Identify Given Values
Extract the refractive indices from the problem statement.
$n_{core} = 1.48$, $n_{cladding} = 1.45$
Step 2
Apply Formula
Square the indices and find the difference.
$$NA = \sqrt{1.48^2 - 1.45^2} = \sqrt{2.1904 - 2.1025}$$
Step 3
Solve
Calculate the final square root.
$$NA = \sqrt{0.0879} \approx 0.296$$
Problem Statement
Find the critical angle at the core-cladding interface for a fiber where the core index is 1.62 and the cladding index is 1.52.
Snell's LawCritical Angle
Strategic Toolkit
$\theta_c = \arcsin\left(\frac{n_{cladding}}{n_{core}}\right)$
Step 1
Set up Ratio
The sine of the critical angle is the ratio of the lower index to the higher index.
$\sin \theta_c = \frac{1.52}{1.62}$
Step 2
Calculate Sine Value
Perform the division.
$\sin \theta_c \approx 0.9383$
Step 3
Inverse Sine
Find the angle whose sine is 0.9383.
$\theta_c = \arcsin(0.9383) \approx 69.8^\circ$
Problem Statement
Calculate the critical angle and angle of entry for an optical fiber with core index 1.50 and cladding index 1.48.
Critical AngleSnell's LawAcceptance Angle
Strategic Toolkit
$$\sin \theta_c = \frac{n_2}{n_1}, \quad n_0 \sin \theta = n_1 \sin \theta'$$
Step 1
Find Critical Angle
Apply the critical angle formula at the core-cladding boundary.
$$\sin \theta_c = \frac{1.48}{1.50} = 0.9867 \implies \theta_c = 80.6°$$
Step 2
Relate Internal Angle to Critical Angle
For light to just undergo TIR, it must hit the core-cladding boundary at $\theta_c$. The complementary angle inside the core is $\theta' = 90° - \theta_c$.
$$\theta' = 90° - 80.6° = 9.4°$$
Step 3
Apply Snell's Law at Fiber Face
Use Snell's law at the air-core interface to find the maximum entry angle $\theta$.
$$\sin \theta = n_1 \sin \theta' = 1.50 \times \sin 9.4° = 1.50 \times 0.1633 \implies \theta = 14.2°$$
Step 4
Interpret the Result
If light enters at angles greater than 14.2°, TIR will not occur and the ray escapes through the cladding.
$\theta_{max} = 14.2°$
Problem Statement
A fiber link is 5 km long. Compare the pulse spread using (a) step-index fiber (33 ns/km) and (b) graded-index fiber (1 ns/km).
Modal DispersionFiber Types
Strategic Toolkit
$$\Delta t = \text{dispersion rate} \times L$$
Step 1
Step-Index Pulse Spread
Multiply the per-km dispersion by the fiber length.
$$\Delta t_{step} = 33 \text{ ns/km} \times 5 \text{ km} = 165 \text{ ns}$$
Step 2
Graded-Index Pulse Spread
Same calculation for the graded-index fiber.
$$\Delta t_{graded} = 1 \text{ ns/km} \times 5 \text{ km} = 5 \text{ ns}$$
Step 3
Compare Results
The graded-index fiber has 33× less pulse spread, allowing much higher data rates over the same distance.
$$\frac{\Delta t_{step}}{\Delta t_{graded}} = \frac{165}{5} = 33$$
Memorize formulas
The Critical Angle
The minimum internal angle for TIR to occur (sine = small/big).
Formula
$\theta_c = \arcsin\left(\frac{n_2}{n_1}\right)$
Numerical Aperture (NA)
The light-gathering 'power' of the fiber.
Formula
$NA = \sqrt{n_{core}^2 - n_{cladding}^2}$
Acceptance Angle
The maximum angle in air to enter the fiber successfully.
Formula
$\theta_{acc} = \arcsin(NA)$
Refractive Index
Ratio of light speed in vacuum to speed in the medium.
Formula
$n = \frac{c}{v}$
Review
1
An optical fiber has core index 1.55 and cladding index 1.50. What is the critical angle at the core-cladding interface?
Ans
75.2°
Review
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Use the critical-angle relation
At the boundary from core to cladding, the critical angle satisfies $\sin \theta_c = n_2/n_1$ because light goes from higher index to lower index.
$$\sin \theta_c = \frac{1.50}{1.55} = 0.9677$$
Evaluate the angle
Take the inverse sine of the ratio to get the threshold for total internal reflection.
$$\theta_c = \arcsin(0.9677) \approx 75.2^\circ$$
2
A single-mode step-index fiber uses a laser source rather than an LED primarily because:
Ans
The ~5 µm core is too small to efficiently capture divergent LED light
Review
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Match the source to the tiny core
A single-mode fiber has a very small core, so the source must be narrow and well-collimated to couple light efficiently into it.
A laser beam is directional enough for a ~5 \mu m core; an LED spreads too widely.
3
In a graded-index fiber, rays traveling along the outer edge arrive at the same time as axial rays because:
Ans
Outer rays travel through lower-index material where light speed is higher, compensating for the longer path
Review
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Use the index gradient
The refractive index is lower near the outer part of the core, so those rays travel faster there and make up for the longer path length.
$$v = \frac{c}{n} \quad \Rightarrow \quad \text{lower } n \text{ gives higher } v$$
4
A fiber has $n_{core} = 1.52$ and $n_{cladding} = 1.41$. The Numerical Aperture and the maximum acceptance angle in air are closest to:
Ans
NA ≈ 0.57, $\theta_{max}$ ≈ 34.6°
Review
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Compute the numerical aperture
For a step-index fiber in air, use $NA = \sqrt{n_1^2 - n_2^2}$ with the core and cladding indices.
$$NA = \sqrt{1.52^2 - 1.41^2} = \sqrt{0.3223} \approx 0.568$$
Convert NA to acceptance angle
In air, the maximum acceptance angle satisfies $\sin \theta_{max} = NA$.
$$\theta_{max} = \arcsin(0.568) \approx 34.6^\circ$$
5
A glass light pipe in air has a critical angle of 42°. When submerged in ethanol ($n = 1.36$), the new critical angle is approximately:
Ans
65.5°
Review
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Recover the glass index from the air case
In air, the critical-angle relation gives $\sin 42^\circ = 1/n_{glass}$, so first find the glass refractive index.
$$n_{glass} = \frac{1}{\sin 42^\circ} = \frac{1}{0.6691} \approx 1.494$$
Recompute the critical angle in ethanol
Replace air with ethanol in the same relation and solve for the new threshold angle.
$$\sin \theta_c' = \frac{1.36}{1.494} = 0.9103 \Rightarrow \theta_c' \approx 65.5^\circ$$
6
A multimode step-index fiber produces a time dispersion of 33 ns/km. Over a 15 km link, the total pulse spread is:
Ans
495 ns
Review
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Multiply dispersion by distance
Total pulse spread is the dispersion rate times the length of the link.
$$33\ \text{ns/km} \times 15\ \text{km} = 495\ \text{ns}$$
7
Which type of optical fiber is best suited for long-distance telecommunications using white light?
Ans
Multimode graded-index
Review
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Choose the fiber that minimizes modal dispersion
White light is multimode by nature, so the best choice is the design that reduces modal spread the most over long distances.
Multimode graded-index fiber
8
In a fiber optic communication system, the light signals are regenerated by repeaters typically placed every:
Ans
30–100 km
Review
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Recall the usual repeater spacing
Even in low-loss fiber, attenuation and scattering eventually weaken the pulse, so repeaters are installed after a moderate transmission distance.
Repeaters are typically spaced about 30–100 km apart.
9
A fiber has $n_{core} = 1.62$ and $n_{cladding} = 1.58$. The critical angle at the core-cladding interface is 77.2°. The maximum entry angle from air is approximately:
Ans
21.0°
Review
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Find the internal angle to the axis
The ray inside the core must make the complementary angle to the surface-normal critical angle before entering the acceptance calculation.
$$\theta' = 90^\circ - 77.2^\circ = 12.8^\circ$$
Apply Snell's law at the fiber face
For entry from air, the acceptance angle satisfies $\sin \theta = n_{core} \sin \theta'$. Substitute the internal angle and solve.
$$\sin \theta = 1.62\sin 12.8^\circ \approx 0.359 \Rightarrow \theta \approx 21.0^\circ$$
10
Which statement about Numerical Aperture (NA) is FALSE?
Ans
NA changes when the fiber is submerged in water instead of air
Review
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Separate fiber property from external medium
NA depends only on the core and cladding indices, while the surrounding medium changes the acceptance angle, not the NA itself.
$$NA = \sqrt{n_{core}^2 - n_{cladding}^2}$$
11
In digital modulation of fiber optic signals, a pulse of light represents:
Ans
Binary 1
Review
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Read the on-off keying convention
Digital fiber communication uses light pulses to encode bits: light on means one binary state and light off means the other.
A pulse of light represents binary 1.
Question 1 of 11
0:00
1
An optical fiber has core index 1.55 and cladding index 1.50. What is the critical angle at the core-cladding interface?