Optical Fibers

Introduction

Optical fiber systems transmit information using light as a carrier. With frequencies on the order of \(10^{14} \ \text{Hz}\), they enable extremely high data-transfer rates.

The fiber consists of a thin, flexible glass core (high refractive index) surrounded by a cladding (lower refractive index). Total internal reflection at the core-cladding boundary confines light within the core. A protective jacket shields the fiber from physical damage.

Principles: Total Internal Reflection

Let the core and cladding refractive indices be \(n_1\) and \(n_2\), respectively, with \(n_1 > n_2\). From Snell's Law at the interface: \[ n_1 \sin i = n_2 \sin r \] Since \(n_1 / n_2 > 1\), it follows that \(\sin r / \sin i > 1\), implying \(r > i\) for small angles.

As \(i\) increases, \(r\) approaches \(90^\circ\). The critical angle \(i_c\) occurs when \(r = 90^\circ\): \[ n_1 \sin i_c = n_2 \sin 90^\circ \quad \Rightarrow \quad \sin i_c = \frac{n_2}{n_1} \] For \(i > i_c\), Snell's law would require \(\sin r > 1\), which is impossible. Hence, no refraction occurs; all light is reflected back into the core. This is total internal reflection, guiding light along the fiber with minimal loss.

Comparison with Copper Cable

1. Bandwidth: Light (carrier) has a much higher frequency, allowing far greater data rates.
2. Attenuation: Signal loss is lower in optical fibers.
3. Security: Tapping into an optical fiber is difficult, enhancing data security.
4. Size & Weight: Fibers are lighter and more compact.
5. Immunity: Unaffected by external electromagnetic interference.
6. Cost: Typically more expensive to manufacture, maintain, and protect.

Applications

• High-speed digital communication systems
• Medical endoscopy
• Sensors (pressure, temperature, etc.)
• Seismic and underwater applications

Example

Problem: An optical fiber has a core index \(n_1 = 1.5\) and cladding index \(n_2 = 1.4\).

1. What is the speed of light in the core?
2. What is the critical angle at the core-cladding interface?
3. What is the maximum angle \(t\) that incoming rays can make with the fiber axis to maintain total internal reflection?

Solution:

1. \( n_1 = \dfrac{c}{v} \quad \Rightarrow \quad v = \dfrac{c}{n_1} = \dfrac{3 \times 10^8}{1.5} = 2 \times 10^8 \ \text{m/s} \)
2. \( \sin i_c = \dfrac{n_2}{n_1} = \dfrac{1.4}{1.5} \approx 0.933 \quad \Rightarrow \quad i_c = \arcsin(0.933) \approx 69^\circ \)
3. For total internal reflection: \( i > i_c \).
   From geometry, \( i = 90^\circ - t \). Thus:
   \( 90^\circ - t > i_c \quad \Rightarrow \quad t < 90^\circ - i_c \approx 21^\circ \).
   The maximum angle \(t\) is therefore less than \(21^\circ\).

References

1. Hecht, J. Understanding Fiber Optics. 2015.
2. Crisp, J. Introduction to Fiber Optics. 2005.
3. Optical Fiber - Wikipedia