OPTICAL

This response addresses your queries regarding optical fibres, drawing solely on the provided sources.

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18. What is the difference between critical angle and acceptance angle? What is acceptance cone?

Critical Angle (${\theta }_c$) The critical angle is an angle of incidence $\theta$ at the core/cladding interface which determines whether Total Internal Reflection (TIR) occurs. For TIR to happen, light must be incident at an angle greater than the critical angle ${\theta }_c$. Light launched into the core and striking the core to cladding interface at an angle greater than the critical angle will be totally reflected back into the core.

Acceptance Angle (${\theta }_0$) The acceptance angle is the maximum external angle (${\theta }_0$) at which a light ray can enter the fibre and still be propagated through total internal reflection. It is the maximum angle of incidence corresponding to the critical angle of reflection at the core/cladding interface. The acceptance angle is coupled into the fibre and will propagate.

Difference The critical angle relates to the conditions necessary for total internal reflection inside the fiber, at the core-cladding boundary. The acceptance angle relates to the maximum angle at which light can be launched into the fiber from the surrounding medium to ensure that it meets the critical angle condition internally.

Acceptance Cone The light rays must fall within a fundamental cone defined by the acceptance angle to be successfully trapped and guided through the fibre. This cone is known as the acceptance cone. The acceptance cone is the light gathering ability of the fibre.

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19. What are the types of optical fibre based on material, no of modes and refractive index.

Optical fibres are classified based on several characteristics:

Based on Material Optical fibres are generally made of either silica or plastic.

Based on Number of Modes Based on the number of modes of propagation they support, optical fibres are classified as:

1. Single Mode Fibre (SMF): Supports only one mode of propagation.
2. Multi-mode Fibre (MMF): Supports more than one mode of propagation.

Based on Refractive Index (R.I.) Profile Based on the refractive index profile, optical fibres are classified as:

1. Step Index Fibre (SIF): The refractive index changes abruptly at the core cladding boundary.
2. Graded Index Fibre (GRIN/GIF): The refractive index of the core varies with distance from the fibre axis, being maximum at the axis and falling off gradually.

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20. Explain step index and graded index optical fibre with their refractive index profile.

Step Index Optical Fibre (SIF)

The step index fibre is the simplest type of optical fibre.

• Structure: It consists of a thin cylindrical structure of transparent glossy material where the core has a uniform refractive index (${\mu }_1$). This core is surrounded by a cladding material which also has a uniform, but slightly lower, refractive index (${\mu }_2$).
• Refractive Index Profile: The refractive index distribution, $\mu (r)$, is characterized by a discontinuity (or a ‘step’) at the core-cladding interface. The R.I. remains constant across the core.
  • $\mu (r)={\mu }_1$ (for core, $0<r<a$)
  • $\mu (r)={\mu }_2$ (for cladding, $r>a$)

Graded Index Optical Fibre (GRIN/GIF)

• Structure: In contrast to SIF, the graded index fibre (GRIN) has its core refractive index varying continuously and smoothly.
• Refractive Index Profile: The core refractive index is maximum at the center of the core (${\mu }_1$) and progressively decreases until it reaches a constant value (${\mu }_2$) at the core-cladding interface. The R.I. of the core decreases nearly in a parabolic manner.
• Mechanism: This variation is achieved by using concentric layers of glass, each having a different refractive index. This structure causes a periodic focusing of light, leading to sinusoidal ray paths. Rays traveling off-axis spend time in lower R.I. regions (where they travel faster), allowing the longer path length of higher order rays to be compensated by a greater average speed. This compensation reduces transit-time dispersion.

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21. What is single mode and multi-mode optical fibre? Explain V number in fibre.

Single Mode Optical Fibre (SMF)

A single mode fibre (SMF) is an optical fibre that supports only one mode of propagation. SMFs generally have a very small core diameter and often require specialized supports.

Multi-mode Optical Fibre (MMF)

A multi-mode fibre (MMF) is an optical fibre that supports more than one mode of propagation. MMFs typically have a larger core diameter and are easier to splice.

V-number (Normalized Frequency)

The V-number, also known as the normalized frequency or sometimes the “cut-off parameter,” is an important measure used to determine the number of modes supported by a fibre.

The mathematical expression for the V-number is given by: $V=\frac{\pi d}{{\lambda }_0}\sqrt{{\mu }_1^2-{\mu }_2^2}$ Alternatively, using the Numerical Aperture ($NA$), where $d$ is the diameter of the core ($d=2a$): $V=\frac{2\pi a}{{\lambda }_0}NA$

Significance of V-number:

• If the V-number is less than 2.405 ($V<2.405$), the fibre supports only one mode, classifying it as a Single Mode Fibre (SMF).
• If the V-number is greater than 2.405 ($V>2.405$), the fibre is a Multi Mode Fibre (MMF) and can support many modes.

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22. What is modes of fibre or modes of propagation on optical fibre?

In an optical fibre, light propagates in the form of an electromagnetic wave. The transmission of light involves various light waves, referred to as modes, traveling along different paths through the fibre.

The sources define modes of propagation as the supported electromagnetic field configurations within the fibre. Each supported mode possesses a particular value of propagation constant, attenuation, and group velocity. The total signal output is a complex combination of all modes traveling through the fibre.

In multi-mode fibres, different modes travel different total lengths of path, leading to intermodal dispersion.

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23. What is grain index fibre (GRIN)? What is normalized frequency (V-number) of an optical fibre?

Graded Index Fibre (GRIN)

GRIN fibre stands for Graded Index Fibre.

• In a GRIN fibre, the core refractive index varies continuously and smoothly.
• The refractive index is maximum at the center of the core and decreases progressively to a constant value at the core-cladding interface.
• The variation is achieved using concentric layers of glass, which results in a parabolic index profile. This profile compensates for the difference in path length traveled by various rays (modes) by allowing off-axis rays (which travel longer distances) to pass through lower R.I. regions where they move faster, minimizing pulse broadening caused by transit-time dispersion.

Normalized Frequency (V-number) of an Optical Fibre

The normalized frequency, or V-number, is a dimensionless parameter that indicates the maximum number of modes supported by a fibre. It is sometimes called the cut-off frequency or “cut off” or “V-number”.

The mathematical formula for the V-number is: $V=\frac{\pi d}{{\lambda }_0}\sqrt{{\mu }_1^2-{\mu }_2^2}=\frac{2\pi a}{{\lambda }_0}NA$ where $a$ is the core radius, ${\lambda }_0$ is the free space wavelength, ${\mu }_1$ and ${\mu }_2$ are the refractive indices of the core and cladding respectively, and $NA$ is the numerical aperture.

If $V<2.405$, the fibre supports only one mode (Single Mode Fibre), and if $V>2.405$, it supports many modes (Multi Mode Fibre).

Based on the sources provided, here are the comprehensive answers to your queries regarding optical fibres.

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24. Derive the expression of Numerical aperture for step index fibre. What is the importance of acceptance angle in fibre optics communication?

Derivation of Numerical Aperture ($NA$) for Step Index Fibre

Numerical Aperture ($NA$) is an important term associated with an optical fibre, providing information about the acceptance angle (${\theta }_0$). We derive the expression using Snell’s Law and the conditions for Total Internal Reflection (TIR).

Assumptions:

1. ${\mu }_0$: Refractive index of the medium surrounding the fibre (usually air, where ${\mu }_0=1$).
2. ${\mu }_1$: Refractive index of the core.
3. ${\mu }_2$: Refractive index of the cladding, where ${\mu }_1>{\mu }_2$.
4. ${\theta }_0$: Acceptance angle (maximum angle of incidence at the entrance).
5. ${\theta }_1$: Angle of refraction in the core.
6. ${\theta }_c$: Critical angle at the core-cladding boundary.

Step 1: Applying the Condition for Total Internal Reflection (TIR) For light to be trapped in the core, the ray must strike the core-cladding interface at an angle greater than the critical angle (${\theta }_c$). The critical angle is defined as: $\sin {\theta }_c=\frac{{\mu }_2}{{\mu }_1}$

Step 2: Relating the Acceptance Angle (${\theta }_0$) to the Critical Angle (${\theta }_c$) Consider the geometry where a light ray enters the fibre at the maximum possible angle, ${\theta }_0$ (at point A, see Figure 5.6.2(a) in the sources). This refracted ray travels through the core and strikes the core-cladding boundary (at point B) at exactly the critical angle ($\varphi ={\theta }_c$).

From the geometry of the fibre cross-section (specifically the right triangle formed by the core axis, the incident ray, and the core boundary): ${\theta }_1+{\theta }_c=90^{\circ }$ $\sin {\theta }_1=\sin (90^{\circ }-{\theta }_c)$ $\sin {\theta }_1=\cos {\theta }_c$

Step 3: Applying Snell’s Law at the input face (Point A) Snell’s law relates the external angle of incidence (${\theta }_0$) to the internal angle of refraction (${\theta }_1$): ${\mu }_0\sin {\theta }_0={\mu }_1\sin {\theta }_1$

Step 4: Substituting $\sin {\theta }_1$ Substitute the expression for $\sin {\theta }_1$ from Step 2 into the Snell’s Law equation: ${\mu }_0\sin {\theta }_0={\mu }_1\cos {\theta }_c$

Step 5: Expressing $\cos {\theta }_c$ in terms of refractive indices Using the trigonometric identity $\cos {\theta }_c=\sqrt{1-{\sin }^2{\theta }_c}$: ${\mu }_0\sin {\theta }_0={\mu }_1\sqrt{1-{\left(\frac{{\mu }_2}{{\mu }_1}\right)}^2}$

${\mu }_0\sin {\theta }_0={\mu }_1\sqrt{\frac{{\mu }_1^2-{\mu }_2^2}{{\mu }_1^2}}$

${\mu }_0\sin {\theta }_0=\sqrt{{\mu }_1^2-{\mu }_2^2}$

Step 6: Defining Numerical Aperture ($NA$) The Numerical Aperture ($NA$) is defined as ${\mu }_0\sin {\theta }_0$. $NA=\sqrt{{\mu }_1^2-{\mu }_2^2}$

If the fibre is surrounded by air (${\mu }_0=1$), the expression simplifies to: $NA=\sin {\theta }_0=\sqrt{{\mu }_1^2-{\mu }_2^2}$

Importance of Acceptance Angle (${\theta }_0$)

The acceptance angle (${\theta }_0$) is crucial in fibre optics communication because:

1. Light Gathering Ability: It is the angle that determines the light gathering ability of the fibre.
2. Definition of Acceptance Cone: The light rays must fall within the cone defined by the acceptance angle to be trapped and guided. This cone is the acceptance cone.
3. Efficiency of Coupling: The maximum angle at which light can enter the fibre and still be propagated by TIR is defined by the acceptance angle. If light sources (like LEDs or lasers) are designed to emit light within this angle, the power coupled into the fibre is maximized.
4. Relationship to NA: The sine of the acceptance angle defines the Numerical Aperture ($NA$) of the fibre. Since $NA$ depends on the refractive indices of the core and cladding, the acceptance angle links the material properties of the fibre to its ability to capture light.

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25. Draw the block diagram of fibre optics communication system and explain.

Block Diagram of a Fibre Optics Communication System

The fibre optic communication system is very similar to that of a traditional communication system.

Diagram Components: (Based on Figure 5.10.1 in sources)

Input $\to$	Drive Circuit $\to$	Light Source $\to$	Optical Fibre $\to$	Photo Detector $\to$	Signal Restorer $\to$	Output

(Note: In a complete system, repeaters might be included along the optical fibre transmission line.)

Explanation of Components

1. Input Signal: This is the initial electric signal containing the message (e.g., voice, data, video).
2. Drive Circuit: This circuit receives the electrical input signal and converts it into current pulses suitable for driving the light source.
3. Light Source: This component generates optical pulses. Typically, this is a Light Emitting Diode (LED) or a laser diode. The device is chosen based on the required efficiency and speed.
4. Optical Fibre: This serves as the transmission medium. The optical pulses are coupled into the fibre and transmitted via Total Internal Reflection. Along the path, the signal may become progressively attenuated and distorted.
5. Repeaters (not explicitly in the simple block but mentioned in text): For long-distance communication, repeaters are used along the transmission line. These units consist of a photodetector, an amplifier, and a light source. They receive the attenuated signal, amplify and reshape it, and relaunch it into the fibre.
6. Photo Detector (Receiver): At the receiving end, the output light signal strikes the photodetector (e.g., a photodiode). The detector performs the reverse function of the light source, converting the incoming optical power into an electric current. Photodetectors are required to have high quantum efficiency, adequate frequency response, and low signal dependence.
7. Signal Restorer (or Amplifier/Filter): The detected current (containing the message) is filtered and amplified. This step extracts the message and ensures the output signal meets the necessary parameters.
8. Output: The recovered electrical signal, which can then be fed to a transducer to convert it into audio or video form if necessary.

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26. What is attenuation coefficient and what are the factors affect this coefficient?

Attenuation Coefficient Definition

Attenuation means the “loss of optical power” in the fibre. This loss occurs as the ray travels along the length ($L$) of the optical fibre.

The attenuation coefficient ($\alpha$) quantifies this loss and is defined mathematically as:

$\alpha =\frac{10}{L}{\log }_{10}\frac{P_{in}}{P_{out}}\frac{dB}{km}$

where:

• $P_{in}$ is the input optical power.
• $P_{out}$ is the output optical power.
• $L$ is the length of the optical fibre (in kilometers).
• The coefficient is typically measured in decibels per kilometer (dB/km).

Factors Affecting Attenuation Coefficient

Attenuation is dependent on the wavelength of the light being transmitted. The primary factors contributing to losses (and thus affecting the attenuation coefficient) include:

1. Absorption: Attenuation is attributed to absorption.
2. Rayleigh Scattering: Attenuation is attributed to Rayleigh scattering.
3. Geometric Effects: Attenuation is also influenced by geometric effects.

The sources note that the wavelength curve shows minimum attenuation at certain bands, which are known as optical windows. These windows are selected for data transmission because the attenuation is minimal:

• Wavelengths between 0.8 $\mu$m to 0.9 $\mu$m.
• Wavelengths between 1.3 $\mu$m to 1.6 $\mu$m.
• Wavelengths between 1.5 $\mu$m to 1.6 $\mu$m.

27. What are the advantages of optical fibre? Explain the use of optical fibre in communication system?

Advantages of Optical Fibre

Optical fibres offer numerous advantages compared to conventional communication systems:

1. High Bandwidth: Optical fibres provide a high bandwidth.
2. Low Loss: Through optical fibre, light passes over long distances, resulting in very little loss, often necessitating repeaters only every few kilometers.
3. Small Diameter and Size: The fibre diameter is very small, leading to light and thin network cables. This small size and weight are particularly advantageous in aircraft cabling.
4. Low Power Requirements: The power required to operate the system is very small compared to conventional electrical systems.
5. Immunity to EMI: Since optical fibres are typically made of silica or plastic, they are immune to electromagnetic interference (EMI).
6. High Security: Fibre optic transmission offers high data security as there is very low crosstalk, and optical signals are well confined within the waveguide, making external tapping difficult. They can also be safely used in high voltage environments.
7. Durability: Optical fibres are less susceptible to parameters like pressure, temperature, twist, and salinity compared to metallic equivalents (except for specially designed fibres).

Use of Optical Fibre in Communication System

Optical fibres form the core transmission medium in a modern communication system, often replacing traditional electrical communication systems.

In a fibre optic communication system:

1. Signal Conversion: The light source (a LED or a laser diode) receives electrical current pulses from a drive circuit and converts them into optical pulses, which carry the information.
2. Transmission: These optical pulses are coupled into the optical fibre and guided through total internal reflection.
3. Repeaters: Over long distances, the signal becomes attenuated and distorted, requiring repeaters (consisting of a photodetector, an amplifier, and a light source) to amplify, filter, reshape, and relaunch the signal into the fibre line.
4. Reception: At the receiving end, a photodetector converts the incoming optical power back into an electric current.
5. Output: The resultant current is filtered and amplified by a signal restorer to extract the message, which can then be converted into audio or video output.
6. Optimal Operation: The device is chosen for high efficiency, and the frequency of operation is chosen to lie within the optical window regions of the fibre to ensure minimum attenuation.

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28. Discuss the fabrication of optical fibre.

The provided source material describes the structure of optical fibres (core, cladding, buffer/sheath), the materials they are made of (silica or plastic), and their classifications (Step Index, Graded Index).

However, the sources do not contain explicit information or details regarding the manufacturing process, such as the chemical deposition methods (like VAD or OVD) or the physical drawing of the optical fibre preform, which constitute the discussion of fibre fabrication.

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29. A glass clad fibre is made with core glass of refractive index 1.5 and the cladding is doped to give a fractional index difference of 0.0005. Determine (i) the cladding index, (ii) the acceptance angle, (iii) NA.

Given: Core refractive index ($n_1$) = 1.5 Fractional index difference ($\Delta$) = 0.0005

This problem follows the methodology shown in Example 5.11.7 in the sources.

(i) Determine the Cladding Index ($n_2$)

The fractional index difference ($\Delta$) is defined as: $\Delta =\frac{n_1-n_2}{n_1}$ Therefore, $n_2=n_1-\Delta n_1=n_1(1-\Delta )$.

$n_2=1.5\times (1-0.0005)$ $n_2=1.5\times 0.9995$ $n_2=1.49925$

(iii) Determine the Numerical Aperture ($NA$)

The Numerical Aperture is related to $n_1$ and $\Delta$ by the expression: $NA=n_1\sqrt{2\Delta }$

$NA=1.5\times \sqrt{2\times 0.0005}$ $NA=1.5\times \sqrt{0.001}$ $NA\approx 1.5\times 0.03162$ $NA\approx 0.04743$

(ii) Determine the Acceptance Angle (${\theta }_0$)

Assuming the fibre is in air (${\mu }_0=1$), the acceptance angle is given by ${\theta }_0={\sin }^{-1}(NA)$.

${\theta }_0={\sin }^{-1}(0.04743)$ ${\theta }_0\approx 2.72^{\circ }$

Answers: (i) Cladding Index ($n_2$) = 1.49925; (ii) Acceptance Angle (${\theta }_0$) $\approx 2.72^{\circ }$; (iii) Numerical Aperture ($NA$) $\approx 0.04743$.

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30. Find the core diameter necessary for single mode operation at 850 $\mu$m in S.I. fibre with $n_1$=1.48 and $n_2$=1.47. What is the numerical aperture and maximum acceptance angle of this fibre.

Given: Wavelength ($\lambda$) = $850\mu m$ ($850\times 10^{-6}m$) Core R.I. ($n_1$) = 1.48 Cladding R.I. ($n_2$) = 1.47

(i) Determine the Numerical Aperture ($NA$)

The Numerical Aperture ($NA$) is calculated using the formula: $NA=\sqrt{n_1^2-n_2^2}$ $NA=\sqrt{(1.48{)}^2-(1.47{)}^2}$ $NA=\sqrt{2.1904-2.1609}$ $NA=\sqrt{0.0295}$ $NA\approx 0.17175$

(ii) Determine the Maximum Acceptance Angle (${\theta }_0$)

The maximum acceptance angle (${\theta }_0$) is determined by the $NA$ (assuming air, $n_0=1$): ${\theta }_0={\sin }^{-1}(NA)$ ${\theta }_0={\sin }^{-1}(0.17175)$ ${\theta }_0\approx 9.89^{\circ }$

(iii) Determine the Core Diameter ($d$) for Single Mode Operation

For single mode operation, the V-number (normalized frequency) must satisfy the condition: $V<2.405$ We must find the maximum diameter $d_{max}$ corresponding to $V=2.405$. The V-number formula is: $V=\frac{\pi d}{\lambda }NA$

Solving for the diameter $d$ at cutoff ($V=2.405$): $d=\frac{V\cdot \lambda }{\pi \cdot NA}$ $d=\frac{2.405\times (850\times 10^{-6}m)}{\pi \times 0.17175}$ $d\approx \frac{2.04425\times 10^{-3}m}{0.54013}$ $d\approx 3.7848\times 10^{-3}m$

The core diameter necessary for single mode operation (maximum allowable diameter) is 3.78 mm.

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31. A certain optical fibre has an attenuation of 3.5 dB/km at 850 nm. If 0.5 mW of optical power is initially launched with fibre, calculate the power level after 4 km.

Given: Attenuation coefficient ($\alpha$) = $3.5\text{ dB/km}$ Input Power ($P_{in}$) = $0.5\text{ mW}$ Length ($L$) = $4\text{ km}$

Step 1: Calculate the Total Loss ($Los{s}_{Total}$) $Los{s}_{Total}=\alpha \times L$ $Los{s}_{Total}=3.5\text{ dB/km}\times 4\text{ km}=14\text{ dB}$

Step 2: Use the Attenuation Formula to find Output Power ($P_{out}$) The attenuation formula is defined as: $Los{s}_{Total}=10{\log }_{10}\left(\frac{P_{in}}{P_{out}}\right)$

$14=10{\log }_{10}\left(\frac{0.5\text{ mW}}{P_{out}}\right)$

$1.4={\log }_{10}\left(\frac{0.5}{P_{out}}\right)$

To remove the logarithm, raise 10 to the power of both sides: $\frac{0.5}{P_{out}}=10^{1.4}$ $\frac{0.5}{P_{out}}\approx 25.1189$

$P_{out}=\frac{0.5\text{ mW}}{25.1189}$ $P_{out}\approx 0.0199\text{ mW}$

The power level after 4 km is approximately 0.0199 mW.

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32. An optical fibre cable 3.0 km long is made up of three 1.0 km length sliced together. The losses due to each length and slice are 5 dB and 1.0 dB respectively. What would be output power if the input power is 5 mW?

Given: Total length = $3.0\text{ km}$ (made of 3 segments of $1.0\text{ km}$ each). Input Power ($P_{in}$) = $5\text{ mW}$. Length Loss ($L_L$) per 1 km segment = $5\text{ dB}$. Slice Loss ($L_S$) per slice = $1.0\text{ dB}$.

Step 1: Calculate Total Length Loss Since there are 3 segments, the total loss due to the fibre material is: $Los{s}_{Length}=3\times L_L=3\times 5\text{ dB}=15\text{ dB}$

Step 2: Calculate Total Splice Loss To connect 3 segments (L1, L2, L3), 2 slices (S1, S2) are needed. $Los{s}_{Splice}=2\times L_S=2\times 1.0\text{ dB}=2.0\text{ dB}$

Step 3: Calculate Total System Loss ($Los{s}_{Total}$) $Los{s}_{Total}=Los{s}_{Length}+Los{s}_{Splice}$ $Los{s}_{Total}=15\text{ dB}+2.0\text{ dB}=17\text{ dB}$

Step 4: Use the Attenuation Formula to find Output Power ($P_{out}$) The attenuation formula is: $Los{s}_{Total}=10{\log }_{10}\left(\frac{P_{in}}{P_{out}}\right)$

$17=10{\log }_{10}\left(\frac{5\text{ mW}}{P_{out}}\right)$

$1.7={\log }_{10}\left(\frac{5}{P_{out}}\right)$

$\frac{5}{P_{out}}=10^{1.7}$ $\frac{5}{P_{out}}\approx 50.1187$

$P_{out}=\frac{5\text{ mW}}{50.1187}$ $P_{out}\approx 0.0997\text{ mW}$

The output power level is approximately 0.0997 mW.