(SEM V) THEORY EXAMINATION 2024-25 OPTICAL COMMUNICATION

Subject Code: BEC057
Maximum Marks: 70
Time: 3 Hours
Paper ID: 310336

Question Paper Overview

SECTION A (2 × 7 = 14 Marks)

(Answer all questions briefly — conceptual and numerical basics)

a. Why does fiber optics promise to be the technology of choice for local area networks (LANs)?
b. Calculate the velocity of light within glass having a refractive index (n) = 1.5.
c. Specify three peak wavelengths for the transparent windows in modern optical fibers.
d. Define Photoluminescence.
e. A laser diode radiates red light of 650 nm wavelength. Calculate the energy of one photon.
f. How does a photodiode convert light into an electrical signal?
g. Differentiate between coherent and non-coherent detection techniques.

SECTION B (Attempt any three × 7 = 21 Marks)

a. Draw a block diagram of a fiber optic communication system and describe the function of each component.
b. Define dispersion in optical fibers. Calculate how much a light pulse spreads after traveling along 5 km of step-index fiber (numerical aperture = 0.275, refractive index of core = 1.487).
c. Using a suitable energy band diagram, explain absorption, spontaneous emission, and stimulated emission in a compound semiconductor.
d. Explain the principle of operation of a Quantum Well Laser Diode (QWLD).
e. With a block diagram, explain Wavelength Division Multiplexing (WDM) in optical fiber communication.

SECTION C (Attempt one part from each question × 7 = 35 Marks)

Q3

(a) Explain the significance of total internal reflection (TIR) in an optical fiber. Given:
Refractive index of core = 1.48, cladding = 1.46.
Find the critical propagation angle and total acceptance angle.
OR
(b) Write short notes on:
i. Dispersion-shifted and dispersion-flattened fibers
ii. Polarization mode dispersion

Q4

(a) Explain the basic techniques used to couple light from an LED into an optical fiber.
Given: n1=1.48,n2=1.46n_1 = 1.48, n_2 = 1.46n1​=1.48,n2​=1.46; SLED radiates 100 μW — calculate power coupled into the fiber.
OR
(b) Draw a graph showing the typical input–output relationship of a laser diode, and explain mechanisms determining the curve.

Q5

(a) With a diagram, explain Edge Emitting LED (ELED) and Surface Emitting LED (SLED).
OR
(b) Write short notes on:
i. Dark Current
ii. Bit Error Rate (BER)

Q6

(a) How does responsivity depend on the wavelength of an optical signal?
Given: Responsivity = 0.85 A/W, input power saturation = 1.5 mW.
Find photocurrent when P=1 mWP = 1 \text{ mW}P=1 mW.
OR
(b) Write advantages of Avalanche Photodiode (APD) over PIN photodiode.

Q7

(a) Define mode coupling. Explain fiber irregularities that cause it.
OR
(b) Draw a functional block diagram of an optical receiver and explain each component.

 Key Topics for Revision

1. Fiber Optic Communication System

Components:

Optical Transmitter (LED/Laser Diode)             Optical Fiber (Transmission Medium)

Optical Receiver (Photodiode, Amplifier, Decoder)

Advantages: High bandwidth, low attenuation, EMI immunity, data security.

2. Velocity of Light in Medium

v=cn=3×1081.5=2×108 m/sv = \frac{c}{n} = \frac{3 \times 10^8}{1.5} = 2 \times 10^8 \, \text{m/s}v=nc​=1.53×108​=2×108m/s 

3. Optical Windows

Transparent regions in silica fiber:

850 nm, 1310 nm, 1550 nm

4. Photoluminescence

Emission of light from a material after absorbing photons — occurs due to electron-hole recombination.

5. Energy of Photon

E=hcλ=6.626×10−34×3×108650×10−9=3.06×10−19 JE = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{650 \times 10^{-9}} = 3.06 \times 10^{-19} \, \text{J}E=λhc​=650×10−96.626×10−34×3×108​=3.06×10−19J 

6. Photodiode Working

Converts optical energy → electrical current.

Mechanism: Incident photons generate electron–hole pairs → photocurrent proportional to light intensity.

Used in optical receivers.

7. Coherent vs Non-Coherent Detection

8. Dispersion

Definition: Pulse broadening due to different propagation speeds.

Types: Modal, chromatic, polarization.

Equation Example:

• Δt=n1×Lc(NA22n12)\Delta t = \frac{n_1 \times L}{c} \left(\frac{NA^2}{2n_1^2}\right)Δt=cn1​×L​(2n12​NA2​)

(Used to calculate pulse spread in Section B Q2b)

9. Emission Mechanisms

Absorption: Electron moves from valence → conduction band.

Spontaneous Emission: Random photon emission.

Stimulated Emission: Triggered by incident photon (basis of LASER).

10. Quantum Well Laser Diode

Uses thin semiconductor layers (quantum wells) for electron confinement.

Advantages: Lower threshold current, higher efficiency, narrower linewidth.

11. Wavelength Division Multiplexing (WDM)

Combines multiple optical signals at different wavelengths (λ₁, λ₂, λ₃...) into a single fiber.

Components: Multiplexer → Fiber → Demultiplexer.

Used in high-capacity data networks.

12. Total Internal Reflection

sin⁡θc=n2n1=1.461.48=0.986⇒θc=80.8°\sin \theta_c = \frac{n_2}{n_1} = \frac{1.46}{1.48} = 0.986 \Rightarrow \theta_c = 80.8°sinθc​=n1​n2​​=1.481.46​=0.986⇒θc​=80.8° NA=n12−n22=0.241⇒Acceptance angle=27.9°\text{NA} = \sqrt{n_1^2 - n_2^2} = 0.241 \Rightarrow \text{Acceptance angle} = 27.9°NA=n12​−n22​​=0.241⇒Acceptance angle=27.9°

→ Enables light confinement inside core.

13. LED Coupling & Fiber Power

Power coupled:

Pcoupled=PLED×NA2n12=100μW×0.24121.482≈2.6 μWP_{coupled} = P_{LED} \times \frac{NA^2}{n_1^2} = 100 \mu W \times \frac{0.241^2}{1.48^2} ≈ 2.6 \, \mu WPcoupled​=PLED​×n12​NA2​=100μW×1.4820.2412​≈2.6μW 

14. ELED vs SLED

15. Responsivity Calculation

Ip=R×P=0.85×1 mW=0.85 mAI_p = R \times P = 0.85 \times 1 \, \text{mW} = 0.85 \, \text{mA}Ip​=R×P=0.85×1mW=0.85mA 

16. Avalanche Photodiode (APD) Advantages

Internal gain via impact ionization.              Higher sensitivity and bandwidth.

Ideal for long-distance optical systems.

17. Mode Coupling & Irregularities

Mode coupling: Transfer of optical power between modes.

Causes: Core diameter variation, bends, stress, refractive index fluctuations.

Effects: Increased loss, modal dispersion.

18. Optical Receiver Block Diagram

Components:                                           Photodetector – Converts optical → electrical.

Amplifier – Boosts weak signal.               Equalizer – Reduces distortion.

Decision Circuit – Reconstructs digital data.

Exam Strategy Tips

Memorize key wavelength bands (850, 1310, 1550 nm).

Practice numerical derivations (velocity, NA, critical angle, responsivity).

Revise emission/absorption energy diagrams.

Draw neat block diagrams for transmitter, receiver, WDM system.

Review dispersion and fiber design improvements (shifted/flattened).