(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
| Type | Description | Example |
|---|---|---|
| Coherent | Uses reference optical signal (local oscillator) | Heterodyne detection |
| Non-Coherent | Direct detection of light intensity | PIN/APD photodiodes |
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(2n12NA2)
(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
sinθc=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=n1n2=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×n12NA2=100μW×1.4820.2412≈2.6μW
14. ELED vs SLED
| Type | Emission | Application |
|---|---|---|
| ELED | Edge emission, high power | Long-distance links |
| SLED | Surface emission, low power | LAN, short links |
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).
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