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JNTU B.Tech IV OPTICAL COMMUNICATIONS Previous Question Paper (September - 2022 - R16)

B.Tech | IV | II | R16 | September - 2022

Answer PaperPaper Code: 138DYMaximum Marks: 75Time: 3 Hours

Answer any five questions. All questions carry equal marks.

Main

115 marks
a

What is numerical aperture and acceptance angle? Derive an expression for numerical aperture and maximum acceptance angle in the case of a step index optical fiber in terms of refractive indices of core and cladding material.

9 marks

Answer

Numerical Aperture (NA) is a dimensionless measure of the light-gathering capacity of an optical fiber, defined as the sine of the maximum acceptance angle. The Acceptance Angle is the maximum angle at which light rays can enter the fiber core from the external medium and still be successfully guided through the core via total internal reflection.

For a step-index fiber with core refractive index n1 and cladding refractive index n2, surrounded by air (n0 = 1), consider a light ray entering the fiber at an angle θa to the axis. By Snell's law at the air-core interface: sin θa = n1 sin θr, where θr is the refraction angle. At the core-cladding interface, the ray must strike at an angle greater than the critical angle θc to undergo total internal reflection, where sin θc = n2 / n1. Since θr = 90° - θc, we have cos θc = sin θr.

NA = sin θa = n1 cos θc = n1 √(1 - sin2 θc) = n1 √(1 - (n2 / n1)2) = √(n12 - n22)

The maximum acceptance angle θa is given by:

θa = arcsin(√(n12 - n22))
b

A step index fiber has a core and cladding refractive indices of 1.48 and 1.46 respectively and supports propagation of an optical signal wavelength 820nm. Calculate core radius, numerical aperture and acceptance angle for single mode fiber.

6 marks

Answer

Given: n1 = 1.48, n2 = 1.46, λ = 820 nm = 0.82 μm. For a single-mode fiber, the normalized frequency V must be less than 2.405.

1. Numerical Aperture (NA):

NA = √(n12 - n22) = √(1.482 - 1.462) = √(2.1904 - 2.1316) = √0.0588 ≈ 0.2425

2. Acceptance Angle (θa):

θa = arcsin(0.2425) ≈ 14.04°

3. Core Radius (a): Using the V-number condition for single-mode operation (V = 2.405):

V = (2πa / λ) * NA
a = (V * λ) / (2π * NA) = (2.405 * 0.82) / (2 * π * 0.2425) ≈ 1.9721 / 1.5237 ≈ 1.29 μm
215 marks
a

Discuss in detail about Mode Coupling and bring out the Salient features of step index fibers and graded index fibers.

10 marks

Answer

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b

List the principal requirements of a good connector.

5 marks

Answer

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315 marks
a

Discuss about the fiber bending losses.

8 marks

Answer

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b

Find the critical radius of curvature at which bending losses occur for a two-step index fiber with a core refractive index of 1.5 with a relative refractive index difference of 0.2 % and an operating wavelength of 1.55 μm.

7 marks

Answer

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415 marks
a

Explain Pulse broadening due to intermodal dispersion.

8 marks

Answer

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b

Describe the types of demountable connector that may be used with single mode fibers.

7 marks

Answer

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515 marks
a

Explain the principle of operation of LED. Enumerate the characteristics of LED.

7 marks

Answer

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b

What is fiber splicing? Discuss about fusion splicing and mechanical splicing.

8 marks

Answer

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615 marks
a

Discuss the principle of operation of the injection laser with suitable diagram.

8 marks

Answer

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b

Write short notes on equilibrium numerical aperture.

7 marks

Answer

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715 marks
a

Discuss the different noise sources and disturbances in the optical detectors.

9 marks

Answer

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b

Explain fundamental receiver operation with neat diagram.

6 marks

Answer

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815 marks
a

Discuss about overall fiber dispersion in single-mode fibers.

8 marks

Answer

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b

Explain about WDM principle of operation.

7 marks

Answer

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This page contains the question paper of OPTICAL COMMUNICATIONS from JNTU for B.Tech ECE IV II, September - 2022, R16.

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