JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech IV Year OPTOELECTRONICS AND LASER INSTRUMENTATION Previous Question Paper (March - 2021 - R16)

B.Tech | IV Year | I Semester | R16 | March - 2021

Answer PaperPaper Code: 137FGMaximum Marks: 75Time: 3 Hours

Answer any Five Questions. All Questions Carry Equal Marks.

Main

115 marks
a

Compare step index fiber and Graded index fiber.

8 marks

Answer

FeatureStep Index FiberGraded Index Fiber
Refractive Index ProfileConstant refractive index in the core (n1)Refractive index decreases parabolically from the center
Pulse DispersionHigh (due to modal dispersion)Low (due to self-focusing effect)
BandwidthLowHigh
Core DiameterLarge (multimode) or Small (single-mode)Generally medium
Signal DistortionSignificant due to different path lengthsReduced due to varying speeds at different radial distances
FabricationEasier to manufactureMore complex to manufacture
b

Define acceptance angle and Numerical aperture of Optical fiber and derive the relationship between acceptance angle and Numerical aperture.

7 marks

Answer

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

Consider a fiber with core refractive index n1 and cladding refractive index n2, surrounded by air (n0 = 1). Let light enter the fiber at an angle θa (acceptance angle) and refract into the core at angle θr. By Snell's Law at the air-core interface:

n0 sin(θa) = n1 sin(θr)

At the core-cladding interface, the light must strike at an angle greater than the critical angle θc to undergo total internal reflection. If φ is the angle of incidence at the core-cladding interface, then φ = 90° - θr. For total internal reflection, sin(φ) ≥ n2 / n1. Since sin(φ) = cos(θr), we have:

cos(θr) = sqrt(1 - sin²(θr)) ≥ n2 / n1

Substituting sin(θr) = sin(θa) / n1 from the first equation:

sqrt(1 - (sin²(θa) / n1²)) ≥ n2 / n1

Squaring both sides and simplifying:

1 - (sin²(θa) / n1²) ≥ n2² / n1²
sin²(θa) ≤ n1² - n2²

Since NA = sin(θa), the relationship is:

NA = sin(θa) = sqrt(n1² - n2²)
215 marks
a

Explain the following characteristics with respect to LASER: i) Stimulated emission ii) Population inversion

7 marks

Answer

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b

Describe different Laser configurations.

8 marks

Answer

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315 marks
a

What are the types of optical Sources? And explain the principle of operation of LED diode.

7 marks

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b

Compare electro-optic and Acousto-optic modulators.

8 marks

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415 marks
a

What is principle of interferometry and describe the working of velocimeter as interferometer.

7 marks

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b

How can you measure the current using optical fibers? Explain.

8 marks

Answer

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515 marks
a

Describe the operation of basic holographic interferometers.

7 marks

Answer

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b

Describe the Laser instruments used for plastic surgery.

8 marks

Answer

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615 marks
a

Describe the different kinds of dispersions happened in optical fibers.

8 marks

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b

Compare CO2 laser and Argon-Ion Lasers.

7 marks

Answer

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715 marks
a

Compare PIN photo diode and Avalanche Photo diode.

8 marks

Answer

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b

How can you measure the Liquid level using optical fibers? Explain.

7 marks

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815 marks
a

Classify the Gas lasers and explain the working principle of any one of them.

7 marks

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b

How can you measure the distance using Lasers? Explain.

8 marks

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About this paper

This page contains the question paper of OPTOELECTRONICS AND LASER INSTRUMENTATION from JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD for B.Tech EIE IV Year I Semester, March - 2021, R16.

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