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JNTU B.Tech I Year ENGINEERING PHYSICS I Previous Question Paper (July - 2021 - R16)

B.Tech | I Year | I-I | R16 | July - 2021

Answer PaperPaper Code: 131AHMaximum Marks: 75Time: 3 hours

Answer any five questions

All questions carry equal marks

Main

115 marks
a

Discuss the theory of Newton rings with relevant diagram in the reflected light.

5 marks

Answer

Newton's rings are an interference pattern created by the reflection of light between a spherical surface of a plano-convex lens and a flat glass plate. When monochromatic light falls normally on the lens-plate system, an air film of varying thickness is formed, which is zero at the point of contact and increases radially outwards. Interference occurs between the light rays reflected from the top and bottom surfaces of this air film. Constructive interference produces bright rings, while destructive interference produces dark rings. The central point of contact is dark because the path difference is effectively zero, but a phase change of π occurs upon reflection at the glass-air interface, resulting in destructive interference.

b

Derive the expressions for the diameters of dark and bright rings.

5 marks

Answer

For a plano-convex lens of radius of curvature R, the thickness t of the air film at a distance r from the center is given by r² = 2Rt. The path difference between reflected rays is 2μt cos(r). For air, μ=1, and for normal incidence, cos(r)=1. Including the phase change of λ/2, the effective path difference is 2t + λ/2.

For dark rings (destructive interference): 2t + λ/2 = (2n+1)λ/2, which simplifies to 2t = nλ. Substituting t = r²/2R, we get r² = nλR. Since diameter D = 2r, D² = 4nλR, or D = 2√(nλR) for n = 0, 1, 2, ...

For bright rings (constructive interference): 2t + λ/2 = (2n+1)λ/2 is incorrect; rather 2t + λ/2 = (2n+1)λ/2 leads to 2t = (2n-1)λ/2. Substituting t = r²/2R, we get D² = 2(2n-1)λR, or D = √(2(2n-1)λR) for n = 1, 2, 3, ...

c

Newton rings are observed in the reflected light of wavelength 5900 Å. The diameter of tenth dark ring is 0.5cm. Find the radius of curvature of the lens used.

5 marks

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215 marks
a

Explain the theory of Fraunhofer diffraction due to ‘n’ slits.

5 marks

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b

Calculate the maximum number of orders possible for plane diffraction grating.

5 marks

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c

Find the highest order that can be seen with the grating having 15000 lines/inches. The wavelength of the light used is 600 nm.

5 marks

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315 marks
a

Discuss the construction of quarter and half wave plates.

5 marks

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b

Explain Malus’s law.

5 marks

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c

Distinguish between polarized and unpolarized light.

5 marks

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415 marks
a

Explain the characteristics of a laser beam.

5 marks

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b

Write any four applications of laser.

5 marks

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c

Describe the construction and working of a semiconductor laser.

5 marks

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515 marks
a

Explain the principle behind the functioning of optical fiber.

5 marks

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b

Derive an expression for the numerical aperture of an optical fiber.

5 marks

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c

Explain the advantages of optical communication system.

5 marks

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615 marks
a

Distinguish between light propagation in Step index and Graded index optical fibers.

5 marks

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b

Define the refractive index difference of an optical fiber. How is it related to numerical aperture?

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c

Write notes on the applications of optical fibers.

5 marks

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715 marks
a

What are Miller Indices? How are they obtained?

5 marks

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b

Derive an expression for the inter-planar spacing in the case of cubic structure.

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c

Calculate the ratio d100 : d110 : d111 for simple cubic lattice.

5 marks

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815 marks
a

Write notes on Point defects in crystals.

5 marks

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b

What are the effects of crystal imperfection on electrical, optical and mechanical properties?

5 marks

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c

Explain Twin boundary and Stacking faults surface imperfection.

5 marks

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This page contains the solved question paper with answers of ENGINEERING PHYSICS I from JNTU for B.Tech CSE I Year I-I, July - 2021, R16.

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