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

B.Tech | I | 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 a phenomenon of interference produced by an air film of variable thickness between a plano-convex lens of large radius of curvature and a plane glass plate. When monochromatic light falls normally on this arrangement, the light reflected from the top and bottom surfaces of the air film undergoes interference. The path difference between the two reflected rays is 2μt cos(r+θ), where μ is the refractive index of the air film (μ=1), t is the thickness of the film, and r is the angle of refraction. For normal incidence, r=0 and θ is negligible, so the path difference is 2t. Due to the reflection at the bottom surface of the air film (denser medium), an additional phase shift of π occurs, equivalent to a path difference of λ/2. Thus, the effective path difference is 2t + λ/2.

b

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

5 marks

Answer

Let R be the radius of curvature of the lens and t be the thickness of the air film at a distance r from the point of contact. From the geometry of the circle, r^2 = t(2R - t). Since t is very small compared to R, r^2 ≈ 2Rt, or t = r^2 / 2R. Substituting this into the interference conditions:

For dark rings, the condition is 2t = nλ. Substituting t, we get 2(r^2 / 2R) = nλ, which simplifies to r^2 = nRλ. The diameter D = 2r, so D^2 = 4nRλ, or D = 2 * sqrt(nRλ) for n = 0, 1, 2...

For bright rings, the condition is 2t = (2n-1)λ/2. Substituting t, we get 2(r^2 / 2R) = (2n-1)λ/2, which simplifies to r^2 = (2n-1)Rλ/2. Thus, the diameter D = 2 * sqrt((2n-1)Rλ/2) 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

Answer

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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

Answer

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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?

5 marks

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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.

5 marks

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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 question paper of ENGINEERING PHYSICS - I from JNTU for B.Tech ECE I I-I, July - 2021, R16.

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