Physical Optics-Polarisation and Modern Optics (IFS (Forests Services) Physics (Mains)): Questions 1 - 5 of 5

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Question number: 1

» Physical Optics » Polarisation and Modern Optics » Principles of Fibre Optics Attenuation

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Appeared in Year: 2012

Essay Question▾

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A step – index silica fiber consists of core of refractive index 1.450 and diameter . If the numerical aperture of the fiber is 0.16, calculate:

(i) the acceptance angle of the fiber,

(ii) refractive index of the cladding, and

(iii) if the fiber is immersed in water of refractive index 1.33, how does the acceptance angle change? (paper -1) (Section – A)

Explanation

  • Here, from question, we can say,

  • The numerical aperture of the optical fiber is given as,

  • where, is acceptance angle of the fiber. is the refracting medium of surrounding of optical fibe

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Question number: 2

» Physical Optics » Polarisation and Modern Optics » Principles of Fibre Optics Attenuation

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Appeared in Year: 2010

Essay Question▾

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With a suitable diagram, deduce an expression for numerical aperture (NA) for ab optical fiber having refractive indices of core and cladding and , respectively and being placed in a medium of index . (Section – A)

Explanation

Quadrilateral poly1Quadrilateral poly1: Polygon A, B, C, DQuadrilateral poly2Quadrilateral poly2: Polygon D, E, F, CQuadrilateral poly3Quadrilateral poly3: Polygon E, F, G, HAngle αAngle α: Angle between W, Z, IAngle βAngle β: Angle between Z, A_1, C_1Angle γAngle γ: Angle between A_1, Z, JSegment aSegment a: Segment [A, B] of Quadrilateral poly1Segment bSegment b: Segment [B, C] of Quadrilateral poly1Segment cSegment c: Segment [C, D] of Quadrilateral poly1Segment dSegment d: Segment [D, A] of Quadrilateral poly1Segment d_1Segment d_1: Segment [D, E] of Quadrilateral poly2Segment eSegment e: Segment [E, F] of Quadrilateral poly2Segment fSegment f: Segment [F, C] of Quadrilateral poly2Segment c_1Segment c_1: Segment [C, D] of Quadrilateral poly2Segment e_1Segment e_1: Segment [E, F] of Quadrilateral poly3Segment f_1Segment f_1: Segment [F, G] of Quadrilateral poly3Segment gSegment g: Segment [G, H] of Quadrilateral poly3Segment hSegment h: Segment [H, E] of Quadrilateral poly3Segment iSegment i: Segment [I, J] Segment jSegment j: Segment [K, L] Segment mSegment m: Segment [M, N] of Quadrilateral poly4Segment nSegment n: Segment [N, O] of Quadrilateral poly4Segment oSegment o: Segment [O, P] of Quadrilateral poly4Segment pSegment p: Segment [P, M] of Quadrilateral poly4Segment qSegment q: Segment [Q, R] of Quadrilateral poly5Segment rSegment r: Segment [R, S] of Quadrilateral poly5Segment sSegment s: Segment [S, M] of Quadrilateral poly5Segment m_1Segment m_1: Segment [M, Q] of Quadrilateral poly5Segment p_1Segment p_1: Segment [P, T] of Quadrilateral poly6Segment tSegment t: Segment [T, U] of Quadrilateral poly6Segment uSegment u: Segment [U, V] of Quadrilateral poly6Segment kSegment k: Segment [W, Z] Segment kSegment k: Segment [W, Z] Segment kSegment k: Segment [W, Z] Segment lSegment l: Segment [Z, A_1] Segment lSegment l: Segment [Z, A_1] Segment lSegment l: Segment [Z, A_1] Segment g_1Segment g_1: Segment [A_1, B_1] Segment g_1Segment g_1: Segment [A_1, B_1] Segment g_1Segment g_1: Segment [A_1, B_1] Segment h_1Segment h_1: Segment [A_1, C_1] Segment i_1Segment i_1: Segment [J, D_1] Point D_1D_1 = (13.78,1.04) Point D_1D_1 = (13.78,1.04) theta ‘text1 = “theta ‘”theta ‘text1 = “theta ‘”thetatext1_1 = “theta “itext1_2 = “i”Coretext1_3 = “Core”Coretext1_3 = “Core”Coretext1_3 = “Core”Coretext1_3 = “Core”Claddingtext1_4 = “Cladding”Claddingtext1_4 = “Cladding”Claddingtext1_4 = “Cladding”Claddingtext1_4 = “Cladding”Claddingtext1_4 = “Cladding”Claddingtext1_4 = “Cladding”Claddingtext1_4 = “Cladding”Claddingtext1_4 = “Cladding”Claddingtext1_5 = “Cladding”Claddingtext1_5 = “Cladding”Claddingtext1_5 = “Cladding”Claddingtext1_5 = “Cladding”Claddingtext1_5 = “Cladding”Claddingtext1_5 = “Cladding”Claddingtext1_5 = “Cladding”Claddingtext1_5 = “Cladding”n_1text1_6 = “n_1 “n_1text1_6 = “n_1 “n_2text1_7 = “n_2”n_2text1_7 = “n_2”n_2text1_8 = “n_2”n_2text1_8 = “n_2”n_0text1_9 = “n_0”n_0text1_9 = “n_0”n_2text1_ {10} = “n_2”n_2text1_ {10} = “n_2”n_1text1_ {11} = “n_1 “n_1text1_ {11} = “n_1 “(theta ‘=90degree-theta)text1_ {12} = ” (theta ‘=90degree-theta) “(theta ‘=90degree-theta)text1_ {12} = ” (theta ‘=90degree-theta) “(theta ‘=90degree-theta)text1_ {12} = ” (theta ‘=90degree-theta) “(theta ‘=90degree-theta)text1_ {12} = ” (theta ‘=90degree-theta) “(theta ‘=90degree-theta)text1_ {12} = ” (theta ‘=90degree-theta) “(theta ‘=90degree-theta)text1_ {12} = ” (theta ‘=90degree-theta) “(theta ‘=90degree-theta)text1_ {12} = ” (theta ‘=90degree-theta) “(theta ‘=90degree-theta)text1_ {12} = ” (theta ‘=90degree-theta) “(theta ‘=90degree-theta)text1_ {12} = ” (theta ‘=90degree-theta) “(theta ‘=90degree-theta)text1_ {12} = ” (theta ‘=90degree-theta) “

A Light Incident in Fiber

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Question number: 3

» Physical Optics » Polarisation and Modern Optics » Principles of Fibre Optics Attenuation

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Appeared in Year: 2010

Short Answer Question▾

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Between multimode and single mode fibers, explain critically how and why single mode fiber is chosen for communication. (Section – A)

Question number: 4

» Physical Optics » Polarisation and Modern Optics » Single Mode Fibres

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Appeared in Year: 2010

Short Answer Question▾

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Why are the two wavelengths and important in single – mode fiber – optical communication system? (Section – A)

Question number: 5

» Physical Optics » Polarisation and Modern Optics » Ruby and He-Ne Lasers

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Appeared in Year: 2013

Essay Question▾

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Explain the lasing action of laser using energy level diagram and describe the operation of laser experimental setup. (Section A)

Explanation

  • The light is produced in He – Ne laser by atomic transitions in the Neon atom. The Helium acts as a buffer gas in it. That means Helium doesn’t directly produce laser light, it only helps the atom of the Neon gas to produce lasing in as manner.

  • Helium absorbed some energy, when energy applied to the He – Ne gas mixture, and it achieve an excited sta

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