No Title
OE594 Guided Wave Optics
Solution to Exam 1
                                                                                                           
Thursday 03/24/11 Name
D119: 1st period
JFW Box #



                    
                                         (1)
(30)
(2)
(20)
(3)
(20)
(4)
(30)
TOTAL
(100)
  1. Give brief answers to the following five questions.

      (a) Define what is meant by a transverse electric mode.
      ANS: The electric field in the light beam is perpendicular to the plane of incidence.
      (b) What makes up a simple waveguide package and what variables are required to solve the guidance equation?
      ANS: The three dielectric materials are the cover, waveguide and substrate. One needs to know the indicies of refraction of the three materials, the vacuum wavelength of the beam and the waveguide thickness.
      (c) In an ideal waveguide system, what does the orthogonality equation tell us?
      ANS: It tells us that the different modes do not communicate with each other.
      (d) What mechanism produces guidance in a waveguide?
      ANS: Total internal reflection.
      (e) If all media in a waveguide system are isotropic and lossless, what can be said of m and e?
      ANS: Both are real and simply constants.
  2. Given
    n1=1
    n2=2.20
    n3=1.80
    at what boundary (cover-waveguide or waveguide-substrate) will total guidance be determined? What is the critical angle for that boundary?
    ANS: Total guidance is achieved when TIR occurs at the waveguide-substrate boundary. The critical angle is
    q23
    =
    sin-1 æ
    è
    n3

    n2
    ö
    ø
    =
    sin-1 æ
    è
    1.80

    2.20
    ö
    ø
    =
    54.9o
  3. If the TE electric field in the cover of an all dielectric waveguide package is given by
    E(x)
    =
    Ce-qx

      (a) Derive the corresponding magnetic field expression H(x)?
      Start with
      Ñ×E
      =
      - B

      t
      =
      -m H

      t
      =
      -imwH
      Solve for the magnetic field and break up into components.
      H
      =
      i

      mw
      Ñ×E
      Hx 
      ^
      i
       
      + Hz 
      ^
      k
       
      =
      i

      mw
      ê
      ê
      ê
      ê
      ê
      ê
      ê
      ^
      i
       
      ^
      j
       
      ^
      k
       

      x

      y

      z
      0
      Ey
      0
      ê
      ê
      ê
      ê
      ê
      ê
      ê
      =
      i

      mw
      æ
      è
      - Ey

      z
       
      ^
      i
       
      + Ey

      x
       
      ^
      k
       
      ö
      ø
      =
      i

      mw
      é
      ë
      -(-ib)Ey 
      ^
      i
       
      + Ey

      x
      ù
      û
      =
      æ
      è
      - b

      mw
      Ey ö
      ø
       
      ^
      i
       
      + æ
      è
      i

      mw
      ö
      ø
      Ey

      x
       
      ^
      k
       
      The phase terms cancel leaving
      Hx(x)
      =
      - b

      mw
      Ey(x)
      =
      - bC

      mw
      e-qx
      Hz(x)
      =
      i

      mw
      Ey

      x
      =
      - iqC

      mw
      e-qx

      (b) From the above fields, calculate the Poynting vector, S.
      The Poynting vector is
      S
      =
      E×H*
      =
      æ
      è
      Ey 
      ^
      j
       
      ö
      ø
      × æ
      è
      Hx* 
      ^
      i
       
      +Hz*  
      ^
      k
       
      ö
      ø
      where the phase terms have cancelled out and the Poynting vector is complex at this point. Expanding,
      S
      =
      EyHz* 
      ^
      i
       
      -EyHx* 
      ^
      k
       
      =
      Ey æ
      è
      iqC

      mw
      e-qx ö
      ø
       
      ^
      i
       
      -Ey æ
      è
      -bC

      mw
      e-qx ö
      ø
       
      ^
      k
       
      =
      Ce-qx é
      ë
      C

      mw
      (iq 
      ^
      i
       
      + b 
      ^
      k
       
      )e-qx ù
      û
      =
      C2

      mw
      æ
      è
      iq 
      ^
      i
       
      +b  
      ^
      k
       
      ö
      ø
      e-2qx
  4. Use your computer program for this problem. Given a waveguide system in which
    n1=1
    n2=2.10
    n3=1.75
    t=1.05 mm
    l = 632.8 nm

      (a) How many modes does the waveguide have?
      ANS: four modes
      (b) What is the effective index for each mode?
      neff0
      =
      2.08
      neff1
      =
      2.03
      neff2
      =
      1.95
      neff3
      =
      1.82



File translated from TEX by TTH, version 3.85.
On 30 Mar 2011, 16:17.