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Physical Optics by 1XjCa6C

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									 Physical Optics
The wave nature of light
     Interference
      Diffraction
     Polarization
             Huygens’ Principle
• Every point on a propagating
  wavefront serves as the
  source of spherical wavelets,
  such that the wavelets at
  sometime later is the
  envelope of these wavelets.
• If a propagating wave has a
  particular frequency and
  speed, the secondary
  wavelets have that same
  frequency and speed.
                                  “Isotropic”
                       Diffraction


                   a




• Diffraction – Bending of light into the shadow region
• Grimaldi - 17th Century observation of diffraction
• Diffraction vs. Refraction?
                                         a
Explanation of Snell’s Law
                                             BD          v1 t
                               s in  1             
                                             AD          AD

                                              AC         v 2t
                               s in  2             
                                              AD         AD


                              s in  1       v1 t       c / n1
                                                    
                              s in  2       v 2t       c /n2




       n 1 s in  1  n 2 s in  2
Superposition of waves




Constructive   Destructive
Interference   Interference
    Conditions for Interference
• To observe interference in light waves, the
  following two conditions must be met:
  1) The sources must be coherent
     • They must maintain a constant phase with respect to
       each other
  2) The sources should be monochromatic
     • Monochromatic means they have a single
       wavelength
Young’s Experiment
Young’s Experiment
          Young’s Experiment
                                                Maxima occur when:


    S2                             m  2              r1  r2  m 
                                   m  1
                                                                     
          d                        m  0                  s in  
                                                                     d
                                   m  1


    S1                             m  2                d s in   m 
                                 maxima

                                                Minima occur when:

                                                                    1 
                        s                           r1  r2   m   
               r2                                                   2
     S2                                     y
                             
d                                                                     1 
     S1            r1
                                                     d s in    m   
                            s  a                                    2
       r1  r2
What the pattern looks like
              Intensity Distribution,
                  Electric Fields
• The magnitude of each
  wave at point P can be
  found
   – E1 = Eo sin ωt
   – E2 = Eo sin (ωt + φ)
   – Both waves have the
     same amplitude, Eo

         2π         2π
   φ         δ         d s in θ
         λ          λ
Intensity Distribution, Resultant
              Field
 • The magnitude of the resultant electric field comes
   from the superposition principle
    – EP = E1+ E2 = Eo[sin ωt + sin (ωt + φ)]
                                                                    A  B       A  B 
                                          s in A  s in B  2 s in         cos        
                                                                      2           2   

 • This can also be expressed as
                        φ               φ 
         EP    2E ocos     s in  ω t    
                         2              2 
    – EP has the same frequency as the light at the slits
    – The magnitude of the field is multiplied by the factor 2
      cos (φ / 2)
 Intensity Distribution, Equation
• The expression for the intensity comes from the
  fact that the intensity of a wave is proportional to
  the square of the resultant electric field magnitude
  at that point
• The intensity therefore is
                     π d s in θ 
                    2                            2  πd  
   I  I m ax   cos               I m ax c o s      y
                         λ                        λL  
       Resulting Interference Pattern
• The light from the two slits
  forms a visible pattern on a
  screen
• The pattern consists of a series
  of bright and dark parallel
  bands called fringes
• Constructive interference
  occurs where a bright fringe
  occurs
• Destructive interference results
  in a dark fringe
                      Example
A He-Ne Laser has a wavelength of 633 nm. Two slits are
  placed immediately in front of the laser and an interference
  pattern is observed on a screen 10 m away. If the first
  bright band is observed 1 cm from the central bright fringe,
  how far apart are the two slits?

At what angle Q is the fourth dark band found?
Thin Films

    Constructive Interference (maxima)

               d ABC  m  n

                          o
                   n 
                          n

     Destructive Interference (minima)

                        1 
           d ABC    m  n
                        2
Chapter 35 - Problem 45


Stealth aircraft are designed to not reflect radar
whose wavelength is 2 cm, by using an anti-
reflecting coating. Ignoring any change in
wavelength in the coating, estimate its
thickness.
           Phase shift on reflection




        External Reflection               Internal Reflection


Now, If one reflection is internal and one reflection is external half
wavelength path differences will result in constructive interference
        Phase Changes Due To
              Reflection
• An electromagnetic wave
  undergoes a phase change
  of 180° upon reflection
  from a medium of higher
  index of refraction than
  the one in which it was
  traveling
   – Analogous to a pulse on
     a string reflected from a
     rigid support
                  Lloyd’s Mirror
• An arrangement for
  producing an interference
  pattern with a single light
  source
• Waves reach point P either
  by a direct path or by
  reflection
• The reflected ray can be
  treated as a ray from the
  source S’ behind the mirror
                     Lloyd’s Mirror

    S
                                             y
                                                  2
a                                                         a s in  m       m 2
         r1  r2                                     
    S1
                     s                           P h a s e s h ift o n re fle c tio n  
                   s  a

                                                                1             ym
                    a s in  m        2m       1    m           a m  a
                                   2                             2              s
    Interference in Thin Films Again
•    Assume the light rays are traveling
     in air nearly normal to the two
     surfaces of the film
•    Ray 1 undergoes a phase change of
     180° with respect to the incident ray
•    Ray 2, which is reflected from the
     lower surface, undergoes no phase
     change with respect to the incident
     wave
•    For constructive interference
      =2t = (m + ½)λn (m = 0, 1, 2 …)
           • This takes into account both the
             difference in optical path length for
             the two rays and the 180° phase
             change
•    For destructive interference
      =2t = mλn (m = 0, 1, 2 …)
            Newton’s Rings



                                           o
                                      
                                           n


Maxima - bright       Minima - dark
              1 
 2d m    m         2d m  m 
              2
   Chapter 37 – Problem 62
Show that the radius of the mth dark Newton’s ring
as viewed from directly above is given by:
                            oRm
                 xm 
                              n
Where R is the radius of curvature of the curved
glass surface and  is the wavelength of the light
used. Assume that the thickness of the air gap is
much less than R at all points and that x<<R
                           Newton’s Rings
                                                                      x  R  d                
                                                                 2            2                      2
                                                             R
                       R
                                                      R  d         
                                                                          2
                                            R                                 R  R d                      2Rd
                                       2         2                                     2    2            2
       R d                        x

                                                              d          2Rd  2Rd
                                                         2            2
                           d
                               
                                                     x
                   x
                                            Minima                                    2d  m  n
                                                                                                 o
                                                                                       2
Newtons rings are a                                                                   xm
                                                                                  2         m
special case of Fizeau fringes.                                                       2R             n
They are useful for testing
surface accuracy of a lens.                                                                  oRm
                                                                                  xm 
                                                                                                 n
              o
                                     Maxima                                          oR    1 
               n                                                 xm                       m  
                                                                                        n     2 
       Wedges – Fringes of Equal Thickness

                                        Minima (destructive)
d
                                                                 o
                                                2t  m  m
                                                                 n
                                    x
                                        Maxima (constructive)
            t  x                                          1 
                                                 2t    m  
                                                            2 

                                                              1 
    Fringes of this type are also              x m        m  
                                                        2     2
    known as Fizeau fringes
                                                             
                                                    x m 
                                                             2
Example: thin film of air
Michelson Interferometer
                           • A lens can be used to form
                 d          fringes of equal inclination
                             (rings)
                           • Tilting the mirrors can cause
                             fringes of equal thickness.
                           • Accurate length
                             measurements are
                             accomplished by fringe
             Compensator     counting as one of the
             Plate           mirrors is moved.


                               2d  m
  Detector




                                        m
                                 d 
                                         2
42.     Monochromatic light is beamed into a Michelson
interferometer. The movable mirror is displaced 0.382
mm, causing the interferometer pattern to reproduce itself
1 700 times. Determine the wavelength of the light. What
color is it?




                                                          7
                                              4.  10
                                                  49           m    449 n m
5.      Young’s double-slit experiment is performed with 589-nm
light and a distance of 2.00 m between the slits and the screen.
The tenth interference minimum is observed 7.26 mm from the
central maximum. Determine the spacing of the slits.

7.       Two narrow, parallel slits separated by 0.250 mm are
illuminated by green light (λ = 546.1 nm). The interference
pattern is observed on a screen 1.20 m away from the plane of
the slits. Calculate the distance (a) from the central maximum to
the first bright region on either side of the central maximum and
(b) between the first and second dark bands.

17.      In Figure 37.5, let L = 120 cm and d = 0.250 cm. The
slits are illuminated with coherent 600-nm light. Calculate the
distance y above the central maximum for which the average
intensity on the screen is 75.0% of the maximum.
 32.      A thin film of oil (n = 1.25) is located on a smooth wet
 pavement. When viewed perpendicular to the pavement, the film
 reflects most strongly red light at 640 nm and reflects no blue
 light at 512 nm. How thick is the oil film?

33.     A possible means for making an airplane invisible to
radar is to coat the plane with an antireflective polymer. If radar
waves have a wavelength of 3.00 cm and the index of refraction
of the polymer is n = 1.50, how thick would you make the
coating?
37.     A beam of 580-nm light passes
through two closely spaced glass plates,
as shown in Figure P37.37. For what
minimum nonzero value of the plate
separation d is the transmitted light
bright?
Mach Zehnder Interferometer

                                                      Detector




•   No compensating plate needed
•   Test cell easily inserted in one leg.
•   No factor of two as in the Michelson interferometer.
•   Difficult to align
                Sagnac Interferometer
s  r
                                      R                            4R     2                          4R        2
v  r                                                                                    tccw 
                               v
                                      2                   t cw 
                                  2
                                                                          v                                v
                                                                   c                              c
                              R                                           2                                    2
                               2
                                           R      v

                                                                              t  t cw  t ccw

                                                                                   8R           8R
                                                                        t                 
                                                                                   2c  v       2c  v

                                                                                 8R                8R
                                                                   t                      
                                                                               2 c  R         2 c  R
                                                                      
                                               8R      1         1       8R             R       R            
                                          t                                      1     1                
                                                2c      R        R                       2c                    
                 Detector                            1        1          2c                       2c
                                                   
                                                         2c        2c 
                                                                       
            Ring laser gyro.
                                                      8 R  2 R    8R 
                                                                          2
                                                                                                         8R 
                                                                                                           2

The rotation effectively shortens the path       t                 2
                                                                                              ct 
 taken by one direction over the other.                 2c  2c      c                                    c
                                   Phase Angle
                                                                 i  1   t               i  2   t 
                  E  E 1  E 2  E 01e                                            E 02 e


                                                                                     i    t 
                                                                 E  E0 e
       E                 E2          E 0 2 s in  2
                      2
                                     E 0 1 s in  1
    1          E1
                                                                          E 0 1 s in  1  E 0 2 s in  2
E 01 c o s  1    E 02 c o s  2                      ta n         
                                                                          E 01 c o s  1  E 02 c o s  2

                                     2                 2
                 E           E 01        E 02             2 E 01 E 02 c o s  2   1 

								
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