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Single Slit Diffraction and Resolution

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					        Diffraction around an edge and
              through an aperture




Huygens’ principle is consistent with
diffraction:

The larger the wavelength with respect to the
aperture size, the more significant the
diffraction
                       Young’s Double Slit
                          Experiment

If light is a wave, interference effects will be seen, where one part
of wavefront can interact with another part.
One way to study this is to do a double-slit experiment:
Two-Source Interference
       Pattern

        Nodal points – blue dots
        Anti-nodal points – red dots
                      Young’s Double Slit
                         Experiment

If light is a wave, there should be an interference pattern
                        Path Difference
In a double-slit experiment, what      1) there is no difference

path difference have the waves         2) half a wavelength
from each slit traveled to give a      3) one wavelength
minimum at the indicated               4) three wavelengths
position?                              5) more than three wavelengths




                                    Intensity
                             Path Difference
                                         1) there is no difference
   In a double-slit experiment, what
                                         2) half a wavelength
   path difference have the waves
                                         3) one wavelength
   from each slit traveled to give a
   minimum at the indicated              4) three wavelengths

   position?                             5) more than three wavelengths




                                                                  
                                                                        2
                                                                               3
                                       Intensity
For Destructive Interference

 = 1/2 , 3/2 , 5/2 , 7/2 , …




                                                                      3/2
                                                                             5/2
                                                                                    7/2
                                                                /2
  = (m + 1/2) 
                   Diffraction by a Single Slit or
                                Disk


                       bright




                       dark



Note: There are
bright and dark
fringes beyond the
shadow. These
resemble
interference fringes
of a double slit.
                                   Diffraction by a Single Slit or
                                                Disk

                Wavelets from the center and edge of the slit                      Contrary to expectation, an interference
                                                                                   pattern is observed because the wave
                                                                Dark Band
                                                                (Interference of
                                                                                   front approaching slit, D, is itself a
         D
                                                      Y         wavelets)
                                                                                   source of wavelets. It is these individual
                        D
    D                                                                              wavelets that will interfere destructively
          Path difference = m/2
                                                                                   at the part of the screen where the dark
                                                                                   band is observed.
                      Wave front going through the
                      slit is made up of many
                      wavelets.                                                    Now from the diagram the angle to the
                               X                                                   first dark band is given by:
                                                                                          sinD = (/2) / (D/2) =  / D
If D is small we can use…..
similar triangles, for small angles                                                Therefore for any dark band
         sin D = tan D = Y / X                                                             sin D = m  / D
 so…..         Y/X           =       m/D                                                       m = 1,2,3…….
 or….            Y =mX/D
                     m = 1,2,3…….
                                  Diffraction by a Single Slit or
                                               Disk
                                                                               If we look at the intensity of the bright
                                                                               bands we find that they are not the
                                                                               same. The brightest band is in the
                                                                               center and the bands seem to get
                                                                               dimmer as the angle increases. The
                                                                               diagram shows how the intensity
                                                                               changes with angle. You can see that
                                                                               as sin approaches  / D the intensity
                                                                               falls to zero as expected from our
                                                                               theory. Also note that the second band
                                                                               is much less intense than the central
 Practice Problem                                                              band and is half as wide.
 Light of wavelength 750 nm passes through a slit 1.0 x 10-3 mm
 wide. How wide is the central maximum a) in degrees and b) in
 centimeters on a screen 20 cm away?


 = 750 x 10-9 m    D = 1.0 x 10-6 m   X = 0.2 m away     m=1
Width of central maximum is 2 or 2Y


 a) Width = 2 = 2 sin-1 (m  / D)      = 2 sin-1 (1) (750 x 10-9m) / (1.0 x 10-6 m )   = 97.20

 b) Width = 2Y = 2 X tan D      = 2 (0.2) tan 48.590    = 0.45 m
                   Check Your Understanding
The diffraction pattern below arises
                                       1) narrow the slit
from a single slit. If we would like
                                       2) widen the slit
to sharpen the pattern, i.e., make
                                       3) enlarge the screen
the central bright spot narrower,
what should we do to the slit width?   4) close off the slit
               Check Your Understanding
The diffraction pattern below arises
                                         1) narrow the slit
from a single slit. If we would like
                                         2) widen the slit
to sharpen the pattern, i.e., make
                                         3) enlarge the screen
the central bright spot narrower,
what should we do to the slit width?     4) close off the slit



The angle at which one finds the first
  minimum is:

            sin  =  /d
                                                         
The central bright spot can be                    d
                                                         
  narrowed by having a smaller angle.
  This in turn is accomplished by
  widening the slit.
             Check Your Understanding
Blue light of wavelength  passes
through a single slit of width d and         1) d/4
forms a diffraction pattern on a screen.     2) d/2
If the blue light is replaced by red light   3) no change needed
of wavelength 2, the original diffraction
                                             4) 2 d
pattern can be reproduced if the slit
width is changed to:                         5) 4 d
              Check Your Understanding
 Blue light of wavelength  passes
 through a single slit of width d and         1) d/4
 forms a diffraction pattern on a screen.     2) d/2
 If the blue light is replaced by red light   3) no change needed
 of wavelength 2, the original diffraction
                                              4) 2 d
 pattern can be reproduced if the slit
 width is changed to:                         5) 4 d



 d sin = m (minima)
If   2 then we must have d 
                                                           
2d for sin  to remain unchanged                       d
                                                           
(and thus give the same diffraction
pattern).
 When violet light of wavelength 415 nm falls on a single slit,
  it creates a central diffraction peak that is 9.20 cm wide
    on a screen that is 2.55 m away. How wide is the slit?




 = 415 x 10-9 m    Y = 4.6 x 10-2 m    X = 2.55 m away         m=1
D=?
      Because X is much bigger than Y, D is small, so we can
      use…..
               Y =mX/D                m = 1,2,3…….

     D = mX/Y         = (1) (415 x 10-9 m) (2.55 m) / (4.66 x 10-2 m)

     D = 2.30 x 10-5 m = 2.30 x 10-2 mm
       Rayleigh Criterion for Resolution of Two
                 Diffracted Images
Two images are said to be just resolved when the central maximum of one image
falls on the first minimum of the diffraction pattern of the other image.

                                               From single slit diffraction
                                                       sinD =  / D

                                                If  << d, sinD ≈ D
                                                       min =  / D
                                                Where min is in radians

                                               The apertures of microscopes and
                                               telescopes are circular. Analysis
                                               shows…

                                                       min = 1.22  / D
               Check Your Understanding
An optical telescope has a 21 cm
                                        1) bigger
mirror diameter. To give the same
                                        2) smaller
angular resolution as an optical
                                        3) the same
telescope the effective diameter of a
radio telescope would have to be….      4) doesn’t matter
               Check Your Understanding
An optical telescope has a 21 cm
                                        1) bigger
mirror diameter. To give the same
                                        2) smaller
angular resolution as an optical
                                        3) the same
telescope the effective diameter of a
radio telescope would have to be….      4) doesn’t matter



The minimum angle of resolution is
  given by:

         min = 1.22  /d
Radio waves have a much longer
  wavelength, so to give the same
  minimum angle of resolution the
  dish diameter has to be much larger
               Check Your Understanding
A scientist wants to observe finer   1) Make the objective bigger
detail on a specimen under a         2) Use smaller wavelengths
microscope. They could….
                                     3) Use oil between the
                                        specimen and the objective
                                     4) All of the above
                Check Your Understanding
 A scientist wants to observe finer      1) Make the objective bigger
 detail on a specimen under a            2) Use smaller wavelengths
 microscope. They could….
                                         3) Use oil between the
                                            specimen and the objective
                                         4) All of the above


The minimum angle of resolution is          Spherical and chromatic
  given by:                                   aberration however limit

          min = 1.22  /d
                                              the size of the objective,
                                              d. Also UV is absorbed
To make min smaller you would have           by glass.
  to make the objective lens bigger or
                                            Electrons can be used
  use a smaller wavelength. Oil has a
                                              because of their wave
  higher n than air so n = air / n.
                                              properties (E = h f)
    Two of Jupiter’s largest moons (it has 16 or more) are separated
   by a distance of 3.0 x 106 km at a time when the planet is 3.3 x 108 km
     from Earth. a) Would a person be able to resolve both moons with the
     unaided eye, assuming night-time pupil diameter of 7.5 mm? b) What
       minimum diameter of mirror would be needed in a radio telescope?

              Use:     light = 550 nm, radio = 21 cm and s = r 


   = 550 x 10-9 m      s = 3.0 x 109 m   r = 3.3 x 1011 m
  d = 7.5 x 10-3 m dmin = ?

        =s/r        = (3.0 x 109 m / (3.3 x 1011 m) = 0.009 rad

a) min = 1.22  / d = 1.22 (550 x 10-9 m / (7.5 x 10-3 m) = 0.000089 rad

      Yes  > min , so a person would be able to resolve both moons

b) dmin = 1.22  /  = 1.22 (21 x 10-2 m) / 0.009 = 28 m

      Now you can see why big dishes are needed for radio telescopes
A camera on a spy satellite orbiting at 200 km has a diameter of 35 cm.
     What is the smallest distance this camera can resolve on the
                          surface of the earth?

                     Use:   light = 550 nm, and s = r 




   = 550 x 10-9 m    s=?      r = 200 x 103 m     d = 35 x 10-2 m

min = 1.22  / d = 1.22 (550 x 10-9 m / (35 x 10-2 m)     = 1.92 x 10-6 rad

   s = r min = (200 x 103 m)(1.92 x 10-6 )     = 0.38 m (38 cm)

  This size of lens would not be good enough to read license plate
  numbers but good enough to resolve individual people.

				
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posted:11/9/2011
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