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					Sight and wave phenomena

     Topic A.4 Diffraction
   For this section, we will be
dealing with Diffraction at a single
                slit
 When a He-Ne beam passes through a single
narrow slit, this was observed on a distant screen
           What happened?
 Recall that when waves meet an aperture or
  barrier, they diffract.
 The image shows the diffraction of the light
  at the slit.
 Take a closer look at the image, what can
  you observe?
     Did you observe 2 things?
 The centre spot is the brightest, and the
  spots get dimmer as they go to the sides
 The centre spot is about twice the length of
  the spots adjacent to it.
Sketch the variation with angle of diffraction
of the relative intensity of light diffracted at a
single slit.
Single Aperture Diffraction Pattern
With a narrower aperture, the overall intensity of the
 light is reduced and the pattern on the screen is
                       wider.
This is how the image and the graph coincide
         Diffraction Patterns
 When plane wavefronts pass through a
  small aperture they spread out
 This is an example of the phenomenon
  called diffraction
 Light waves are no exception to this
 However, when we look at the diffraction pattern
  produced by light we observe a fringe pattern,
 that is, on the screen there is a bright central
  maximum with "secondary" maxima either side of
  it.
 There are also regions where there is no
  illumination and these minima separate the
  maxima.
 This intensity pattern arises from the fact
  that each point on the slit acts, in
  accordance with Huygen's principle, as a
  source of secondary wavefronts.
 It is the interference between these
  secondary wavefronts that produces the
  typical diffraction pattern.
Derive the formula θ = λ / b for the position
of the first minimum of the diffraction pattern
produced at a single slit.
Consider the diagram below
 In particular we consider the light from
  one edge of the slit to the point P where
  this point is just one wavelength further
  from the lower edge of the slit than it is
  from the upper edge.
 The secondary wavefront from the upper
  edge will travel a distance λ/2 further
  than a secondary wavefront from a point
  at the centre of the slit.
 Hence when these wavefronts arrive at
  P they will be out of phase and will
  interfere destructively.
 The wavefronts from the next point below
  the upper edge will similarly interfere
  destructively with the wavefront from the
  next point below the centre of the slit.
 In this way we can pair the sources across
  the whole width of the slit.
 If the screen is a long way from the slit
  then the angles θ1 and θ2 become nearly
  equal.
 From the previous figure we see
  therefore that there will be a minimum
  at P if
 sin θ1 = λ / b
 where b is the width of the slit.
 However, both angles are very small, and
 θ1≈ θ2 ≈ θ, so we can write that
 θ =λ/b

 With this, we have linked the size of the
  slit to the angle and wavelength of the
  wave.
 What is the relationship between the size
  of the image formed on the screen and
  the wavelength?
                          tan θ ≈ y/D
                          tan θ ≈ sin θ ≈ θ ≈ y/D




b




    There is a mistake in the diagram, can you spot it?
      If the distance of the screen to the slit is D, the
    actual width of the maximum along the screen can
                       be calculated:
                                            y

                  θ
                           D

   tan θ = y/D
   For very small values of θ, we approximate
   tan θ ≈ θ
   Therefore θ ≈ y/D , since θ = λ / b
   We have y = λ D / b
Solve problems involving single-slit
diffraction.
                   Example
 Light from a laser is used to form a single
  slit diffraction pattern. The width of the slit is
  0.10 mm and the screen is placed 3.0 m
  from the slit. The width of the central
  maximum is measured as 2.6 cm. What is
  the wavelength of the laser light?
                  Answer
 Since the screen is a long way from the slit
  we can use the small angle approximation.
 The central max = 2.6 => y = 1.3 x 10-2m
 Given D = 3.0m, b = 0.10 x 10-3m
 Using y = λ D / b
 λ = (1.3 x 10-2) x(0.10 x 10-3) / 3.0
 λ = 430 x 10-9 or 430 nm

				
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posted:5/9/2013
language:English
pages:23