Topic 11 11.3 Diffraction
Diffraction
Single edge <ul><li>This is the diffraction pattern produced around the edge of a razor blade </li></ul>
Narrow Slit
Circular Aperture <ul><li>The diffraction fringe pattern produced by a circular aperture. </li></ul>
Diffraction Patterns <ul><li>When plane wavefronts pass through a small aperture they spread out  </li></ul><ul><li>This i...
<ul><li>However, when we look at the diffraction pattern produced by light we observe a fringe pattern, </li></ul><ul><li>...
<ul><li>This intensity pattern arises from the fact that each point on the slit acts, in accordance with Huygen's principl...
Derivation of Equation <ul><li>We can deduce a useful relationship from a simple argument. </li></ul><ul><li>In this argum...
 
<ul><li>The source is placed at the principal focus of lens 1 and the screen is placed at the principal focus of lens 2. <...
<ul><li>To obtain a good idea of how the single slit pattern comes about we consider the next diagram </li></ul>
 
<ul><li>In particular we consider the light from one edge of the slit to the point P where this point is just one waveleng...
<ul><li>The wavefronts from the next point below the upper edge will similarly interfere destructively with the wavefront ...
<ul><li>If the screen is a long way from the slit then the angles  θ 1  and  θ 2  become nearly equal. </li></ul><ul><li>F...
<ul><li>However, both angles are very small, equal to  θ  say, so we can write that </li></ul><ul><li>θ   =  λ  / b </li><...
<ul><li>To obtain the position of the next maximum in the pattern we note that the path difference is 3/2  λ . </li></ul><...
Example <ul><li>Light from a laser is used to form a single slit diffraction pattern. The width of the slit is 0.10 mm and...
Answer <ul><li>Since the screen is a long way from the slit we can use the small angle approximation such that the f in d ...
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11.3

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11.3

  1. 1. Topic 11 11.3 Diffraction
  2. 2. Diffraction
  3. 3. Single edge <ul><li>This is the diffraction pattern produced around the edge of a razor blade </li></ul>
  4. 4. Narrow Slit
  5. 5. Circular Aperture <ul><li>The diffraction fringe pattern produced by a circular aperture. </li></ul>
  6. 6. Diffraction Patterns <ul><li>When plane wavefronts pass through a small aperture they spread out </li></ul><ul><li>This is an example of the phenomenon called diffraction </li></ul><ul><li>Light waves are no exception to this </li></ul>
  7. 7. <ul><li>However, when we look at the diffraction pattern produced by light we observe a fringe pattern, </li></ul><ul><li>that is, on the screen there is a bright central maximum with &quot;secondary&quot; maxima either side of it. </li></ul><ul><li>There are also regions where there is no illumination and these minima separate the maxima. </li></ul>
  8. 8. <ul><li>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. </li></ul><ul><li>It is the interference between these secondary wavefronts that produces the typical diffraction pattern. </li></ul>
  9. 9. Derivation of Equation <ul><li>We can deduce a useful relationship from a simple argument. </li></ul><ul><li>In this argument we deal with something that is called Fraunhofer diffraction, </li></ul><ul><li>that is the light source and the screen are an infinite distance away form the slit. </li></ul><ul><li>This can be achieved with the set up shown in the next figure. </li></ul>
  10. 11. <ul><li>The source is placed at the principal focus of lens 1 and the screen is placed at the principal focus of lens 2. </li></ul><ul><li>Lens 1 ensures that parallel wavefronts fall on the single slit and lens 2 ensures that the parallel rays are brought to a focus on the screen. </li></ul><ul><li>The same effect can be achieved using a laser and placing the screen some distance from the slit. </li></ul>
  11. 12. <ul><li>To obtain a good idea of how the single slit pattern comes about we consider the next diagram </li></ul>
  12. 14. <ul><li>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. </li></ul><ul><li>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. </li></ul><ul><li>Hence when these wavefronts arrive at P they will be out of phase and will interfere destructively. </li></ul>
  13. 15. <ul><li>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. </li></ul><ul><li>In this way we can pair the sources across the whole width of the slit. </li></ul>
  14. 16. <ul><li>If the screen is a long way from the slit then the angles θ 1 and θ 2 become nearly equal. </li></ul><ul><li>From the previous figure we see therefore that there will be a minimum at P if </li></ul><ul><li>λ = b sin θ 1 </li></ul><ul><li>where b is the width of the slit. </li></ul>
  15. 17. <ul><li>However, both angles are very small, equal to θ say, so we can write that </li></ul><ul><li>θ = λ / b </li></ul><ul><li>This actually gives us the half-angular width of the central maximum. </li></ul><ul><li>We can calculate the actual width of the maximum along the screen if we know the focal length of the lens focussing the light onto the screen. If this is f then we have that </li></ul><ul><li>θ = d / f </li></ul><ul><li>Such that </li></ul><ul><li>d = f λ / b </li></ul>
  16. 18. <ul><li>To obtain the position of the next maximum in the pattern we note that the path difference is 3/2 λ . </li></ul><ul><li>We therefore divide the slit into three equal parts, two of which will </li></ul><ul><li>produce wavefronts that will cancel and the other producing wavefronts that reinforce. The intensity of the second maximum is therefore much less than the intensity of the central maximum. </li></ul><ul><li>(Much less than one third in fact since the wavefronts that reinforce will have differing phases). </li></ul>
  17. 19. Example <ul><li>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? </li></ul>
  18. 20. Answer <ul><li>Since the screen is a long way from the slit we can use the small angle approximation such that the f in d = f λ / b becomes 3.0m. (i.e. f is the distance from the slit to the screen) </li></ul><ul><li>The half width of the centre maximum is 1.3cm so we have </li></ul><ul><li>λ = (1.3 x 10 -2 ) x(0.10 x 10 -3 ) / 3.0 </li></ul><ul><li>λ = 430 x 10 -9 or 430 nm </li></ul>

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