11.3 2013

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IBO Diploma Physics Topic 11.3

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11.3 2013

  1. 1. Topic 1111.3 Diffraction
  2. 2. Single edge  This is the diffraction pattern produced around the edge of a razor blade
  3. 3. Narrow Slit
  4. 4. DiffractionSketch the variation with angle ofdiffraction of the relative intensity oflight diffracted at a single slit. Hyperlink
  5. 5. Circular Aperture  The diffraction fringe pattern produced by a circular aperture.
  6. 6. 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
  7. 7.  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.
  8. 8.  This intensity pattern arises from the fact that each point on the slit acts, in accordance with Huygens principle, as a source of secondary wavefronts. It is the interference between these secondary wavefronts that produces the typical diffraction pattern.
  9. 9. Derivation of Equation We can deduce a useful relationship from a simple argument. In this argument we deal with something that is called Fraunhofer diffraction, that is the light source and the screen are an infinite distance away form the slit. This can be achieved with the set up shown in the next figure.
  10. 10.  The source is placed at the principal focus of lens 1 and the screen is placed at the principal focus of lens 2. 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. The same effect can be achieved using a laser and placing the screen some distance from the slit.
  11. 11. HyperlinkD i ffr a c tio n a t a si n g le a p e rt u r e S ing le slit d is ta n t s c r e e n in ten s ity ac ro s s s c re e n
  12. 12. Effect of slit width Single Aperture Diffraction Pattern Single Aperture Diffraction Pattern: Narrower Aperture
  13. 13. Effect of wavelength θ =λ/b θ =2λ/b
  14. 14.  To obtain a good idea of how the single slit pattern comes about we consider the next diagram
  15. 15.  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.
  16. 16.  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.
  17. 17.  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 λ = b sin θ1 where b is the width of the slit.
  18. 18.  However, both angles are very small, equal to θ say, so we can write that θ =λ/b This actually gives us the half-angular width of the central maximum. 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 θ =d/f Such that d=fλ/b
  19. 19.  To obtain the position of the next maximum in the pattern we note that the path difference is 3/2 λ. We therefore divide the slit into three equal parts, two of which will 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. (Much less than one third in fact since the wavefronts that reinforce will have differing phases).
  20. 20. 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?
  21. 21. Answer 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) The half width of the centre maximum is 1.3cm so we have 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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