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# Diffraction New

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### Diffraction New

1. 1. Diffraction
2. 2. Diffraction <ul><li>Diffraction occurs when waves are forced to travel through a gap. </li></ul><ul><li>We see diffraction in harbour mouths when the waves spread out once they pass through the narrow opening. </li></ul><ul><li>We also hear diffraction as sound spread out once it travels through a doorway. </li></ul><ul><li>Waves generally diffract best when the gap they pass through is about the same as the wavelength of the wave. </li></ul>
3. 3. <ul><li>In diagram a the waves are only weakly diffracted as the gap is many times greater than the wavelength. </li></ul><ul><li>In diagram b the wave length is the same as the gap and the waves are strongly diffracted. </li></ul>a b
4. 4. <ul><li>With the EM spectrum the shorter the wavelength of the waves the less they generally diffract. </li></ul><ul><li>As a result you can easily pick up long wave radio in mountainous regions as they diffract more easily around the mountains, but FM is hard to pick up as it diffracts less easily. </li></ul>
5. 5. Diffraction from a single slit. <ul><li>If we place a single slit in front of a coherent light source such as a laser, an interference pattern is formed. </li></ul>laser Single slit screen Dark fringes
6. 6. <ul><li>The interference pattern from a single slit looks similar to the pattern from a double slit but there are subtle differences. The width of the slit determines the exact appearance of the fringes. </li></ul><ul><li>The light fringes are much wider than those seen in Young's double slits, the dark fringes only appear in the areas with complete destructive interference. </li></ul>
7. 7. As you will notice the bright fringes in a single slit pattern are many times wider than the dark fringes. (The formula used to predict the position of the dark fringes is the same as the diffraction grating formula, except that it is used to predict minima rather than maxima, this is not specified in the spec though)
8. 8. Additional notes: Huygens principle If we take situation a to be a wave before it passes through a slit, then the wave front is linear. In this model the wave front is a made up of a series of wavelets, each of which will have a point source at some point behind them. As these wavelets (and therefore their point sources) pass through a narrow opening they will diffract, to give a continuous curved wave front made up of wavelets. However as the wavelets are no longer in one line, they will start to interfere with each other despite their coherent nature.
9. 9. <ul><li>We using only 2 wavelets to explain this phenomena, in reality even in a narrow slit there will still be many “wavelets” passing through. </li></ul><ul><li>The example given is an over simplification of the principles but demonstrates how interference occurs. </li></ul><ul><li>(it is similar, but inverse, to the diffraction grating theory) </li></ul>Additional notes: reasons for minima in single slit patterns Wavelet point sources slit The two waves are completely out of phase (  rad), due to their path difference of  /2. This means that they cancel and there is a dark fringe. The next dark fringe will be at 3  /2 and so on. First beam has path of n  2 nd beam has path of n  +  /2
10. 10. Diffraction gratings
11. 11. <ul><li>A diffraction grating is basically a optical component with a regular pattern, which splits (diffracts) light into several beams travelling in different directions. </li></ul><ul><li>The diffraction gratings that we will be using are essentially a piece of glass with several thousand parallel lines etched on it’s surface. </li></ul>When a diffraction grating is placed in front of pan-chromatic light it will cause the colours to separate giving a spectrum.
12. 12. <ul><li>If we place a diffraction grating in front of a coherent monochromatic light source it produces an interference pattern. </li></ul><ul><li>It differs from both single and double slit patterns in that the maxima are small and relatively widely spaced. </li></ul>
13. 13. <ul><li>The order of diffraction (n) </li></ul><ul><li>The beam that travels straight is of the 0 th order (n=0), the first of the diffracted beams is the 1 st order (n=1), the second is the 2 nd order (n=2) and so on. </li></ul>n=0 n=1 n=2 n=-1 n=-2
14. 14. <ul><li>The formula we use to predict the maxima is: </li></ul><ul><li>n  = d sin  </li></ul><ul><li>Where: </li></ul><ul><li>n = the order of the diffracted beam. </li></ul><ul><li> = the wavelength of the light. </li></ul><ul><li> = the angle of the maxima (n) </li></ul><ul><li>d = the separation of the lines on the grating. </li></ul><ul><li>(the gratings have the number of lines/m stated on them. To work out d you have to calculate 1/(lines/m) </li></ul><ul><li>(some times values are in lines/inch) </li></ul>
15. 15. Derivation of n  = d sin  (Required) Consider a diffraction grating with slit spacing d . For the beam leaving a and b to form the first maxima on the screen they must be in phase  path difference of  . Distance bc must therefore be  (or n  as 1 st order). Distance ab must = d Angle b â c =  Using trig. Sin  = opp/hyp (opp = n  , hyp = d)  Sin  = n  /d (or n  = d sin  )
16. 16. 2 nd order (further explanation) c The next point at which we will get a maxima is when the path difference is sufficient for the waves to be in phase. This will next happen when the path difference bc is 2  . In this case sin  = 2  /d Each successive maxima will have a path difference of n  (where n is the order of the diffraction fringe). We can predict the number of fringes to appear as the maximum value of sin  is 1, by rearranging our formula we can calculate the maximum value of n .
17. 17. <ul><li>1) A diffraction grating has a second order diffraction of 15 degrees, the light incident is 450nm in wave length. </li></ul><ul><li>What is the slit spacing? </li></ul><ul><li>2) A diffraction grating has 300 lines per mm, the second order maxima is at 18.9 degrees, what is the wavelength of the light? </li></ul><ul><li>3) What is the maximum order for a diffraction grating of 500,000 lines per m if the wavelength of the light is 644nm? </li></ul><ul><li>4) Light of 5.70E-7 m is incident on a grating with a spacing of 2.00E-6 </li></ul><ul><li>what is the angle of a) the first maxima and b) the second maxima? </li></ul>
18. 18. <ul><li>1) A diffraction grating has a second order diffraction of 15 degrees, the light incident is 450nm in wave length. </li></ul><ul><li>What is the slit spacing? </li></ul><ul><li>n  =d sin  d= n  /sin  </li></ul><ul><li>d = (2x450E-9) / (sin15) </li></ul><ul><li>d = 3.48E-6 </li></ul><ul><li>A diffraction grating has 300 lines per mm, the second order maxima is at 18.9 degrees, what is the wavelength of the light? </li></ul><ul><li>n  =d sin   =d sin  /n </li></ul><ul><li> = ((1/3.00E5) sin18.9)/2 </li></ul><ul><li> = 5.40E-7 </li></ul>
19. 19. <ul><li>What is the maximum order for a diffraction grating of 500,000 lines per m if the wavelength of the light is 644nm? </li></ul><ul><li>sin   1  n  d/  </li></ul><ul><li>n = 2.00E-6 / 6.44E-7 </li></ul><ul><li>n = 3.11  n max =3 </li></ul><ul><li>4) Light of 5.70E-7 m is incident on a grating with a spacing of 2.00E-6 </li></ul><ul><li>what is the angle of a) the first maxima and b) the second maxima? </li></ul><ul><li>sin  =n  /d </li></ul><ul><li>a) sin  = 5.70E-7/2.00E-6 = 16.6 deg </li></ul><ul><li>b) sin  = (2 x 5.70E-7)/2.00E-6 = 34.8 deg </li></ul>
20. 20. Determination of the  of sodium light by diffraction. <ul><li>We can determine the wavelength of a light source using a diffraction grating, you will attempt to determine the wavelength of sodium light using a diffraction grating and a spectrometer. </li></ul><ul><li>The spectrometer has fixed collimator and slit through which the light will enter, the slit ensures that the light is coherent. The light will fall on the grating along its plane normal. A movable eye piece will allow you to calculate the angles of the maxima. </li></ul><ul><li>To determine the wavelength of the light you will have to determine the angles of n = 1 and n = -1 and halve the value, before calculating  from this value. </li></ul>
21. 21. <ul><li>To determine the wavelength of the light you will have to determine the angles of n = 0, n = 1 and n = -1. </li></ul><ul><li>Use these angles to determine the wavelength of the light. </li></ul><ul><li>(angle n=1 n=-1)/2 will give you a third value to check. </li></ul>n = 0 n = 1 2  Sodium lamp collimator Eye piece Grating Scale (deg)
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