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Diffraction Applications Physics 202 Professor Lee Carkner Lecture 28
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PAL #27 What is in the first side pattern in a double slit set-up with a = 0.08 mm and d = 0.25 mm and l = 650 nm? a sin = and a sin = 2 sin = /a and sin = 2 /a sin 2 = (2)(650 X 10 -9 )/0.08 X 10 -3 =1.625 X 10 -2
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PAL #27 What interference maxima are between the two angles? d sin 1 =m 1 and d sin 2 = m 2 m 1 = (0.25 X 10 -3 )(8.125 X 10 -3 )/650 X 10 -9 = 3.13 m 2 = (0.25 X 10 -3 )(1.625 X 10 -2 )/650 X 10 -9 = 6.25
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PAL #27 Middle interference fringe is m = 5 = ( a/ sin = [( )(0.08 X 10 -3 ) /(650 X 10 -9 )] (0.013) = 5.026 rad = ( d/ ) sin = [( )(0.25 X 10 -3 ) /(650 X 10 -9 )] (0.013) = 15.71 rad I = I m cos 2 (sin / ) 2 = I m (1)(0.0358) = 0.036 I m
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Diffraction Gratings For double slit interference the maxima are fairly broad If we increase the number of slits (N) to very large numbers (1000’s) the individual maxima (called lines) become narrow A system with large N is called a diffraction grating and is useful for spectroscopy
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Maxima From Grating
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Diffraction Grating
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Grating Path Length
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Location of Lines d sin = m The m=0 maxima is in the center, and is flanked by a broad minima and then the m=1 maxima etc. For polychromatic light each maxima is composed of many narrow lines (one for each wavelength the incident light is composed of)
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Grating Orders
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Line Width The half-width (angular distance from the peak to zero intensity) of a line is given by: where N is the number of slits and d is the distance between 2 slits
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Line Profile
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Spectroscopy A hot material (e.g. a gas) composed of atoms will emit light due to electrons changing energy levels Gratings are used to separate light into its constituent wavelengths in order to identify this light as spectral lines This is how the chemical compositions of stars are determined
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Spectroscope This will produce a series of orders, each order containing lines (maxima) over a range of wavelengths The wavelength of a line corresponds to its position angle We measure with a optical scope mounted on a vernier position scale
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Spectroscope
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Spectral Type This is how the temperature of stars is determined Examples: Stars like the Sun (T~5500 K) can be identified by the Ca h and k doublet which is only produced at moderate temperatures
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Using Spectroscopy What properties do we want our spectroscope grating to have? How can we achieve this?
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Dispersion The size of the spectrum (from short to long ) produced by a grating is a function of the dispersion, D: The dispersion can also be written: For larger m and smaller d the resulting spectra takes up more space
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Dispersed Spectra
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Resolving Power The most important property of a grating is the resolving power, a measure of how well closely separated lines (in ) can be distinguished R = av / For example, a grating with R = 10000 could resolve 2 blue lines ( = 450 nm) that were separated by 0.045 nm
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Dispersion and Resolving Power
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Resolving Power of a Grating R = Nm Gratings with large N are the best for spectral work
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