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AUGUSTANA PH 202 - Diffraction Applications

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Diffraction ApplicationsPAL #25Slide 3Slide 4Slide 5Diffraction GratingsMaxima From GratingLocation of LinesElemental LinesSpectroscopeUsing SpectroscopyLine WidthDispersionResolving PowerResolving Power of a GratingSlide 16Spectral TypeNext TimeSlide 19Slide 20Slide 21Slide 22Diffraction ApplicationsPhysics 202Professor Lee CarknerLecture 26PAL #25 The first side pattern is between the m=1 and m=2 diffraction minima:a sin 1 =  and a sin 2 = 2sin  =  /a and sin  = 2 /asin 1 = 650 X 10-9 / 0.08 X 10-3 = 8.125 X 10-3sin 2 = (2)(650 X 10-9)/0.08 X 10-3 =1.625 X 10-2PAL #25What interference maxima are between the two angles? m1 = d sin 1 / and m2 = d sin 2 /m1 = (0.25 X 10-3)(8.125 X 10-3)/650 X 10-9= m2 = (0.25 X 10-3)(1.625 X 10-2)/650 X 10-9= We should see 3 bright fringes (m = 4,5,6) in the first side diffraction envelopePAL #25Middle interference fringe is m = 5sin  = (5)( )/(d) =  = (a/ sin  = [()(0.08 X 10-3 ) /(650 X 10-9 )] (0.013) = 5.026 rad = (d/) sin  = I = Imcos2 (sin /)2 = Im (1)(0.0358) = 0.036 ImPAL #25Screen is 2 meters away, what is at point 4.3 cm from the center? Diffraction patterny/D = m/a, m = ya/D = (4.3X10-2)(0.08X10-3)/(2)(650X10-9) = 2.65Diffraction Gratings If light of 2 different wavelengths passes through, each will produce a maxima, but they will tend to blur together This makes lines from different wavelengths easier to distinguishA system with large N is called a diffraction grating and is useful for spectroscopyMaxima From GratingLocation of Lines d sin  = mwhere d is the distance between any two slits (or rulings) on the grating For polychromatic light, each maxima is composed of many narrow lines (one for each wavelength the incident light is composed of)Elemental Lines When electrons move between these energy levels, they can produce light at a specific wavelength The pattern of spectral lines can identify the elementSpectroscope 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Can also take an image of the patternUsing Spectroscopy We want to be able to resolve lines that are close together How can we achieve this?Line WidthThe narrower the lines, the easier to resolve lines that are closely spaced in wavelength hw =  /(Nd cos )where N is the number of slits and d is the distance between 2 slitsDispersion D =    D = m / d cos For larger m and smaller d the resulting spectra takes up more spaceResolving PowerThe most important property of a grating is the resolving power, a measure of how well closely separated lines (in ) can be distinguishedR = av/ For example, a grating with R = 10000 could resolve 2 blue lines (= 450 nm) that were separated by 0.045 nmResolving Power of a Grating R = Nm Looking at higher orders helps to resolve linesSpectral TypeThe types of elements present in a star and the transitions they make depends on the temperature Examples:Very cool stars (T~3000 K) can be identified by the presence of titanium oxide which cannot exist at high temperaturesNext TimeFinal examMonday, 9-11 am, SC304Bring pencil and calculator4 equation sheets providedCovers 2/3 optics, 1/3 rest of courseIn a double slit diffraction pattern, what happens to the number of interference maxima in the first side pattern if you double the width of each slit?a) Increasesb) Decreasesc) Stays the sameIn a double slit diffraction pattern, what happens to the number of interference maxima in the first side pattern if you double the distance between the slits?a) Increasesb) Decreasesc) Stays the sameIn a double slit diffraction pattern, what happens to the number of interference maxima in the first side pattern if you double the wavelength?a) Increasesb) Decreasesc) Stays the sameIn a double slit diffraction pattern, what could to do to maximize the number of fringes in the central pattern?a) Increase a, increase db) Decrease a, decrease dc) Increase a, decrease dd) Decrease a, increase de) You can’t change the number of fringes in the central


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AUGUSTANA PH 202 - Diffraction Applications

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