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Diffraction Applications

Diffraction Applications. Physics 202 Professor Lee Carkner Lecture 26. PAL #25. The first side pattern is between the m=1 and m=2 diffraction minima: a sin q 1 = l and a sin q 2 = 2 l sin q 1 = l /a and sin q 2 = 2 l /a

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Diffraction Applications

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  1. Diffraction Applications Physics 202 Professor Lee Carkner Lecture 26

  2. PAL #25 • The first side pattern is between the m=1 and m=2 diffraction minima: a sin q1 = l and a sin q2 = 2l sin q1 = l /a and sin q2 = 2l /a sin q1 = 650 X 10-9 / 0.08 X 10-3 = 8.125 X 10-3 sin q2 = (2)(650 X 10-9)/0.08 X 10-3 =1.625 X 10-2

  3. PAL #25 • What interference maxima are between the two angles? m1 = d sin q1 /l and m2 = d sin q2 /l 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 envelope

  4. PAL #25 • Middle interference fringe is m = 5 • sin q = (5)(l)/(d) = • a = (pa/l) sin q = [(p)(0.08 X 10-3 ) /(650 X 10-9 )] (0.013) = 5.026 rad • b = (pd/l) sin q = • I = Imcos2b (sin a/a)2 = Im (1)(0.0358) = 0.036 Im

  5. PAL #25 • Screen is 2 meters away, what is at point 4.3 cm from the center? • Diffraction pattern • y/D = ml/a, m = ya/Dl = (4.3X10-2)(0.08X10-3)/(2)(650X10-9) = 2.65

  6. 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 spectroscopy

  7. Maxima From Grating

  8. Location of Lines d sin q = ml • 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)

  9. 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 element

  10. 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 q • We measure q with a optical scope mounted on a vernier position scale • Can also take an image of the pattern

  11. Using Spectroscopy • We want to be able to resolve lines that are close together • How can we achieve this?

  12. Line Width • The narrower the lines, the easier to resolve lines that are closely spaced in wavelength D qhw = l /(Nd cos q) • where N is the number of slits and d is the distance between 2 slits

  13. Dispersion D = Dq/Dl D = m / d cos q • For larger m and smaller d the resulting spectra takes up more space

  14. Resolving Power • The most important property of a grating is the resolving power, a measure of how well closely separated lines (in l) can be distinguished R = lav/Dl • For example, a grating with R = 10000 could resolve 2 blue lines (l = 450 nm) that were separated by 0.045 nm

  15. Resolving Power of a Grating R = Nm • Looking at higher orders helps to resolve lines

  16. Spectral 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

  17. 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 course

  18. In 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? • Increases • Decreases • Stays the same

  19. In 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? • Increases • Decreases • Stays the same

  20. In a double slit diffraction pattern, what happens to the number of interference maxima in the first side pattern if you double the wavelength? • Increases • Decreases • Stays the same

  21. In a double slit diffraction pattern, what could to do to maximize the number of fringes in the central pattern? • Increase a, increase d • Decrease a, decrease d • Increase a, decrease d • Decrease a, increase d • You can’t change the number of fringes in the central pattern

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