1 / 22

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

zuzana
Download Presentation

Diffraction Applications

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related