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Incident Wave wavelength l

2. Question. Light interfering from 10 equally spaced slits initially illuminates a screen. Now we double the number of slits, keeping the spacing constant. What happens to the net power on the screen?a. stays the sameb. doubles c. increases by 4. If we double the number of slits, we expect the net power on the screen to double. How does it do this

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Incident Wave wavelength l

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    1. 1 http://w3.physics.uiuc.edu/flexner/lpages/flash/PhasorDiagrams.html http://w3.physics.uiuc.edu/flexner/lpages/flash/PhasorDiagrams.html

    2. 2 Question Light interfering from 10 equally spaced slits initially illuminates a screen. Now we double the number of slits, keeping the spacing constant. What happens to the net power on the screen? a. stays the same b. doubles c. increases by 4

    3. 3 N-Slit Interference – Summary The Intensity for N equally spaced slits is given by: Can show basic idea with UK slit program http://w3.physics.uiuc.edu/flexner/lpages/flash/PhasorDiagrams.html Can show basic idea with UK slit program http://w3.physics.uiuc.edu/flexner/lpages/flash/PhasorDiagrams.html

    4. 4

    5. 5 Gratings We have set up some discharge tubes with various gases. Notice that the colors of the various discharges are quite different. In fact the light emitted from the highly excited gases is composed of many “discrete wavelengths.” You can see this with the plastic “grating”.

    6. 6 Diffraction Gratings Diffraction gratings rely on N-slit interference. They simply consist of a large number of evenly spaced parallel slits. Recall that the intensity pattern produced by light of wavelength ? passing through N slits with spacing d is given by:

    7. 7 How effective are diffraction gratings at resolving light of different wavelengths (i.e. separating closely-spaced ‘spectral lines’)? Concrete example: Na lamp has a spectrum with two yellow “lines” very close together: ?1 = 589.0 nm, ?2 = 589.6 nm (Dl = 0.6 nm) Are these two lines distinguishable using a particular grating? Diffraction Gratings

    8. 8 Resolution

    9. 9 Diffraction Gratings

    10. 10 Example Angular splitting of the Sodium doublet: Consider the two closely spaced spectral (yellow) lines of sodium (Na), ?1 = 589 nm and??2 = 589.6 nm, mentioned earlier. If light from a sodium lamp illuminates a diffraction grating with 4000 slits/cm, what is the angular separation of these two lines in the second-order (m=2) spectrum?

    11. 11 Question Assuming we fully illuminate the grating from the previous problem (d = 2.5 ?m), how big must it be to resolve the Na lines (589 nm, 589.6 nm)? Spend some time on theseSpend some time on these

    12. 12 So far in the N-slit problem we have assumed that each slit is a point source. Point sources radiate equally in all directions. Real slits have a non-zero extent – - a “slit width” a. The transmission pattern depends on the ratio of a to ?. In general, the smaller the slit width, the more the wave will diffract! demo with laser and slit firstdemo with laser and slit first

    13. 13 Slit of width a. Where are the minima? The first minimum is at an angle such that the light from the top and the middle of the slit destructively interfere: The second minimum is at an angle such that the light from the top and a point at a/4 destructively interfere: Single-Slit Diffraction

    14. 14

    15. 15 Question Which of the following would broaden the diffraction peak? a. reduce the laser wavelength b. reduce the slit width c. move the screen further away demo with laser and two slit widthsdemo with laser and two slit widths

    16. 16 Single-slit Diffraction — Summary The intensity of a single slit has the following form:

    17. 17

    18. 18 Light of wavelength ? is incident on an N-slit system with slit width a and slit spacing d. 1. The intensity ? as a function of y at a viewing screen located a distance L from the slits is shown to the right. What is N? (L >> d, y, a) Question

    19. 19 Appendix: N-Slit Interference Intensity for N equally spaced slits is easily found from phasor analysis:

    20. 20 Appendix: Single-slit Diffraction To analyze diffraction, we treat it as interference of light from many sources (i.e. the Huygens wavelets that originate from each point in the slit opening). Model the single slit as M point sources with spacing between the sources of a/M. We will let M go to infinity on the next slide. The phase difference ? between first and last source is given by ??????? da????? a sin? / ????? a????

    21. 21 Single-slit Diffraction — the math We have turned the single-slit problem into the M-slit problem that we already solved in this lecture. However, as we let M ???, the problem becomes much simpler because the polygon becomes the arc of a circle. The radius of the circle is determined by the relation between angle and arc length: ? = Ao/R.

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