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# lecture4 - Lecture 4 Diffraction Spectroscopy y d L Spectra...

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Lecture 4, p 1 Lecture 4: Diffraction & Spectroscopy y L d θ Spectra of atoms reveal the quantum nature of matter Take a plastic grating from the bin as you enter class.

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Lecture 4, p 2 Today s Topics * Derivations in Appendix (also in Young and Freeman, 36.2 and 36.4) Multiple-slit Interference * Diffraction Gratings Spectral Resolution Optical Spectroscopy Single-Slit Diffraction * Interference + Diffraction
Lecture 4, p 3 Review of 2-Slit Interference Only the phase difference matters. Phase difference is due to source phases and/or path difference. In a more complicated geometry (see figure on right) , one must calculate the total path from source to screen. If the amplitudes are equal Use trig identity: A = 2A 1 cos( φ /2). • Phasors: Phasors are amplitudes. Intensity is the square of the phasor length.

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Lecture 4, p 4 Multiple Slit Interference The positions of the principal maxima occur at φ = 0, ± 2 π , ± 4 π , ... where φ is the phase between adjacent slits. θ = 0, ± λ /d , ± 2 λ /d, ... The intensity at the peak of a principal maximum goes as N 2 . 3 slits: A tot = 3A 1 I tot = 9I 1 . N slits: I N = N 2 I 1 . Between two principal maxima there are N-1 zeros and N-2 secondary maxima The peak width 1/N . The total power in a principal maximum is proportional to N 2 (1/N) = N. 0 2 π - 2 π I 4 0 16I 1 N=4 0 2 π - 2 π I 5 0 25I 1 N=5 0 2 π - 2 π I 3 0 9I 1 N=3 - λ / d 0 λ / d φ θ φ θ - λ / d 0 λ / d - λ / d 0 λ / d φ θ
Lecture 4, p 5 N-Slit Interference The Intensity for N equally spaced slits is given by: Derivation (using phasors) is in the supplementary slides. 2 1 sin( /2) sin( /2) N N I I φ φ Λ Ξ = Μ Ο Ν Π * * Your calculator can probably graph this. Give it a try. y L d θ φ is the phase difference between adjacent slits. You will not be able to use the small angle approximations unless d >> λ . As usual, to determine the pattern at the screen, we need to relate φ to θ or y = L tan θ : dsin d and 2 y L φ δ θ θ θ π λ λ λ = =

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Lecture 4, p 6 Example Problem 0 θ min λ /d θ In an N-slit interference pattern, at what angle θ min does the intensity first go to zero? (In terms of λ , d and N).
Lecture 4, p 7 Solution 0 θ min λ /d θ In an N-slit interference pattern, at what angle θ min does the intensity first go to zero? (In terms of λ , d and N). But φ min = 2 π (d sin θ min )/ λ 2 π d θ min / λ = 2 π /N. Therefore, θ min λ /Nd. As the number of illuminated slits increases, the peak widths decrease! General feature: Wider slit features à narrower patterns Narrower slit features à wider patterns .... 2 1 sin( / 2) sin( / 2) N N I I φ φ Λ Ξ = Μ Ο Ν Π has a zero when the numerator is zero. That is, φ min = 2 π /N. Exception: When the denominator is also zero. That s why there are only N-1 zeros.

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Note:The simple calculations we have done only hold in the far-field (a.k.a. Fraunhofer limit), where L >> d 2 / λ .
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