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# Lect04 - 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) circle6 Single-Slit Diffraction * circle6 Multiple-slit Interference * circle6 Diffraction Gratings circle6 Spectral Resolution circle6 Optical Spectroscopy circle6 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
Phasors A 1 α α A φ A 1 Lets find the resultant amplitude of two waves using phasors. Suppose the amplitudes are the same. Represent each wave by a vector with magnitude (A 1 ) and direction ( φ ). One wave has φ = 0. Isosceles triangle: α=φ/2 . So, This is identical to our previous result ! More generally, if the phasors have different amplitudes A and B: C 2 = A 2 + B 2 + 2AB cos φ 1 2 cos 2 = φ A A Here φ is the external angle. φ A C B See the supplementary slide. See text: 35.3, 36.3, 36.4. See Physics 212 lecture 20. Phasors make it easier to solve other problems later.

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Phasors for 2 Phasors for 2 - - Slits Slits circle6 Plot the phasor diagram for different φ : A 1 A 1 φ A φ = 45 ° 8 / λ δ = A 1 A 1 φ A φ = 90 ° 4 / λ δ = A 1 A 1 φ A φ = 135 ° 8 / 3 λ δ = A 1 A 1 φ = 180 ° φ 2 / λ δ = A 1 A 1 A φ = 360 ° λ δ = A 1 A 1 A φ = 0 0 = δ A 1 A 1 φ A φ = 225 ° 8 / 5 λ δ = A 1 A 1 φ A φ = 270 ° 4 / 3 λ δ = A 1 A 1 φ A φ = 315 ° 8 / 7 λ δ = I 0 4I 1 φ 0 2 π -2 π θ λ /d - λ /d y /d)L -( λ /d)L *Small-angle approx. assumed here
Lecture 4, p 7 Q: What happens when a plane wave meets a small aperture? A: The result depends on the ratio of the wavelength λ to the size of the aperture, a: Huygens’ principle A Consequence of Superposition λ >> a Similar to a wave from a point source. This effect is called diffraction . λ << a The transmitted wave is concentrated in the forward direction, and at near distances the wave fronts have the shape of the aperture. The wave eventually spreads out. We will next study what happens when waves pass through one slit. We will use Huygens’ principle (1678): All points on a wave front ( e.g. , crest or trough) can be treated as point sources of secondary waves with speed, frequency, and phase equal to the initial wave. Wavefront at t=0 Wavefront at later time

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Lecture 4, p 8
Lecture 4, p 9 So far in the multiple-slit interference problems we have assumed that each slit is a point source.

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