Lect04 - Le cture4 Diffraction S ctroscopy pe 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) Single-Slit Diffraction * Multiple-slit Interference * Diffraction Gratings Spectral Resolution Optical Spectroscopy 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
Lecture 4, p 5 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 6 So far in the multiple-slit interference problems 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 λ . In general, the smaller the slit width, the more the wave will diffract. screen Diffraction profile I 1 Small slit: I 1 Large slit: Single-slit Diffraction
Single-Slit Diffraction Slit of width a. Where are the minima? Use Huygens’ principle: treat each point across the opening of the slit as a wave source. The first minimum is at an angle such that the light from the top and the middle of the slit destructively interfere . This works, because for every point in the top half, there is a corresponding point in the bottom half that cancels it. θ a/2 δ min sin 2 2 a λ δ θ = = min sin a = P Incident Wave (wavelength λ ) y L a The second minimum is at an angle such that the light from the top and a point at a/4 destructively interfere: min,2 sin 4 2 a = = min,2 2 sin a = Location of nth-minimum : min, sin n n a =

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Lecture 4, p 9 ACT 1 Which of the following would broaden the diffraction peak?

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Lect04 - Le cture4 Diffraction S ctroscopy pe y d L Spectra...

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