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**Unformatted text preview: **Lecture 25: Diffraction Gratings and Polarization • Last lecture we started to discuss what happens when light passes through a large number of evenly-spaced small openings – this is called a diffraction grating • Found that the angles at which maxima form from a diffraction grating are the same as those from a double-slit: • Note that there’s always a maximum at θ = 0 – but the positions of all the other maxima depend on the wavelength of the light • This makes the diffraction grating an ideal tool to analyze the spectrum of light – different wavelengths will form maxima at different locations on a screen d sin θ βριγητ = μ λ; μ = 0, ±1, ±2Κ Resolving Power • With an infinite number of grating lines, the maxima would become infinitely narrow – meaning the maxima for different wavelengths would be distinct, no matter how close those wavelengths are • But in real life the light hits a finite number of slits, so the maxima still have non-zero width – so if two wavelengths are close enough their maxima will overlap, making it impossible to tell them apart • The resolving power of a grating is defined as the fractional difference in wavelength that can just barely be detected by the grating: R = λ 1 + λ 2 2 λ 2- λ 1 = λ αωγ ∆λ • To determine the resolving power for a given grating, remember how the intensity pattern from interference changes as more slits are added: Three-slit intensity-0.06-0.04-0.02 0.02 0.04 0.06 y Five-slit intensity-0.06-0.04-0.02 0.02 0.04 0.06 y Double-slit intensity-0.06-0.04-0.02 0.02 0.04 0.06 y • From these examples, we see that • In terms of the phase difference of the light from the slits, we know that maxima happen when and the nearest minimum is when distance from peak to nearest minimum distance between peaks ≈ 1 νυμ βεροφσλιτσ φ = 2 p d sin q l = 2 p m φ = 2 p m + 2 p N • We can define the limit of resolution for the grating as when the maximum from one wavelength is at the same position as the nearest minimum for another wavelength:...

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