Lecture24 - Lecture 24: More on Diffraction Intensity of...

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Lecture 24: More on Diffraction
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Intensity of the Diffraction Pattern Last lecture we started to derive the intensity at any angle due to diffraction through a single slit In this case, there are an infinite number of waves adding at each point on the screen (from Huygen’s Principle) if we consider every small portion of this slit as a source of waves, the phasor diagram for a given point on the screen would look like: The total phase shift β is related to the path length difference from the top and bottom of the slit: β= 2 p l a sin q E R Phasors lie on arc of a circle
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Define the total length of the phasor chain as E o We can use geometry to find E R : The arc length is given by E o = β R From the triangle, we see that: R β sin 2 = Ε Ρ / 2 = 2 ο
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Which means the total electric field is: so the intensity varies as: E R = ο β σιν 2 I = ΧΕ Ρ 2 = Χ σιν 2 2 Ι μ αξ = Χ Ε [ ] 2 Ι=Ι μ αξ 2 σιν 2 2 = Ι μ αξ λ πασινθ σιν 2
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Note what happens when the slit becomes much bigger than the wavelength: The angular spread due to diffraction becomes smaller that’s why the ray approximation works well in many cases a = 25 λ a = 100
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Back to the Double Slit When we studied double-slit interference earlier, the equations said we should see alternating dark and bright regions, with all the bright regions having the same intensity Experimentally, we did see the bright and dark regions, but
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This note was uploaded on 08/24/2010 for the course PHYS 142 taught by Professor Staff during the Fall '08 term at University of Arizona- Tucson.

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Lecture24 - Lecture 24: More on Diffraction Intensity of...

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