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UVM PHYS 012 - Single Slit Diffraction and Resolving Power
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Physics 012 1st Edition Lecture 29 Outline of Last Lecture I. Single Thin Film Interferencea. Path length difference + phase difference = condition for constructive or destructive interferenceb. Problem: Light of wavelength 469 nm enters a layer of gasoline (ng = 1.4) on top of a layer of water (nw = 1.33) at an angle. What is the minimum non-zero value of t so the light coming from gasoline is bright?i. 2t + λg/2 = λg/2, 3λg/2, 5λg/2, etc.ii. t = m λg/2 = [(1)λair]/[2ng] = 168 nmII. Double Thin Film Interferencea. Ray 2 travels extra distance of ~2t and undergoes 2 phase changes.b. For constructive interference: 2t + λw/2 + = λw, 2 λw, 3 λw, etc.III. Diffraction – Single SlitThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.a. λ >> wb. Central maximum is twice as wide as secondary maxima.c. For dark fringes: wsinθ = mλOutline of Current Lecture IV. Single Slit Diffractiona. Wsinθ = mλb. λ/W: ratio determining the amount of diffraction coming from a slit; larger number means more diffraction (bending of light)V. Resolving Power (Resolution)a. sinθmin = 1.22


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UVM PHYS 012 - Single Slit Diffraction and Resolving Power

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