Experiment 5 Overview

Experiment 5 Overview - Single Slit Diffraction Laser beam...

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Unformatted text preview: Single Slit Diffraction Laser beam λ screen I(x) slit θ x I(0) D a Intensity distribution on the screen: ⎛ sin α ⎞ I (θ ) = I (0)⎜ ⎟ ⎝α⎠ 2 Eq. (1) ⎛π a ⎞ ⎟ sin θ λ⎠ ⎝ α =⎜ The minima of this diffraction pattern occur at: h α min = ±π , ± 2π , ± 3π , K If D >> a, we can approximate α as: πa x α≅ λD slit geometry Single Slit Diffraction ⎛ sin (πax / λD ) ⎞ I ( x) = I (0)⎜ ⎟ ⎜ (πax / λD ) ⎟ ⎠ ⎝ I(θ) 2 1 0.8 a = 40 μm 0.6 0.4 0.2 -0.03 The diffraction angle θdiff is defined as the half-width of the main diffraction peak: ⎛ πa ⎞ ⎟ sin θ diff ⎝λ⎠ -0.02 -0.01 θ diff ≅ 0.03 θ a = 80 μm 0.6 0.4 0.2 -0.03 -0.02 -0.01 0.01 0.02 0.03 1 0.8 a = 160 μm 0.6 λ a 0.02 0.8 α min = π , so π = ⎜ and 0.01 1 0.4 0.2 -0.03 -0.02 -0.01 0.01 0.02 0.03 λ = 0.6328 μm N Slit Diffraction ⎛ sin (πax / λD ) ⎞ 2 I ( x) = N I (0)⎜ ⎟ ⎜ (πax / λD ) ⎟ ⎠ ⎝ 2 ⎛ sin ( Nπdx / λD ) ⎞ ⎜ ⎟ ⎜ N sin (πdx / λD ) ⎟ ⎠ ⎝ 2 4 I(θ) 3.5 3 2.5 Note that: N=2 2 1.5 sin( Nx ) = ±N x→mπ sin( x ) where m = 0, ± 1, ± 2,K 1 L 0.5 -0.2 -0.1 0.1 0.2 θ 8 Hence I(x=0) ~ N2 6 N=3 4 2 -0.2 -0.1 0.1 0.2 16 14 12 10 N=4 8 6 4 2 -0.2 -0.1 0.1 0.2 λ= 0.6328 μm a = 10 μm d = 30 μm Fourier Transform of Sinc[γx]2 ⎛ Sin[γ x] ⎞ f ( x) = ⎜ ⎜ γx ⎟ ⎟ ⎝ ⎠ g(ω) = FT[f(x)] = 2 with γ= πa , Dλ FourierTransform@Sin@g xD ^2 ê Hγ xL ^2, x, ωD α≅ πa x Dλ π# " ##### HH2 γ − ωL Sign@2 γ − ωD − 2 ω Sign@ωD + H2 γ + ωL Sign@2 γ + ωDL 2 4 γ2 Mathematica g(ω) 1π γ2 0 0 ω0 = 2γ ω = 2πf We can relate ω0 to the diffraction angle because αmin = π there, yielding xmin = λD/a, and finally 1/f0 = xmin, where ω0 = 2πf0. Hence the linear frequency at which the FT goes to zero is exactly equal to the inverse of the distance to the first minimum in the single slit diffraction pattern! This result can be used to estimate the slit width . ...
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