Weekly_003.SOLN

# Weekly_003.SOLN - Phys 274 c Eric B Szarmes Weekly Homework...

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Phys 274 – c Eric B. Szarmes 7 February 2011 Weekly Homework 003 SOLUTIONS Problem 1 – Single-slit diffraction The given integral is straightforward to evaluate. For integration over the slit coordinate z we obtain y s ( x ) = Z a/ 2 - a/ 2 cos( kx - kθz ) dz = - 1 sin( kx - kθz ) a/ 2 - a/ 2 = - 1 h sin ( kx - kθa 2 ) - sin ( kx + kθa 2 ) i = - λ 2 πθ h sin ( kx - πθa λ ) - sin ( kx + πθa λ ) i , (1) where we substituted k = 2 π/λ in a few places. We may now use the trigonometric identity sin α - sin β = 2 sin ( α - β 2 ) cos ( α + β 2 ) , (2) with α = kx - πθa/λ and β = kx + πθa/λ , to rewrite this result as y s ( x ) = - λ 2 πθ · 2 sin ( - πθa λ ) cos ( kx ) = λ πθ sin ( πθa λ ) cos ( kx ) , (3) which is indeed of the form y s ( x ) = A ( θ ) cos( kx ) , (4) with amplitude A ( θ ) given by A ( θ ) = λ πθ sin ( πθa λ ) . (5) For λ = 0 . 6 μ m and a = 100 μ m, the graph of the squared-amplitude A 2 ( θ ) (i.e. the intensity ) is 1

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Problem 2 – Q4S.8 and Q4S.11 a) The deBroglie relation is λ = h p = hc pc . (6) Now, we are given the relativistic relation between momentum p and energy E in eq. Q4.17b, p 2 c 2 = ( E - mc 2 )( E + mc 2 ) ,
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