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Unformatted text preview: N 2 j =1 e i (2 j 1) α = sin( Nα ) 2 sin α Fig. 8.— The case N = 7 and b λ . N 2 j =1 F j = 2 b sin β β × sin( Nα ) 2 sin α = b sin β β × sin( Nα ) sin α ψ P = E L r e i ( kr ωt ) b sin β β × sin( Nα ) sin α and the intensity is I ∝ ψ * P ψ P = E L r b 2 × sin β β 2 diffraction off wide slit × sin( Nα ) sin α 2 interference of N point sources which is the result we were seeking. Notice that when N = 1 we reduce to the result on a single wide slit, and when N = 2 we reduce to the case of two wide slits. We get maxima from the interference part when α = mπ , or, equivalently, π a λ sin θ = mπ , i.e., a sin θ = mλ . Notice, that as N is integral, when α = mπ so does Nα . Therefore when the denominator vanishes, so does the numerator, and at the limit sin( Nα ) / sin α → N as α → mπ . But for large N , Nα equals mπ more often then α , so that the numerator vanishes also when the denominator does not. This gives us N 1 minima between principal maxima, or N 2 secondary maxima. Consider now the diffraction part, namely the sinc 2 β function. Recall that β := 1 2 kb sin θ = π b λ sin θ . When b λ , the sinc function does not vary by much and effectively is constant 84 Fig. 9.— The case N = 7 and b λ ....
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 Spring '09
 LIORBURKO
 Physics, Wavelength, Sin

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