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In single slit diffraction, why (according to the equation for $\theta$) do successive dark spots exist at whole number intervals ($n=1,2,3, \ldots$) and not half? The correct path difference for destructive interference is $1/2$ a wavelength, so why in $\theta = n(\lambda)/(\mbox{slit width})$ do we use $n = 1,2,3,\ldots$?

Vishnu
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tmanning
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1 Answers1

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You need to split the slit by small zones and sum up field from the zones. If the first and the last zones has destructive interference with each other than we have big total intensity. If the phase make a full turn we have a dark line. Therefore, angles of the dark lines are $\theta_n = n\lambda/b,\ \,n = 1,2,3,\dots$

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ybeltukov
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  • I understand splitting the slit into separate zones - usually these are referred to as rays - are we talking about the same thing? What do you mean by "if the phase makes a full turn"? I don't understand the circle...I can draw diagrams with different numbers of rays, I fully understand the derivation, but in drawing the diagrams, I can prove myself wrong as easily as I can prove myself right. Why does destructive occur at whole numbers of lambda divided be slit width? – tmanning Jan 21 '14 at 13:35
  • @tmanning It is a so-called phasor diagram. It is a simple method to obtain results without integration. Does links 1, 2, 3 are helpful? – ybeltukov Jan 21 '14 at 17:02