In discussing how a planar wave propagates, we could turn our earlier argument around and say that for each point \(\mathrm{Q}\) in Figure 34.2 there is a point \(\mathrm{S}\) somewhere on the wavefront that radiates toward \(\mathrm{P}\) along a path exactly one wavelength longer than that from \(Q\), and therefore there should be a nonzero intensity at \(\mathrm{P}\). What is wrong with this argument?

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Figure 34.2 The reason we don't usually see diffraction for light beams, provided the beam is very much wider than the wavelength of the light waves. Q and R are on same wavefront, so wavelets are in phase. P Distance to P differs by half a wavelength, so... 1/2 -wavelet R Q' R' ...at P waves are 180 out of phase and interfere destructively. wavefront Similarly, for Q', there is a point R' that causes cancellation at P.