Question: Consider a single-slit diffraction pattern. The center of the central maximum, where the intensity is I0, is located at = 0. (a) Let + and
(a) Let θ+ and θ_ be the two angles on either side of θ = 0 for which I = ½ I0. ∆θ = | θ+ – θ_| is called the full width at half maximum, or FWHM, of the central diffraction maximum. Solve for ∆θ when the ratio between alit width a and wavelength λ is (i) a/λ. = 2; (ii) a/λ = 5; (iii) a/ λ = 10.
(b) The width of the central maximum can alternatively be defined as 2θ0, where θ0 is the angle that locates the minimum on one side of the central maximum. Calculate 2θ0 for each case considered in part (a) and compare to λ.
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a IDENTIFY and SET UP The intensity in the diffraction patter is given by Eq 365 I 1 sin 32 ... View full answer
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