Chapter 36, Problem 15P

SSM WWW The full width at half-maximum (FWHM) of a central diffraction maximum is defined as the angle between the two points in the pattern where the intensity is one-half that at the center of the pattern. (See Fig. 36-8b.) (a) Show that the intensity drops to one-half the maximum value when sin² α = α²/2. (b) Verify that α = 1.39 rad (about 80°) is a solution to the transcendental equation of (a) (c) Show that the FWHM is ∆θ = 2 sin^–1(0.443λ/a), where a is the slit width. Calculate the FWHM of the central maximum for slit width (d) 1.00λ, (e) 5.00λ, and (f) 10.0λ.