일반물리학 2 — Diffraction: Problem 49

49 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 \(\sin^2 \alpha = \alpha^2/2\). (b) Verify that \(\alpha = 1.39 \text{ rad}\) (about \(80^\circ\)) is a solution to the transcendental equation of (a). (c) Show that the FWHM is \(\Delta\theta = 2 \sin^{-1}(0.443\lambda/a)\), where \(a\) is the slit width. Calculate the FWHM of the central maximum for slit width (d) \(2.00\lambda\), (e) \(7.00\lambda\), and (f) \(12.0\lambda\).