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. (a) Show that the intensity drops to one-half the maximum value when sin2 a = a2/2. (b) Verify that a =1.39 rad (about 80°) is a solution to the transcendental equation of (a). (c) Show that the FWHM is u= 2 sin-1(0.443l/a), where a is the slit width. Calculate the FWHM of the central maximum for slit width (d) 1.00l, (e) 5.00l, and (f) 10.0l.
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. (a) Show that the intensity drops to one-half the maximum value when sin2 a = a2/2. (b) Verify that a =1.39 rad (about 80°) is a solution to the transcendental equation of (a). (c) Show that the FWHM is u= 2 sin-1(0.443l/a), where a is the slit width. Calculate the FWHM of the central maximum for slit width (d) 1.00l, (e) 5.00l, and (f) 10.0l.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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. (a) Show that the intensity drops to
one-half the maximum value when sin2 a = a2/2. (b) Verify that a =1.39 rad (about 80°) is a solution to the transcendental equation of
(a). (c) Show that the FWHM is u= 2 sin-1(0.443l/a), where a is the
slit width. Calculate the FWHM of the central maximum for slit width
(d) 1.00l, (e) 5.00l, and (f) 10.0l.
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