Problem 4 | The full width at half maximum (FWHM) of the 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 of the maximum value when sin² a = a²/2. (b) Verify that a = 1.39 radians (about 80°) is a solution to the transcendental equation of part (a). (c) Show that the FWHM is A0 = 2 sin-¹ (0.443x/a). (d) Calculate the FWHM of the central maximum for slits whose widths are 1.0, 5.0, and 10 wavelengths.
Problem 4 | The full width at half maximum (FWHM) of the 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 of the maximum value when sin² a = a²/2. (b) Verify that a = 1.39 radians (about 80°) is a solution to the transcendental equation of part (a). (c) Show that the FWHM is A0 = 2 sin-¹ (0.443x/a). (d) Calculate the FWHM of the central maximum for slits whose widths are 1.0, 5.0, and 10 wavelengths.
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Transcribed Image Text:Problem 4 | The full width at half maximum (FWHM) of the 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 of the maximum value when sin² a = a²/2. (b) Verify that
a = 1.39 radians (about 80°) is a solution to the transcendental equation of part (a). (c) Show that the FWHM
is A0 = 2 sin ¹ (0.443\/a). (d) Calculate the FWHM of the central maximum for slits whose widths are 1.0,
5.0, and 10 wavelengths.
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