College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Question

Please answer 'check your understanding 4.2'.  I submitted previously and it was rejected b/c they said it was a 'writing assignment', but it's not a writing assignment, it's a numerical problem.

**4.2 Check Your Understanding** For the experiment in Example 4.2, at what angle from the center is the third maximum and what is its intensity relative to the central maximum? --- **Example 4.2** **Intensity in Single-Slit Diffraction** Light of wavelength 550 nm passes through a slit of width 2.00 µm and produces a diffraction pattern similar to that shown in Figure 4.9. (a) Find the locations of the first two minima in terms of the angle from the central maximum and (b) determine the intensity relative to the central maximum at a point halfway between these two minima. **Strategy** The minima are given by Equation 4.1, \(D \sin \theta = m\lambda\). The first two minima are for \(m = 1\) and \(m = 2\). Equation 4.4 and Equation 4.2 can be used to determine the intensity once the angle has been worked out. **Solution** **a.** Solving Equation 4.1 for \(\theta\) gives us \(\theta_m = \sin^{-1}(m\lambda/D)\), so that \[ \theta_1 = \sin^{-1}\left(\frac{(1)(550 \times 10^{-9} \, \text{m})}{2.00 \times 10^{-6} \, \text{m}}\right) = + 16.0^\circ \] and \[ \theta_2 = \sin^{-1}\left(\frac{(2)(550 \times 10^{-9} \, \text{m})}{2.00 \times 10^{-6} \, \text{m}}\right) = + 33.4^\circ. \] **b.** The halfway point between \(\theta_1\) and \(\theta_2\) is \[ \theta = (\theta_1 + \theta_2)/2 = (16.0^\circ + 33.4^\circ)/2 = 24.7^\circ. \] **Equation 4.2** gives \[ \beta = \frac{\pi D}{\lambda} \sin \theta = \frac{\pi (2.00 \times 10^{-6} \, \text{m})\sin(24.
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Transcribed Image Text:**4.2 Check Your Understanding** For the experiment in Example 4.2, at what angle from the center is the third maximum and what is its intensity relative to the central maximum? --- **Example 4.2** **Intensity in Single-Slit Diffraction** Light of wavelength 550 nm passes through a slit of width 2.00 µm and produces a diffraction pattern similar to that shown in Figure 4.9. (a) Find the locations of the first two minima in terms of the angle from the central maximum and (b) determine the intensity relative to the central maximum at a point halfway between these two minima. **Strategy** The minima are given by Equation 4.1, \(D \sin \theta = m\lambda\). The first two minima are for \(m = 1\) and \(m = 2\). Equation 4.4 and Equation 4.2 can be used to determine the intensity once the angle has been worked out. **Solution** **a.** Solving Equation 4.1 for \(\theta\) gives us \(\theta_m = \sin^{-1}(m\lambda/D)\), so that \[ \theta_1 = \sin^{-1}\left(\frac{(1)(550 \times 10^{-9} \, \text{m})}{2.00 \times 10^{-6} \, \text{m}}\right) = + 16.0^\circ \] and \[ \theta_2 = \sin^{-1}\left(\frac{(2)(550 \times 10^{-9} \, \text{m})}{2.00 \times 10^{-6} \, \text{m}}\right) = + 33.4^\circ. \] **b.** The halfway point between \(\theta_1\) and \(\theta_2\) is \[ \theta = (\theta_1 + \theta_2)/2 = (16.0^\circ + 33.4^\circ)/2 = 24.7^\circ. \] **Equation 4.2** gives \[ \beta = \frac{\pi D}{\lambda} \sin \theta = \frac{\pi (2.00 \times 10^{-6} \, \text{m})\sin(24.
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