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It is possible to calculate the intensity in the single-
Chapter 36, Problem 73CP(choose chapter or problem)
CALC It is possible to calculate the intensity in the single-slit Fraunhofer diffraction pattern without using the phasor method of Section 36.3. Let \(y^{\prime}\) represent the position of a point within the slit of width a in Fig. 36.5a, with \(y^{\prime}=0\) at the center of the slit so that the slit extends from \(y^{\prime}=-a / 2\) to \(y^{\prime}=a / 2\). We imagine dividing the slit up into infinitesimal strips of width \(d y^{\prime}\), each of which acts as a source of secondary wavelets. (a) The amplitude of the total wave at the point O on the distant screen in Fig. 36.5a is \(E_{0}\). Explain why the amplitude of the wavelet from each infinitesimal strip within the slit is \(E_{0}\left(d y^{\prime} / a\right)\), so that the electric field of the wavelet a distance x from the infinitesimal strip is \(d E=E_{0}\left(d y^{\prime} / a\right) \sin (k x-\omega t)\). (b) Explain why the wavelet from each strip as detected at point P in Fig. 36.5 a can be expressed as
\(d E=E_{0} \frac{d v^{\prime}}{a} \sin \left[k\left(D-y^{\prime} \sin \theta\right)-\omega t\right]\)
where D is the distance from the center of the slit to point P and \(k=2 \pi / \lambda\). (c) By integrating the contributions dE from all parts of the slit, show that the total wave detected at point P is
\(\begin{aligned} E &=E_{0} \sin (k D-\omega t) \frac{\sin [k a(\sin \theta) / 2]}{k a(\sin \theta) / 2} \\ &=E_{0} \sin (k D-\omega t) \frac{\sin [\pi a(\sin \theta) / \lambda]}{\pi a(\sin \theta) / \lambda} \end{aligned}\)
(The trigonometric identities in Appendix B will be useful.) Show that at \(\theta=0\), corresponding to point O in Fig. 36.5a, the wave is \(E=E_{0} \sin (k D-\omega t)\) and has amplitude \(E_{0}\), as stated in part (a). (d) Use the result of part (c) to show that if the intensity at point O is \(I_{0}\), then the intensity at a point P is given by Eq. (36.7).
Questions & Answers
QUESTION:
CALC It is possible to calculate the intensity in the single-slit Fraunhofer diffraction pattern without using the phasor method of Section 36.3. Let \(y^{\prime}\) represent the position of a point within the slit of width a in Fig. 36.5a, with \(y^{\prime}=0\) at the center of the slit so that the slit extends from \(y^{\prime}=-a / 2\) to \(y^{\prime}=a / 2\). We imagine dividing the slit up into infinitesimal strips of width \(d y^{\prime}\), each of which acts as a source of secondary wavelets. (a) The amplitude of the total wave at the point O on the distant screen in Fig. 36.5a is \(E_{0}\). Explain why the amplitude of the wavelet from each infinitesimal strip within the slit is \(E_{0}\left(d y^{\prime} / a\right)\), so that the electric field of the wavelet a distance x from the infinitesimal strip is \(d E=E_{0}\left(d y^{\prime} / a\right) \sin (k x-\omega t)\). (b) Explain why the wavelet from each strip as detected at point P in Fig. 36.5 a can be expressed as
\(d E=E_{0} \frac{d v^{\prime}}{a} \sin \left[k\left(D-y^{\prime} \sin \theta\right)-\omega t\right]\)
where D is the distance from the center of the slit to point P and \(k=2 \pi / \lambda\). (c) By integrating the contributions dE from all parts of the slit, show that the total wave detected at point P is
\(\begin{aligned} E &=E_{0} \sin (k D-\omega t) \frac{\sin [k a(\sin \theta) / 2]}{k a(\sin \theta) / 2} \\ &=E_{0} \sin (k D-\omega t) \frac{\sin [\pi a(\sin \theta) / \lambda]}{\pi a(\sin \theta) / \lambda} \end{aligned}\)
(The trigonometric identities in Appendix B will be useful.) Show that at \(\theta=0\), corresponding to point O in Fig. 36.5a, the wave is \(E=E_{0} \sin (k D-\omega t)\) and has amplitude \(E_{0}\), as stated in part (a). (d) Use the result of part (c) to show that if the intensity at point O is \(I_{0}\), then the intensity at a point P is given by Eq. (36.7).
ANSWER:Solution 73CP