CALC It is possible to calculate the intensity in the

Chapter 36, Problem 36.73

(choose chapter or problem)

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’ represent the position of a point within the slit of width a in Fig. 36.5a, with y’=0 at the center of the slit so that the slit extends from \(y^{\prime}=-a / 2 \text { to } y^{\prime}=a / 2\) We imagine dividing the slit up into infinitesimal strips of width dy’ 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 E0 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. 36.5a can be expressed as

\(d E=E_{0} \frac{d y^{\prime}}{a} \sin \left[k\left(D-y^{\prime} \sin \theta\right)-\omega t\right]\)

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

\(E=E_{0} \sin (k D-\omega t) \frac{\sin [k a(\sin \theta) / 2]}{k a(\sin \theta) / 2}\)

\(=E_{0} \sin (k D-\omega t) \frac{\sin [\pi a(\sin \theta) / \lambda]}{\pi a(\sin \theta) / \lambda}\)

(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 E0 as stated in part (a). (d) Use the result of part (c) to show that if the intensity at point O is I0 then the intensity at a point P is given by Eq. (36.7).