It is possible to calculate the intensity in the singleslit 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. $36.5 \mathrm{a}$, 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. $36.5 \mathrm{a}$ 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
$$
d E=E_{0} \frac{d y^{\prime}}{a} \sin \left[k\left(D-y^{\prime} \sin \theta\right)-\omega t\right]
$$
where $D$ is the distance from the center of the slit to point $P$ and $k=2 \pi / \lambda .$ (c) By integrating the contributions $d E$ 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 $\mathrm{D}$ will be useful.) Show that at $\theta=0^{\circ}$, corresponding to point $O$ in Fig. $36.5 \mathrm{a}$, 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).