Chapter 36, Problem 36.68CP

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′ 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′ = −a/2 to y′ − a/ 2. We imagine dividing the slit up into infinitesimal strips of width d y′ 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 E₀(dy′/a)_, so that the electric field of the wavelet a distance x from the infinitesimal strip is dE = E₀(dy′/a) sin (kx – ωt). (b) Explain why the wavelet from each strip as detected at point P in Fig. 36.5a can be expressed as

$dE=E_0\frac{d{y}^{\prime }}{a}\sin [k(D-y^{\prime }\sin \theta )-\omega t]$

Where D is the distance from the center of the slit to point P and k =2πλ.. (c) By integrating the contributions dE from all parts of the slit, show that the total wave detected at point P is

$\begin{array}{l}E=E_0\sin (kD-\omega t)\frac{\sin [ka(\sin \theta )/2]}{ka(\sin \theta )/2}\\ E_{\text{0}}\sin (kD-\omega t)\frac{\sin [\pi a(\sin \theta )/\lambda ]}{\pi a(\sin \theta )/\lambda }\end{array}$

(The trigonometric identities in Appendix B will be useful.) Show that at θ = 0°, corresponding to point O in Fig. 36.5a, the wave is E = E₀ sin(kD − ωt) 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).