Diffraction from Rectangular Aperture

Suppose that the aperture is rectangular. In other words,

$$f(u,v) = \left\{ \begin{array}{lll} 1&\mbox{\hspace{0.5cm}}& ... ... ~$v_1<v<v_2$}\\ [0.5ex] 0&&\mbox{otherwise}\end{array}\right.,$$ (10.118)

where $u_1<u_2$ and $v_1< v_2$. In this case, the two-dimensional integrals in Equations (10.103) and (10.104) are separable. Using some standard trigonometric identities (see Section B.3), we obtain

$$f_c(u',v') = \frac{1}{2}\,C(u',u_1,u_2)\,C(v',v_1,v_2) -\frac{1}{2}\,S(u',u_1,u_2)\,S(v',v_1,v_2),$$ (10.119)

$$f_s(u',v') = \frac{1}{2}\,S(u',u_1,u_2)\,C(v',v_1,v_2) + \frac{1}{2}\,C(u',u_1,u_2)\,S(v',v_1,v_2),$$ (10.120)

where

$$C(z',z_1,z_2) = C(z_2-z')+C(z'-z_1),$$ (10.121)

$$S(z',z_1,z_2) = S(z_2-z')+S(z'-z_1).$$ (10.122)

Here, the functions

$$C(z) =\int_0^z \cos\left(\frac{\pi}{2}\,t^{\,2}\right)dt,$$ (10.123)

$$S(z) = \int_0^z \sin\left(\frac{\pi}{2}\,t^{\,2}\right)dt$$ (10.124)

are known as Fresnel integrals (Abramowitz and Stegun 1965). Note that $C(-z)=-C(z)$, $S(-z)=-S(z)$, and $C(\infty)=S(\infty) = 1/2$. The Frensel integrals are plotted in Figure 10.16.

Figure 10.16: The Fresnel integrals. The solid curve shows $C(z)$, whereas the dashed curve shows $S(z)$.

Suppose that the aperture is of infinite extent in the $v$-direction, which implies that $v_1=-\infty$ and $v_2=\infty$. In this case, the aperture becomes a rectangular slit running parallel to the $v$-axis, and extending from $u=u_1$ to $u=u_2$. It is easily seen that $C(v',-\infty,\infty)=S(v',-\infty,\infty)=1$. Hence, Equations (10.119) and (10.120) yield

$$f_c(u') = \frac{1}{2}\,C(u',u_1,u_2)-\frac{1}{2}\,S(u',u_1,u_2)$$ (10.125)

$$f_s(u') = \frac{1}{2}\,S(u',u_1,u_2) + \frac{1}{2}\,C(u',u_1,u_2).$$ (10.126)

Furthermore, Equation (10.110) gives

$$\frac{{\cal I}(u')}{{\cal I}_0} = f_c^{\,2}(u')+f_s^{\,2}(u').$$ (10.127)