= Solution
Let the slit run over $-w/2<x<w/2$ and let its coherent illumination have constant complex <amplitude> $A_0$. In a direction making angle $\theta$ to the normal, the relative propagation <phase> from position $x$ is $-kx\sin\theta$, with $k=2\pi/\lambda$. Summing the secondary <amplitudes> gives
$$
\mathcal A(\theta)\propto A_0\int_{-w/2}^{w/2}e^{-ikx\sin\theta}\,dx
=A_0w\frac{\sin\beta}{\beta},\qquad
\beta=\frac{\pi w\sin\theta}{\lambda}.
$$
The <intensity> is proportional to the squared modulus, hence within the usual scalar <Fraunhofer diffraction> approximation
$$
\boxed{I(\theta)=I(0)\left(\frac{\sin\beta}{\beta}\right)^2.}
$$
The central value is defined by the continuous limit $\sin\beta/\beta\to1$. The nonzero minima satisfy $w\sin\theta=m\lambda$ for integer $m\ne0$. This expression neglects the slowly varying obliquity factor and uses a uniform slit; it describes the narrow-angle <diffraction> pattern for $w\gg\lambda$.
Back to article page
