Electromagnetic waves > Propagation of electromagnetic waves > Problem 12.2.4

Statement

$12.2.4$. Find the angles that determine the directions of radiation minima when a plane wave is incident perpendicularly on a slit of width $b$. The wavelength $\lambda < b$.

Solution

According to the Huygens-Fresnel principle, each element of the slit area becomes a source of a spherical wave. The amplitude of oscillations at a certain point at a distance $r \gg b$ is the sum of the amplitudes of these waves.

Let $\Delta(x)$ be the path difference at point $A$ between the wave emitted from the center of the slit and the wave emitted from a point with coordinate $x$:
$$
\Delta(x)=r-r(x)=r-(r^2+x^2-2xr\cos\theta)^{\frac{1}{2}}
$$
Using standard approximations,
$$
\Delta(x)\approx r-r\left(1-2x\cos\theta\right)^{\frac{1}{2}}\approx x\cos\theta
$$
Let the phase of the wave emitted from the middle of the slit (by a segment of width $dx$) be 0; its equation is:
$$
\psi (0, t)=\psi_0dx=A_0e^{i(\omega t-kr)}
$$
Then
$$
\psi (x, t)=A_0e^{i(\omega t-k(r+\Delta))}=\psi_0e^{-ikx\cos\theta}dx
$$
The amplitude at point $A$ is:
$$
A\propto\int_{-b/2}^{b/2}e^{-ixk\cos\theta}dx\propto\frac{e^{ik\frac{b}{2}\cos\theta}-e^{-ik\frac{b}{2}\cos\theta}}{ik\cos\theta}
$$
Recall that $\sin u = \frac{e^{i u} - e^{-i u}}{2i}$:
$$
A\propto 2\frac{\sin\left(k\frac{b}{2}\cos\theta\right)}{k\cos\theta}
$$
Condition for minima:
$$
A=0 \ \to k\frac{b}{2}\cos\theta=\pi m, \quad m\in Z
$$
The wavenumber is $k=\frac{2\pi}{\lambda}$, so the minima correspond to angles:
$$
b\cos\theta=\lambda m,\quad m\in Z
$$
Savchenko uses the angle to the normal, so in his notation the same is:
$$
\sin \alpha_k = \frac{k\lambda}{b},\quad k\in Z
$$

Answer

$$
\boxed{\sin \alpha_k = \frac{k\lambda}{b},\quad k\in Z}
$$