Electromagnetic waves > Propagation of electromagnetic waves > Problem 12.2.12

Statement

$12.2.12.$

Calculate the amplitude a of an elementary secondary Huygens-Fresnel wave.
(The amplitude a is proportional to the amplitude A of the primary wave that came from the element ∆S, the area of this element, and is inversely propor-
tional to r, i.e. $a =cA∆S$
. To determine c, compare the plane wave amplitude
r
at some point and the Fresnel amplitude at the same point, when the plane
wave front is taken as an auxiliary surface.)

Solution

Huygens–Fresnel Principle

Each point of a wavefront becomes an emitter of monochromatic secondary spherical waves. The total disturbance at a point P is obtained by integrating the contributions from all elements dS of the wavefront. For an element dS, the contribution is:

$dU_P = \frac{c A \, e^{ikr}}{r} \, dS$

where A is the amplitude of the primary wave at the wavefront, r is the distance from the element to P,
$k = 2\pi/\lambda$

and c is the constant to be determined.

Consider a plane wave propagating in the z-direction. At t = 0, the wavefront coincides with the plane z = 0. We want to calculate the amplitude at a point P located on the z-axis at a distance R from the plane. The expected amplitude of the plane wave at P is:

$U_{\text{plane}}(P) = A e^{ikR}$.

We use the plane z = 0 as an auxiliary surface. Each element dS in polar coordinates

$(\rho,\phi) has dS = \rho d\rho d\phi$.

The distance to P is
$ r = \sqrt{\rho^2 + R^2}$.

$U(P) = \iint_{\text{plane}} \frac{c A e^{ikr}}{r} dS= cA \int_0^{2\pi} d\phi \int_0^\infty \frac{e^{ik\sqrt{\rho^2+R^2}}}{\sqrt{\rho^2+R^2}} \rho d\rho$.

The angular integral gives $2\pi$.

We make the substitution

$u = \sqrt{\rho^2+R^2} ⇒ \rho d\rho = u du$. The limits: $ \rho = 0 \rightarrow u = R, \rho \to \infty \rightarrow u \to \infty$.

$U(P) = 2\pi c A \int_R^\infty \frac{e^{iku}}{u} u du = 2\pi c A \int_R^\infty e^{iku} du$.

The improper integral is evaluated by introducing a convergence factor

$ e^{-\epsilon u} and taking \epsilon \to 0^+$:

$\int_R^\infty e^{iku} du = \lim_{\epsilon\to 0^+} \int_R^\infty e^{i(k+i\epsilon)u} du
= \lim_{\epsilon\to 0^+} \frac{1}{i(k+i\epsilon)} \left[ e^{i(k+i\epsilon)u} \right]_R^\infty
= \frac{i}{k} e^{ikR}$

Therefore:

$U(P) = 2\pi c A \cdot \frac{i}{k} e^{ikR}$

We equate this amplitude to that of the plane wave

$A e^{ikR}$:

$2\pi c A \frac{i}{k} e^{ikR} = A e^{ikR}$

Canceling A e^{ikR} (assumed non-zero), we obtain:

$2\pi c \frac{i}{k} = 1 \quad\Rightarrow\quad c = \frac{k}{2\pi i} = \frac{1}{i\lambda}$

since $k = 2\pi/\lambda$. It can also be written as:

$c = -\frac{i}{\lambda}$.

Answer

The elementary amplitude of a Huygens–Fresnel secondary wave is:

$\boxed{a = \frac{1}{i\lambda} \frac{A \Delta S}{r} e^{ikr} = -\frac{i}{\lambda} \frac{A \Delta S}{r} e^{ikr}}$

The proportionality constant c is $1/(i\lambda) (or equivalently -i/\lambda)$
The factor $1/i = -i$ implies a phase shift of $\pi/2$
between the secondary wave and the primary one