Detailed Diffraction Theory

Fresnel diffraction (near-field): the observation screen is close enough that the curvature of the wavefronts matters. The Fresnel diffraction integral is:

$E(P) = \frac{i}{\lambda}\iint \frac{E(Q)}{r_{QP}}\, e^{ikr_{QP}}\, dS$

Fraunhofer diffraction (far-field): the observation screen is far enough that the phase variation across the aperture can be approximated as linear. This occurs when:

$R \gg \frac{a^2}{\lambda}$

Where $a$ is the aperture size and $R$ is the distance to the screen.

For a point $P$ at distance $R$ from an aperture, the Fresnel zones are annular regions where the path length from $P$ differs by $\lambda/2$. The $n$-th Fresnel zone has inner radius:

$r_n = \sqrt{n\lambda R + \frac{n^2\lambda^2}{4}} \approx \sqrt{n\lambda R}$

Zone plate. A Fresnel zone plate blocks alternate zones, producing a focused beam. It acts as a lens with focal length $f = r_1^2/\lambda$.

For a semi-infinite plane ($x > 0$), the Fresnel integral gives the intensity at a point $P$:

$I(P) = \frac{I_0}{2}\left[\left(C(u) + \frac{1}{2}\right)^2 + \left(S(u) + \frac{1}{2}\right)^2\right]$

Where $C(u)$ and $S(u)$ are the Fresnel integrals and $u = x\sqrt{2/(\lambda R)}$ is the Fresnel number. At the geometric shadow edge ($u = 0$): $I/I_0 = 1/4$ (not zero!), demonstrating the failure of geometric optics.

The Fresnel integrals are defined as:

$C(u) = \int_0^u \cos\left(\frac{\pi t^2}{2}\right) dt, \quad S(u) = \int_0^u \sin\left(\frac{\pi t^2}{2}\right) dt$

Properties of Fresnel Integrals:

1. $C(0) = 0$, $S(0) = 0$.
2. $C(\infty) = S(\infty) = 1/2$.
3. $C(-u) = -C(u)$, $S(-u) = -S(u)$ (odd functions).
4. $(C(u) + iS(u))^2 = \int_0^u e^{i\pi t^2/2} dt$ (Cornu spiral representation).

The Cornu spiral is a parametric plot of $(C(u), S(u))$, which is useful for graphical determination of Fresnel diffraction amplitudes.

Babinet’s principle states that the diffraction pattern from an opaque obstacle is complementary to that from an aperture of the same shape: the sum of the two patterns equals the pattern with no obstruction at all.

$E_{\mathrm{obstacle}} + E_{\mathrm{aperture}} = E_{\mathrm{unobstructed}}$

Example 12.1. The diffraction pattern from a small circular disk has a bright spot at the center (Poisson spot), which is the complement of the Airy pattern from a circular aperture. This was originally considered a paradox but confirms the wave theory of light.

For a rectangular aperture of dimensions $a \times b$, the Fraunhofer diffraction pattern is:

$E(\theta_x, \theta_y) = E_0 \frac{\sin(\pi a \sin\theta_x/\lambda)}{\pi a \sin\theta_x/\lambda} \cdot \frac{\sin(\pi b \sin\theta_y/\lambda)}{\pi b \sin\theta_y/\lambda}$

The intensity is:

$I(\theta_x, \theta_y) = I_0 \left[\frac{\sin(\pi a \sin\theta_x/\lambda)}{\pi a \sin\theta_x/\lambda}\right]^2 \left[\frac{\sin(\pi b \sin\theta_y/\lambda)}{\pi b \sin\theta_y/\lambda}\right]^2$

This is a product of two sinc² functions. The first zeros occur at $\sin\theta_x = \pm\lambda/a$ and $\sin\theta_y = \pm\lambda/b$, giving a rectangular pattern of side lobes.

For a circular aperture of diameter $D$, the Fraunhofer diffraction pattern is:

$E(\theta) = E_0 \frac{2J_1(\pi D \sin\theta/\lambda)}{\pi D \sin\theta/\lambda}$

Where $J_1$ is the Bessel function of the first kind of order 1. The intensity (Airy pattern) is:

$I(\theta) = I_0 \left[\frac{2J_1(\pi D \sin\theta/\lambda)}{\pi D \sin\theta/\lambda}\right]^2$

The first dark ring occurs at $\sin\theta \approx 1.22\lambda/D$. The Rayleigh criterion for resolution states that two point sources are resolvable when the central maximum of one coincides with the first minimum of the other: $\theta_{\mathrm{min}} \approx 1.22\lambda/D$.

Example 12.2. For a telescope with $D = 10$ cm observing at $\lambda = 500$ nm: $\theta_{\mathrm{min}} \approx 1.22 \times 500 \times 10^{-9} / 0.1 = 6.1 \times 10^{-6}$ rad.

The Huygens-Fresnel principle states that every point on a wavefront acts as a source of spherical secondary wavelets, and the amplitude at any point beyond is the superposition of all these wavelets. Mathematically:

$U(P) = \frac{i}{\lambda} \iint_{\Sigma} U(Q) \frac{e^{ikr}}{r} \cos\theta\, dS$

Where $\Sigma$ is the aperture surface, $r = |P - Q|$, and $\cos\theta$ is the obliquity factor.

Kirchhoff’s integral theorem provides a rigorous mathematical formulation of the Huygens-Fresnel principle:

$U(P) = \frac{1}{4\pi} \iint_{\Sigma} \left[U \frac{\partial}{\partial n}\left(\frac{e^{ikr}}{r}\right) - \frac{e^{ikr}}{r} \frac{\partial U}{\partial n}\right] dS$

The Kirchhoff boundary conditions assume $U = 0$ and $\partial U/\partial n = 0$ on the opaque portion of the screen, though these assumptions are not strictly consistent — a limitation addressed by the Sommerfeld radiation theory.

A thin lens with focal length $f$ adds a quadratic phase factor to the incident field:

$t_{\mathrm{lens}}(x, y) = \exp\left(-i\frac{k}{2f}(x^2 + y^2)\right)$

In the Fresnel approximation, the field at the back focal plane of a lens is the Fourier transform of the field at the front focal plane:

$U_f(u, v) = \frac{i}{\lambda f} \iint U_{\mathrm{in}}(x, y) e^{-i\frac{2\pi}{\lambda f}(xu + yv)}\, dx\, dy$

This is the fundamental principle behind optical Fourier processing and 4f imaging systems.

Problem 1. Calculate the Fraunhofer diffraction pattern of a single slit of width $a$.

Solution. For a slit along the $y$-axis of width $a$, the field at angle $\theta$ is:

$E(\theta) = E_0 \int_{-a/2}^{a/2} e^{-ikx\sin\theta}\, dx = E_0 a \frac{\sin(\pi a \sin\theta/\lambda)}{\pi a \sin\theta/\lambda}$

The intensity is $I(\theta) = I_0 \mathrm{sinc}^2(\pi a \sin\theta/\lambda)$. The first minimum is at $\sin\theta = \lambda/a$, giving an angular width $\Delta\theta \approx 2\lambda/a$ for small angles. $\blacksquare$

Problem 2. Find the fringe spacing in a double-slit Fraunhofer pattern with slit separation $d$.

Solution. For two slits of width $a$ separated by $d$, the pattern is:

$I(\theta) = 4I_0 \mathrm{sinc}^2\left(\frac{\pi a\sin\theta}{\lambda}\right) \cos^2\left(\frac{\pi d\sin\theta}{\lambda}\right)$

The $\cos^2$ term gives interference fringes at $\sin\theta = m\lambda/d$ for integer $m$. The fringe spacing in the small-angle limit is $\Delta y = \lambda R / d$ at distance $R$. $\blacksquare$