Diffraction (Fresnel Class)

Diffraction is the non-rectilinear deviation of optical waves into the geometrical shadow when encountering obstacles or apertures. In Fresnel diffraction, the source, the diffracting obstacle, or the observation screen (or all three) are located at finite distances, resulting in non-planar (spherical) wavefronts.

1. Construction of Fresnel Half-Period Zones

Consider a spherical monochromatic wavefront $W$ of wavelength $\lambda$ propagating from a point source $S$. Let $P$ be an observation point at distance $b$ from the wavefront pole $O$. Fresnel divided the spherical wavefront into annular concentric zones such that the optical distance from the boundaries of successive zones to the observation point $P$ increases sequentially by half a wavelength ($\lambda/2$): $$d_1 = b + \frac{\lambda}{2}, \quad d_2 = b + \frac{2\lambda}{2}, \quad \dots, \quad d_n = b + \frac{n\lambda}{2}$$

2. Radius and Area of the $n$-th Half-Period Zone

Let $r_n$ be the radius of the circle bounding the $n$-th zone. By Pythagoras' theorem in the triangle formed by pole $O$, zone boundary, and observation point $P$: $$(b + \frac{n\lambda}{2})^2 = b^2 + r_n^2 \implies b^2 + n b \lambda + \frac{n^2 \lambda^2}{4} = b^2 + r_n^2$$ Neglecting the second-order term $\frac{n^2 \lambda^2}{4} \ll n b \lambda$: $$r_n = \sqrt{n b \lambda}$$ The area of the $n$-th zone is: $$A_n = \pi (r_n^2 - r_{n-1}^2) = \pi [n b \lambda - (n-1) b \lambda] = \pi b \lambda$$ Remarkably, to first-order approximation, **every Fresnel half-period zone possesses the identical surface area** $A_n \approx \pi b \lambda$.

3. Resultant Optical Amplitude at Observation Point $P$

The amplitude $m_n$ contributed by the $n$-th zone depends on:

1. Directly on zone area $A_n$ (constant).
2. Inversely on average distance $b + \frac{n\lambda}{2}$ (slowly decreasing).
3. The inclination factor (obliquity factor) $K(\theta_n) = \frac{1}{2}(1 + \cos \theta_n)$, which decreases monotonically from $1$ at $\theta = 0$ to $0$ at $\theta = \pi$.

Therefore, the amplitudes form a monotonically decreasing sequence: $$m_1 > m_2 > m_3 > m_4 > \dots > m_n$$ Because consecutive zones differ in path length by $\lambda/2$, their wavelets arrive with a phase difference of $\pi$ radians ($e^{i\pi} = -1$). The resultant amplitude $R$ is an alternating sum: $$R = m_1 - m_2 + m_3 - m_4 + m_5 - \dots \pm m_n$$ Regrouping: $$R = \frac{m_1}{2} + \left(\frac{m_1}{2} - m_2 + \frac{m_3}{2}\right) + \left(\frac{m_3}{2} - m_4 + \frac{m_5}{2}\right) + \dots$$ Since the amplitude decreases smoothly, $m_2 \approx \frac{m_1 + m_3}{2}$. Every bracketed term vanishes: $$R = \frac{m_1}{2}$$ This proves Fresnel's profound theorem: **the total optical amplitude produced by an entire unobstructed wavefront is equal to exactly half the amplitude contributed by the first half-period zone alone**! The intensity is $I = R^2 = \frac{m_1^2}{4}$.

A Fresnel zone plate is a diffractive optical element that demonstrates the reality of half-period zones by selectively blocking or phase-shifting alternating zones, functioning as a powerful focusing lens.

1. Physical Principle and Amplitude Amplification

In an unobstructed wavefront, alternating zones cancel each other out ($R = m_1 - m_2 + m_3 - m_4 \dots = \frac{m_1}{2}$). If an opaque mask is constructed to obstruct all even-numbered zones ($2, 4, 6, \dots$), leaving only odd zones ($1, 3, 5, \dots$) transparent, the transmitted wavelets arrive at $P$ in phase: $$R_{\text{zp}} = m_1 + m_3 + m_5 + \dots + m_{2N-1} \approx N m_1$$ The intensity at the focal point becomes: $$I_{\text{zp}} = (N m_1)^2 = N^2 m_1^2 = 4 N^2 \left(\frac{m_1^2}{4}\right) = 4 N^2 I_{\text{unobstructed}}$$ For a zone plate with $N = 50$ open zones, the intensity at the focal point is amplified by a factor of $4(50)^2 = 10,000$ relative to the unobstructed beam!

2. Focal Length Formula

Recall that the outer radius of the $n$-th zone is given by $r_n^2 = n b \lambda$. Rearranging: $$f_1 = b = \frac{r_n^2}{n \lambda} = \frac{r_1^2}{\lambda}$$ This defines the primary focal length $f_1$.

3. Multiple Focal Lengths

Unlike a classical refractive lens which has a single focal length, a zone plate behaves as a diffractive optic with multiple foci: $$f_m = \frac{r_1^2}{m \lambda}, \quad m \in \{\pm 1, \pm 3, \pm 5, \dots\}$$ Even orders ($m = 2, 4, \dots$) vanish due to internal destructive interference within each newly subdivided zone. The negative focal lengths correspond to virtual diverging foci.

The calculation of Fresnel diffraction for non-circular apertures requires the evaluation of Fresnel integrals, which can be visualized geometrically using the Cornu spiral (clothoid).

1. Fresnel Integrals Definition

The Cartesian coordinates of the Cornu spiral are given by the normalized Fresnel integrals: $$C(v) = \int_0^v \cos\left(\frac{\pi t^2}{2}\right) dt, \quad S(v) = \int_0^v \sin\left(\frac{\pi t^2}{2}\right) dt$$ where $v$ is a dimensionless arc-length parameter defined along the diffracting wavefront. As $v \to \infty$, the integrals converge to asymptotic limits: $$C(\infty) = S(\infty) = \frac{1}{2}, \quad C(-\infty) = S(-\infty) = -\frac{1}{2}$$ The asymptotic focal points of the spiral are $Z_+ = (0.5, 0.5)$ and $Z_- = (-0.5, -0.5)$.

2. Geometric Determination of Resultant Amplitude

The resultant complex optical amplitude between wavefront limits $v_1$ and $v_2$ is represented directly by the chord vector connecting points $v_1$ and $v_2$ on the Cornu spiral: $$\mathbf{A}(v_1, v_2) = [C(v_2) - C(v_1)] + i [S(v_2) - S(v_1)]$$ The optical intensity is proportional to the square of the chord length: $$I \propto |\mathbf{A}|^2 = [C(v_2) - C(v_1)]^2 + [S(v_2) - S(v_1)]^2$$

3. Diffraction at a Semi-Infinite Opaque Straight Edge

Consider a semi-infinite opaque screen covering the region $y < 0$. The upper edge is at $y = 0$.

• Inside the Geometrical Shadow ($v < 0$): One limit is the edge $v$, the other is $-\infty$. As the observer moves deeper into shadow ($v \to -\infty$), the chord length shrinks monotonically toward zero without oscillating: $$I(v) \propto [C(v) - (-0.5)]^2 + [S(v) - (-0.5)]^2 \to 0$$
• At the Geometrical Edge ($v = 0$): The vector connects $(0, 0)$ to $(0.5, 0.5)$: $$I(0) \propto (0.5)^2 + (0.5)^2 = 0.25 + 0.25 = 0.5 = \frac{1}{4} I_0$$ The intensity at the geometric boundary edge is exactly **$25\%$ of the unobstructed incident intensity**.
• In the Illuminated Region ($v > 0$): The vector connects $Z_-$ to points along the spiral that wind around $Z_+$, producing a sequence of alternating diffraction fringes with decaying amplitude before settling to $I_0$. The first bright fringe exceeds the incident intensity by $1.37 I_0$.

The diffraction of light by a circular aperture and an opaque circular disc represents one of the most celebrated triumphs of wave optics over corpuscular theory.

1. Circular Aperture Axial Intensity

Consider a circular aperture of radius $a$ illuminated at normal incidence by a plane wave. At an on-axis observation point $P$ at distance $b$: The number of exposed Fresnel zones is: $$n = \frac{a^2}{b \lambda}$$

• When $n$ is an odd integer ($n = 1, 3, 5, \dots$): The exposed zones interfere constructively to yield an intensity maximum: $$I = (m_1 - m_2 + \dots + m_n)^2 \approx m_1^2 = 4 I_0$$
• When $n$ is an even integer ($n = 2, 4, 6, \dots$): The exposed zones cancel in pairs, producing an on-axis intensity minimum: $$I \approx 0$$

As the observation screen is moved along the optical axis (varying $b$), the center of the pattern alternates between bright and dark spots!

2. The Celebrated Poisson-Arago Bright Spot (Circular Obstacle)

In 1818, Siméon Denis Poisson used Fresnel's wave theory to derive a seemingly absurd prediction intended to disprove the wave hypothesis: that the exact center of the geometrical shadow of an opaque circular disc illuminated by a point source must be bright. François Arago performed the experiment immediately and verified the phenomenon: **a bright central spot is always observed in the center of the shadow**. Mathematical Proof: Let an opaque disc obstruct the first $k$ half-period zones. The remaining unobstructed wavefront consists of zones from $k+1$ to $\infty$. The resultant amplitude at the central point is: $$R = m_{k+1} - m_{k+2} + m_{k+3} - \dots = \frac{m_{k+1}}{2}$$ The intensity at the center is: $$I_{\text{center}} = \frac{m_{k+1}^2}{4} \approx I_0$$ Regardless of the radius of the disc (provided the disc edge is smooth on the scale of $\lambda$), the wavelets diffracting over the circular perimeter travel identical optical distances to the on-axis point, arriving strictly in phase and forming a bright focal spot.