17-Phys-A7 Optics · December 2018

Question 3 of 10: Fresnel/Fraunhofer diffraction and aperture patterns

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 3: Fresnel/Fraunhofer diffraction and aperture patterns (8 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — diffraction; Fresnel vs. Fraunhofer. Diffraction is the bending/spreading of a wave as it passes an obstacle or aperture whose size is comparable to the wavelength — a direct consequence of Huygens' principle: every point on a wavefront inside the aperture radiates a secondary spherical wavelet, and the field beyond is their superposition. Similarities: both are governed by the same Huygens–Fresnel diffraction integral over the aperture; both predict spreading, fringes/lobes, and a diffraction-limited resolution set by aperture size vs. $\lambda$. Differences: Fresnel (near-field) diffraction applies when the source or screen is at a finite distance comparable to the aperture's Fresnel number $N_F=a^2/(\lambda z)\gtrsim1$; the wavefronts across the aperture are curved (quadratic phase term retained) and the pattern changes shape with distance. Fraunhofer (far-field) diffraction is the $N_F\ll1$ limit (or equivalently, a lens placed right after the aperture with the screen at its focal plane, as in parts (b)/(c) here): the aperture is illuminated by effectively plane wavefronts, the diffraction integral reduces to a Fourier transform of the aperture function, and the pattern's shape is fixed, only its overall scale changing with distance/focal length.

Part (b) — circular aperture: Airy pattern.

Given. Circular aperture radius $a=2$ mm (diameter $D=4$ mm), $\lambda=2\ \mu\text{m}$, lens $f=1$ m placed immediately after the aperture, screen at the focal plane — the lens-at-the-aperture, screen-at-$f$ geometry is exactly the Fraunhofer condition identified in part (a).

Find. The diffraction pattern (Airy disk) radius on the screen.

Approach. A circular aperture's Fraunhofer pattern is the Airy pattern: a bright central disk surrounded by concentric rings of rapidly falling intensity, first dark ring at angle $\theta_1=1.22\lambda/D$; the lens maps angle to focal-plane position via $r=f\theta_1$ (paraxial).

1. First dark-ring radius. $r_1=\dfrac{1.22\,\lambda f}{D}=\dfrac{1.22(2\times10^{-6})(1)}{4\times10^{-3}}=\boxed{6.10\times10^{-4}\ \text{m}=0.610\ \text{mm}}$ (Airy-disk diameter $1.22$ mm).

Part (c) — rectangular aperture: separable sinc pattern.

Given. Rectangular aperture $4\ \mu\text{m}$ (along $z$) $\times\,2\ \mu\text{m}$ (along $y$), $\lambda=1\ \mu\text{m}$, $f=1$ m.

Find. The diffraction pattern's central-lobe half-widths along $Y$ and $Z$.

Approach. A rectangular aperture's Fraunhofer pattern separates into the product of two 1-D $\mathrm{sinc}^2$ patterns, one per aperture dimension, each with first zero at $\theta=\lambda/b$ (aperture width $b$ in that direction) — mapped to the screen via $y=f\theta$, exactly as for the single slit in part (b) of the diffraction family. Check: at these micron-scale apertures the first-zero half-angle $\lambda/b$ reaches $0.25$–$0.5$ rad, beyond the strict paraxial regime; the standard exam-level $y=f\lambda/b$ estimate is used here as the intended illustrative answer, consistent with the "no detailed computations required, sketch and justify" framing of this sub-part.

1. Half-width along $Z$ (set by the $4\ \mu\text{m}$ dimension). $\Delta z=\dfrac{f\lambda}{b_z}=\dfrac{(1)(1\times10^{-6})}{4\times10^{-6}}=\boxed{0.25\ \text{m}=250\ \text{mm}}.$
2. Half-width along $Y$ (set by the narrower $2\ \mu\text{m}$ dimension). $\Delta y=\dfrac{f\lambda}{b_y}=\dfrac{(1)(1\times10^{-6})}{2\times10^{-6}}=\boxed{0.50\ \text{m}=500\ \text{mm}}.$ The pattern is elongated along $Y$ — the narrower aperture dimension diffracts more, the reciprocal relationship at the heart of Fraunhofer diffraction.