17-Phys-A7 Optics · Undated paper

Question 3 of 10: Fresnel vs. Fraunhofer diffraction from a rectangular aperture

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

Question 3: Fresnel vs. Fraunhofer diffraction from a rectangular aperture (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.

Given. Rectangular aperture $10\ \mu\text{m}\times200\ \mu\text{m}$, normal-incidence plane wave, $\lambda=1\ \mu\text{m}$.

a) Fresnel (near-field) diffraction. Fresnel diffraction is the regime observed at a finite distance from the aperture (or with a converging/diverging illuminating wave), where the Fresnel number $N_F=a^2/(\lambda z)\gtrsim1$ ($a$ = aperture half-size, $z$ = observation distance). The wavefront curvature reaching different points of the observation plane cannot be neglected, so the diffraction pattern still recognisably outlines the aperture’s own geometric shadow, but with a fine ripple of bright/dark fringes decorating the edges of that shadow (and, on-axis behind a circular obstacle, a bright Poisson/Arago spot). As $z$ increases, $N_F$ falls and the pattern gradually morphs into the Fraunhofer pattern of part (b).

b) Fraunhofer (far-field) diffraction. Fraunhofer diffraction is the far-field limit, $N_F\ll1$ (observation distance $z\gg a^2/\lambda$, or equivalently a lens used to bring infinity to its focal plane) — the geometric shadow shape is lost entirely and what remains is the aperture’s spatial Fourier transform. For a rectangular aperture of width $w$ (10 μm) and length $\ell$ (200 μm) this is the separable product of two $\text{sinc}^2$ functions, one along each aperture edge’s normal direction: a bright central lobe flanked by much weaker side lobes, forming a cross (plus-sign) pattern whose arms run perpendicular to the two pairs of aperture edges. Because the aperture is narrow in $x$ (10 μm) and long in $y$ (200 μm), diffraction is inversely proportional to aperture size in each direction — the pattern spreads far in the direction perpendicular to the narrow (10 μm) dimension and stays narrow perpendicular to the long (200 μm) dimension, i.e. the long axis of the diffraction cross is perpendicular to the long axis of the slit.

c) Fraunhofer pattern of an equilateral triangular aperture (answering the c/d choice). A rectangular aperture has two pairs of parallel edges, so its Fraunhofer pattern is a simple 4-arm cross (two perpendicular streak directions). An equilateral triangle has three edges at $60^\circ$ to one another and no edge parallel to any other, so each edge contributes its own streak of diffraction minima running perpendicular to that edge, through the central maximum, on both sides of the centre — three streak-lines, spaced $60^\circ$ apart, giving a six-armed star (hexagram) pattern rather than a four-armed cross. Of the images provided, the pattern showing six thin, roughly evenly-spaced streaks radiating from a bright centre (labelled B in the source) is the one that matches this reasoning, and is selected here; the other six-armed candidate offered (C) has its streaks bunched rather than evenly spaced through $60^\circ$, which is not what three symmetric edge-normal directions produce.

d) Angular deviation to the first minimum, reduced slit.

Given. New slit width $a=1\ \text{mm}$ (length now $1\ \text{cm}\gg a$, so the slit is effectively one-dimensional); illumination wavelength $\lambda=1\ \text{mm}$ (as printed).

Find. $\theta$, the angle (in the $y$–$z$ plane of the source figure) to the first off-axis Fraunhofer minimum.

Approach. Single-slit Fraunhofer minima occur at $a\sin\theta=m\lambda$; the first minimum is $m=1$.

1. Single-slit minimum condition, first order. $\sin\theta=\dfrac{\lambda}{a}=\dfrac{1\ \text{mm}}{1\ \text{mm}}=1.000\ \Rightarrow\ \theta=\boxed{90^\circ}.$