Describe diffraction of waves at a single aperture, relate the degree of spreading to the ratio of wavelength to aperture width, and recognise the single-slit pattern

What this dot point is asking

SEAB wants you to describe diffraction at a single aperture, to relate the amount of spreading to the ratio of wavelength to aperture width, and to recognise the single-slit diffraction pattern with its broad central maximum. Diffraction is a defining wave property and sets the ultimate limit on the resolution of optical instruments.

The answer

What diffraction is

Diffraction is the spreading of a wave as it passes through an aperture or around an obstacle. Every wave diffracts: water waves spread after a harbour gap, sound bends around a doorway, and light spreads through a narrow slit. The wavefronts curve at the edges, sending energy into the geometric shadow region.

The wavelength-to-width ratio

The amount of spreading depends on how the aperture width $b$ compares with the wavelength $\lambda$:

• if $b \gg \lambda$, the wave passes almost straight through with little spreading,
• if $b \approx \lambda$, the spreading is large and the wave fans out widely.

This is why sound (long wavelength) diffracts noticeably around everyday objects while light (very short wavelength) usually appears to travel in straight lines: ordinary openings are vastly wider than the wavelength of light.

The single-slit pattern

When monochromatic light passes through a single narrow slit, it produces a pattern with a bright, wide central maximum flanked by progressively dimmer secondary maxima, separated by dark minima. The first minimum occurs at an angle given by:

where $b$ is the slit width. The central maximum is twice as wide as the secondary maxima and contains most of the energy. A narrower slit (smaller $b$) widens the central maximum and spreads the whole pattern.

Diffraction and resolution

Because every aperture diffracts, the image of a point source through a lens or aperture is a small diffraction pattern rather than a point. Two nearby sources can only be told apart if their diffraction patterns do not overlap too much. This diffraction limit is why larger telescope apertures (larger $b$ relative to $\lambda$) give sharper images: they diffract less and resolve finer detail.

Examples in context

Example 1. Hearing around a corner. You can hear someone speaking around a doorway even when you cannot see them, because sound's wavelength (around a metre) is comparable to the door width and diffracts strongly. Light's wavelength is far smaller than the door, so it barely diffracts and the person stays out of sight.

Example 2. Telescope resolving power. A telescope's ability to separate two close stars is limited by diffraction at its aperture. A larger mirror diameter $b$ gives a smaller diffraction angle $\sim \lambda/b$, so big telescopes resolve finer detail. This is the practical reason observatories build ever-larger apertures.

Try this

Q1. State the condition under which a wave diffracts most strongly at an aperture. [1 mark]

• Cue. When the aperture width is comparable to the wavelength of the wave.

Q2. A single slit of width $0.20\ \text{mm}$ is illuminated by light of wavelength $640\ \text{nm}$. Find the angle of the first minimum. [2 marks]

• Cue. $\sin\theta = \lambda/b = \dfrac{640 \times 10^{-9}}{0.20 \times 10^{-3}} = 3.2 \times 10^{-3}$, so $\theta = 0.18^\circ$.

Q3. Explain how the single-slit diffraction pattern changes when the slit is made narrower. [2 marks]

• Cue. The ratio $\lambda/b$ increases, so the central maximum widens and the whole pattern spreads out, while becoming dimmer as less light passes.