On a straight highway at night, a truck comes towards you from far away. At first you see a single light. As it comes nearer, that light stretches a little, and then splits into two headlights. The lights were always two, about one and a half metres apart. Something in your eye could not separate them.
Every point becomes a small disc
A lens or a mirror of diameter \(D\) is a round opening, and light diffracts at its edge. So a lens never images a star as a point. It images it as a small bright disc surrounded by faint rings, the Airy disc. For a round opening the first dark ring lies at the angle \[ \sin\theta_1 = \frac{1.22\,\lambda}{D}. \] The factor \(1.22\) replaces the \(1\) of a slit because the opening is round, not straight-sided.
The central disc holds about \(84\,\%\) of the light, and the first bright ring under \(2\,\%\) of its peak. So the image of a point is, for practical purposes, a small disc of angular radius \(1.22\lambda/D\). A wider lens makes the disc smaller.
When are two discs two?
Two stars close together in the sky give two such discs side by side. If the discs are far apart, we see two stars. If they overlap almost completely, we see one blurred star. Lord Rayleigh proposed a simple line between the two cases in 1879.
Two point sources are just resolved when the central maximum of one's diffraction pattern falls on the first minimum of the other's. For a round opening of diameter \(D\), their angular separation is then \(\theta_R = 1.22\lambda/D\).
At Rayleigh's criterion, the sum of the two patterns has two peaks with a dip between them. For round openings the dip falls to \(74\,\%\) of the peaks. The eye can just notice that. Move the stars apart in the figure below, and watch the dip.
In the figure above, at half the limit, the two images merge into one peak. At \(0.8\,\theta_R\), the top of the sum is flat, and still one peak. At \(\theta_R\) a dip to \(74\,\%\) appears between two peaks. At \(2\theta_R\) the centre falls to zero, and the stars are clearly two.
The smallest angle that can be resolved is the limit of resolution. Its inverse is the resolving power. A small limit of resolution means a large resolving power, and finer detail seen.
By Rayleigh's criterion, two point sources seen through a round lens are just resolved when
The limit of a telescope
A telescope's objective is its round opening, so its limit of resolution is \[ \theta_R = \frac{1.22\,\lambda}{D}. \] Its resolving power is the inverse, \(1/\theta_R = D/1.22\lambda\). Take an objective \(12.2\,\text{cm}\) across and light of \(500\,\text{nm}\). Then \(\theta_R = 1.22 \times 500\,\text{nm}/12.2\,\text{cm}\), which is \(5 \times 10^{-6}\,\text{rad}\). Its resolving power is \(2 \times 10^5\) per radian.
Magnifying power and resolving power are different things. The chapter on optical instruments found \(M = f_o/f_e\) for a telescope. A stronger eyepiece enlarges the image, but it cannot put back detail that diffraction at the objective has already blurred. Only a wider objective, or a shorter wavelength, resolves more.
So astronomers build large objectives for two reasons. A wider objective collects more light, in proportion to its area, so fainter stars become visible. It also has a smaller \(1.22\lambda/D\), so close stars are seen as two.
Two stars are \(10^{-5}\,\text{rad}\) apart. What is the smallest diameter of objective that can resolve them in light of \(500\,\text{nm}\)?
The stars are resolved if \(1.22\lambda/D \le 10^{-5}\,\text{rad}\).
\(D \ge 1.22\lambda/10^{-5}\).
\(D \ge (1.22 \times 500 \times 10^{-9})/10^{-5} = 6.1 \times 10^{-2}\,\text{m}\).
The objective must be at least \(6.1\,\text{cm}\) across.
The Devasthal telescope near Nainital has a mirror \(3.6\,\text{m}\) across. What is its limit of resolution in light of \(500\,\text{nm}\), in units of \(10^{-7}\,\text{rad}\)?
On the ground, the shimmering air above a telescope blurs every star to about one second of arc, \(5 \times 10^{-6}\,\text{rad}\). So a large telescope on a mountain rarely reaches its diffraction limit. Its size still gathers light, and modern mirrors bend many times a second to undo the shimmer.
Radio telescopes need enormous dishes
A radio telescope obeys the same limit, but its wavelength is long. A dish \(100\,\text{m}\) across, working at \(\lambda = 1\,\text{m}\), has \(\theta_R = 1.22/100 = 1.22 \times 10^{-2}\,\text{rad}\), about \(0.7°\). An optical telescope only \(10\,\text{cm}\) across, at \(500\,\text{nm}\), resolves \(6.1 \times 10^{-6}\,\text{rad}\). The huge dish is \(2000\) times coarser.
Near Narayangaon, north of Pune, thirty dishes stand in the fields, each \(45\,\text{m}\) across. They form the Giant Metrewave Radio Telescope, run by the National Centre for Radio Astrophysics. One dish alone, at a wavelength of \(1\,\text{m}\), resolves only \(1.22/45 \approx 0.027\,\text{rad}\), more than \(1.5°\). But the dishes are spread over about \(25\,\text{km}\), and their signals are combined. Together they act like one dish \(25\,\text{km}\) wide, resolving about \(5 \times 10^{-5}\,\text{rad}\).
The limit of a microscope
A microscope looks at two points close together on a slide, not at two distant stars. What matters is the least distance between two points that can still be seen as two. Let the objective collect a cone of light of half-angle \(\theta\) from the object, in a medium of index \(\mu\). Then \[ d_{\min} = \frac{1.22\,\lambda}{2\mu\sin\theta}. \] The quantity \(\mu\sin\theta\) is the numerical aperture of the objective.
An objective of numerical aperture \(1.22\), used with light of \(500\,\text{nm}\), resolves \(d_{\min} = 1.22 \times 500/2.44\), which is \(250\,\text{nm}\). Its resolving power is \(1/d_{\min} = 4 \times 10^6\,\text{m}^{-1}\). No light microscope does much better than about \(200\,\text{nm}\).
The formula shows three ways to resolve finer detail. Use a shorter wavelength, blue or ultraviolet. Fill the space between the slide and the objective with oil of \(\mu \approx 1.5\), which is called oil immersion. And bring a wide objective close to the slide, so that \(\theta\) is large. With \(\sin\theta = 0.9\) at \(500\,\text{nm}\), a dry objective resolves \(339\,\text{nm}\); in oil it resolves \(226\,\text{nm}\).
An objective collects a cone of half-angle \(\theta\) with \(\sin\theta = 0.9\). With oil of \(\mu = 1.5\) between it and the slide, and light of \(500\,\text{nm}\), what is the least distance it can resolve, in nm?
Which of these changes let a microscope resolve finer detail?
The electron microscope takes the first way to the extreme. The unit on Modern Physics shows that a moving electron behaves as a wave. Electrons accelerated through \(10\,\text{kV}\) have a wavelength of about \(0.012\,\text{nm}\), some forty thousand times shorter than green light. So an electron microscope resolves detail far finer than any light microscope can, down to single molecules.
When two headlights become two
The pupil of the eye is a round opening, so the eye has a limit of resolution too. Take a pupil \(2.44\,\text{mm}\) across and light of \(600\,\text{nm}\). Then \[ \theta_R = \frac{1.22 \times 600\,\text{nm}}{2.44\,\text{mm}} = 3 \times 10^{-4}\,\text{rad}, \] about one minute of arc.
Two headlights \(1.5\,\text{m}\) apart, at a distance \(L\), are \(1.5/L\) apart in angle. They are just resolved when \(1.5/L = 3 \times 10^{-4}\), so \(L = 5000\,\text{m}\). The truck on the highway shows one light while it is more than \(5\,\text{km}\) away. Closer than that, the single light splits into two.
At night the pupil of your eye is \(3\,\text{mm}\) across. Take light of \(500\,\text{nm}\). Two headlights are \(1.5\,\text{m}\) apart. At what greatest distance can you see them as two, in km?
With a pupil of \(2.44\,\text{mm}\) and light of \(600\,\text{nm}\), estimate the smallest distance between two features on the Moon that the naked eye could separate. The Moon is \(3.84 \times 10^5\,\text{km}\) away.
George Biddell Airy was the Astronomer Royal of Britain. Looking at bright stars through good telescopes, astronomers always saw a tiny disc with faint rings, never a point. In 1835 Airy worked out the diffraction of light by a round object glass. He showed that the disc and its rings come from the lens itself, and that the disc shrinks as the lens widens. The bright central disc still carries his name.
- Believing that a stronger eyepiece lets a telescope separate closer stars.Resolution is set by diffraction at the objective, \(1.22\lambda/D\). Magnification only enlarges what the objective has already blurred.
- Using \(\lambda/D\) for a round lens.A round opening has its first dark ring at \(1.22\lambda/D\). The factor 1.22 comes from its round shape.
- Mixing up the limit of resolution and the resolving power.The limit is the least resolvable angle or distance; the resolving power is its inverse. A smaller limit is a larger resolving power.