Physics · Optics · Year 1

Resolving power

Why does a large telescope separate a double star while a small one sees a single blob?

The situation

If on an evening you look at the sky away from the city lights, next to Mizar in the handle of the Plough in Ursa Major you can see faint Alcor — according to legend, in ancient times this was a test of keen eyesight. In a small telescope Mizar itself turns out to be two stars, and the star ε in Lyra looks like a “double-double” in a good telescope. But even in the best instrument a star is not a point, it is a small disc surrounded by rings. What determines this disc, and why does a large telescope separate close stars while a small one does not?

Topic

Resolving power, Fraunhofer diffraction. The bigger the telescope, the sharper the image: the Airy disc and the Rayleigh criterion

Key ideas

Airy disc
Rayleigh criterion
Fresnel zones

Formulas

I = I_{0} \frac{\sin^{2} β}{β^{2}} {(\frac{\sin N γ}{N \sin γ})}^{2}, β = \frac{π a \sin θ}{λ}, γ = \frac{π d \sin θ}{λ}

Fraunhofer diffraction at N slits: minima $a \sin θ = m λ$ , principal maxima $d \sin θ = m λ$ .

\sin θ_{R} = 1.22 \frac{λ}{D}

Circular aperture (objective lens, pupil of the eye): the first dark ring of the Airy disc — the smallest angle that can be resolved by the Rayleigh criterion.

d_{m i n} = \frac{0.61 λ}{N A}

The resolution limit of a microscope; $N A$ is the numerical aperture of the objective (up to ≈ 1.4–1.5 with oil immersion).

r_{n} = \sqrt{n λ b}, f = \frac{r_{1}^{2}}{λ}

Radii of the Fresnel zones (plane wave, screen at distance $b$ ) and the focal length of a zone plate.

R = \frac{λ}{δ λ} = m N, \frac{d θ}{d λ} = \frac{m}{d \cos θ}

The resolving power and angular dispersion of a diffraction grating; $N$ is the number of illuminated lines.

Theory

The Huygens–Fresnel principle and the vector diagram

Every point of a wavefront is a source of secondary waves; the oscillation at the observation point is their sum, taking their phases into account (the Huygens–Fresnel principle). It is convenient to draw this sum: the oscillation from each small piece is an arrow, and the path difference turns it; the arrows are joined one after another (the vector diagram). The length of the resultant arrow is the amplitude, and its square is the intensity. If the chain closes into a polygon, the point is dark (a minimum).

Fraunhofer diffraction: one slit and many slits

A plane wave passing through a slit gives a Fraunhofer pattern far away (or at the focus of a lens). For one slit the minima are at $a \sin θ = m λ$ : the narrower the slit, the wider the central maximum ( $2 λ f / a$ ). With $N$ slits the waves reinforce in the directions $d \sin θ = m λ$ (the principal maxima); between them there are $N - 1$ minima and $N - 2$ small maxima; the single-slit pattern wraps them like an “envelope”. As $N$ grows, the principal maxima become narrower — this is the diffraction grating.

The Airy disc and the Rayleigh criterion

A circular aperture (a telescope objective, the pupil of the eye) images a point source as a central disc surrounded by rings — the Airy disc (≈ 84 % of the light is in the central disc). The first dark ring is at $\sin θ = 1.22 λ / D$ . When the maximum of one falls on the first minimum of the other, two points are “just resolved” (the Rayleigh criterion): the dip between them is ≈ 73 %. So a large $D$ and a small $λ$ give good resolution. The pupil of the eye is 2–8 mm, but in practice the eye resolves ≈ 1′ — with a wide pupil it is limited by the imperfections (aberrations) of the eye's optics, and also by the density of cells in the retina. Large ground-based telescopes are limited by atmospheric turbulence (≈ 1″) — which is why adaptive optics and space telescopes are built. In a microscope $d_{m i n} = 0.61 λ / N A$ : in visible light details smaller than ≈ 0.2 µm cannot be resolved.

Fresnel zones: hole, disc, zone plate, edge

If the screen is near (Fresnel number $m = R^{2} / λ b ≳ 1$ ), we divide the wavefront into Fresnel zones: waves from neighbouring zones have a path difference of $λ / 2$ — they cancel each other. If a hole uncovers an odd number of zones, the centre is bright (up to 4I₀); if even, dark. But at the centre of the shadow of an opaque disc there is always a bright spot (the Poisson–Arago spot). A zone plate that blocks the even zones focuses like a lens: $f = r_{1}^{2} / λ$ (exactly this kind of “lens” is used in X-ray microscopes). In the vector diagram the zones go round a circle (the obliquity factor slowly turns it towards the centre — a spiral). For a straight edge and a slit the Cornu spiral (Fresnel integrals) is used: at the boundary of the geometric shadow $I = I_{0} / 4$ , and on the lit side the brightest fringe is ≈ 1.37I₀.

The diffraction grating and spectral instruments

At a grating $d \sin θ = m λ$ : different colours go at different angles (the angular dispersion is $m / (d \cos θ)$ ). To resolve two close lines, the angle between their maxima must be larger than the width of a line — the Rayleigh criterion gives $R = λ / δ λ = m N$ . For the sodium doublet ( $Δ λ = 0.597$ nm) $R \approx 987$ is needed. Spectral instruments are divided into prism types (dispersion of the refractive index; $R = b d n / d λ$ , where $b$ is the base of the prism), diffraction and interference (Fabry–Pérot) types. X-ray diffraction in a crystal lattice (the Bragg condition) is covered in the “Infrared, UV and X-rays” experiment, and holography in “Laser physics and holography”.

Model

Scalar diffraction theory: the light is monochromatic, the Fresnel calculations use the paraxial approximation, and the obliquity factor is ignored. In the eyepiece and screen images the brightness is compressed with a square root so that faint rings are visible. The slits are drawn enlarged on the slide; the Fresnel obstacle and the grating are shown at true size in the scene.

Did you know?

Augustin Fresnel submitted a paper on the wave theory of light to the 1818 competition of the French Academy of Sciences. Siméon Poisson deduced from it that there should be a bright spot at the centre of a disc's shadow and considered this absurd — but François Arago showed exactly this spot in an experiment.
The mirror of the Hubble Space Telescope has a diameter of 2.4 m: in visible light it can resolve an angle of ≈ 0.05″. Ground-based telescopes are usually limited by the atmosphere to ≈ 1″.

Inside the experiment

The experiment starts from a real-life situation: you predict the result, change the controls, run the experiment and draw a conclusion. Inside are missions, theory, a quiz and a guided tour in which the camera moves and explains the key points.

More experiments in this section

Luminous flux and illuminance — Grade 11
Infrared, UV and X-rays — Grade 11
Scattering of light — Year 1
Crystal optics — Year 1
Interferometers and LIGO — Year 1
Nonlinear optics: the green laser — Year 2