Radio telescopes and interferometers

Builds on: Sections 5.2 and 5.4
Reading time: 30 min
Exercises: 4
Olympiad problems: 1

The Giant Metrewave Radio Telescope near Pune in India has 30 dishes, each 45 m across. At a frequency of 200 MHz, where the radio waves are 1.5 m long, a single dish blurs the sky into patches about 2° across, four times the width of the full Moon. Working together across the 25 km over which they are spread, the dishes can pick out detail about 12″ across. In this section we see why a radio telescope must be so large, how two dishes far apart can behave like one enormous one, and how long a source takes to pass through the beam of a dish that is held still.

5.7.1A single dish

Before you readMake a guess first

Hydrogen gas in the Milky Way gives out radio waves 21 cm long. How large would a dish have to be to see detail as fine as the eye sees, about 1′, at that wavelength?

A radio dish works like the mirror of an optical telescope. Its curved surface reflects radio waves to a receiver at its focus, and it obeys the same diffraction limit of (5.7). Radio waves are thousands to millions of times longer than the waves of visible light, so even the largest dishes see only a blurred sky. The patch of sky to which a dish responds is called its beam, and its width at half the greatest response, the full width at half maximum, is about $λ / D$ . Olympiad problems often take the beam width as $1.22 λ / D$ , the resolution of the dish, and the difference does not matter for estimates.

5.7.2Two dishes together

Two dishes a distance $B$ apart, called the baseline, both receive waves from the same distant source. If the source is not overhead, the waves reach the dish further from it later, after travelling an extra distance $B \sin θ$ , where $θ$ is the angle of the source from the zenith, measured in the plane that contains the two dishes. When the signals of the two dishes are added, they reinforce each other when that extra path is a whole number of wavelengths and cancel when it is a whole number and a half. As the source moves across the sky, the combined signal rises and falls in a row of fringes, and a pair of dishes used this way is an interferometer.

ResultA two-element interferometer(5.11)

\begin{aligned} Δ & = B \sin θ \\ θ_{fringe} & \approx \frac{λ}{B} \end{aligned}

$Δ$	the extra distance the waves travel to the dish further from the source
$B$	the baseline, the distance between the two dishes
$θ$	the angle of the source from the zenith, in the plane of the two dishes and the zenith
$θ_{fringe}$	the angle between neighbouring fringes near the zenith, in radians

Two sources closer together than about

λ / B

give fringes that cannot be told apart, so

λ / B

is the resolution of the interferometer. It is set by the baseline. The size of the dishes sets only how much of the sky the pair looks at.

Figure 5.5Fringes from two dishesinteractive

Angle θBaseline

Extra path to the far dish
Combined signal
Spacing of the fringes, λ/B

The figure shows two dishes P and Q a baseline

B

apart, receiving waves from a source at an angle

θ

from the zenith, with the tilt of the waves drawn five times larger than

θ

so that it can be seen. The dashed wavefront passes through the dish nearer the source, and the heavy line is the extra path

B \sin θ

that the same wave still has to travel to reach the other dish. The graph below shows the combined signal of the two dishes against

θ

, with the present angle marked. The sliders set

θ

and the baseline in wavelengths. With a baseline of the fringes are about 2.9° apart, and at the extra path is half a wavelength and the signal vanishes. With the fringes are three times closer together.

Many dishes linked in pairs give many baselines of different lengths and directions, and as the Earth turns, each baseline changes its direction against the sky. Combining all of them builds up a picture as sharp as a single dish as wide as the largest baseline would give. This method is called aperture synthesis. In very long baseline interferometry, dishes on different continents record their signals with precise atomic clocks, and the recordings are combined afterwards. The Event Horizon Telescope used this method at a wavelength of 1.3 mm with baselines nearly as long as the Earth is wide to image the shadow of the black hole in the galaxy M87.

Example 5.9A telescope as wide as the Earth

What resolution does an interferometer reach at a wavelength of 1.3 mm with a baseline as long as the Earth's diameter, 12 742 km? Give the answer in microarcseconds, millionths of an arcsecond.

Solution

By (5.11),

\frac{λ}{B} = \frac{1.3 \times 10^{- 3}}{1.2742 \times 10^{7}} = 1.02 \times 10^{- 10} rad,

and multiplying by 206 265 gives $2.1 \times 10^{- 5}$ ″, or 21 microarcseconds.

AnswerThe resolution is about 21 microarcseconds, enough to make out an object about 4 cm across on the Moon.

5.7.3A source drifting through a beam

A radio dish held still while the sky turns is the simplest way to observe, called a drift scan. A source passes through the beam just as a star crosses the field of an eyepiece in the drift method of Section 5.3. A source at declination $δ$ moves ${15.04}^{″} \cos δ$ each second, or 1° in 3.99 minutes of time divided by $\cos δ$ . A beam of width $θ$ in degrees, crossed along its middle, holds the source for $3.99 min \times θ / \cos δ$ .

Common errorsSection 5.7

The resolution of an interferometer is set by the distance between its dishes. The size of each dish sets only the width of the patch of sky it sees.
The rule $λ / D$ needs the wavelength. A frequency $ν$ must first be turned into $λ = c / ν$ .
A source near the celestial pole moves slowly, so it stays in a fixed beam longer by the factor $1 / \cos δ$ .

ExercisesSection 5.74 questions

5.7.1
What is the resolution of a radio dish 100 m across at a wavelength of 21 cm, in arcminutes?
arcminutes
5.7.2
Two dishes 1 km apart observe at a wavelength of 21 cm. How far apart are their fringes near the zenith, in arcseconds?
arcseconds
5.7.3
The dishes of the Giant Metrewave Radio Telescope are spread over about 25 km. What resolution can they reach at a frequency of 200 MHz?
arcseconds
5.7.4
The baseline of an interferometer is doubled at the same wavelength. What happens to its fringes?

Olympiad problemsPast IOAA problems, restated1 problem

2016A source drifting through the beam of a GMRT dish10 points

Adapted from IOAA 2016 (Bhubaneswar), theory problem 5. We have restated it in our own words and kept the original data. The solution was last checked on 20 October 2026.

The Giant Metrewave Radio Telescope is at latitude 19°06′ N. One of its dishes, 45.0 m across, is held still pointing at a zenith distance of 39°42′ on the northern meridian, so that a point source that crosses the meridian passes along a diameter of its beam. For how long does the source stay within the full width at half maximum of the beam of the dish at 200 MHz? Take the width of the beam to be the resolution of the dish, $1.22 λ / D$ .

I solved it without the solution

AnswerThe source stays within the half-power width of the beam for about 18 minutes.