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Telescopes across the spectrum

Astronomical sources emit at every wavelength, but the atmosphere is only transparent in two broad bands, so some telescopes have to be flown or put in orbit. This topic covers where each type of telescope must sit, the resolving power limit θ ≈ λ/D, and why CCDs replaced the eye and the photographic plate as detectors.

Builds on Diffraction and the single slit and Telescopes and image formation.

IN THIS TOPIC

  • Compare radio, infrared, ultraviolet and X-ray telescopes with optical ones in structure, siting and use.
  • Use the Rayleigh criterion θ ≈ λ/D, in radians, to compare resolving powers.
  • Compare collecting powers through diameter squared, and the CCD with the eye.

COMMON MISCONCEPTION

Observatories sit on mountaintops to be closer to the stars.

Where to stand

Stars, galaxies and everything between them radiate right across the electromagnetic spectrum, and each band tells a different story. Cold dust glows in the infrared. Hot gas emits in X-rays, while cool hydrogen emits weakly at radio wavelengths. The catch is the atmosphere. Air is transparent to visible light and to radio waves, and opaque to almost everything else, so only two broad windows open clear to sea level, with partial infrared access from high, dry sites.

What the atmosphere lets through: visible and radio reach the ground, some infrared a high dry summit; ultraviolet and X-rays need orbitX-rayultravioletvisibleinfraredradiotwo broad sea-level windows; orbit for the blocked bandswavelength rises left to right; a high dry site buys some infrared
FIG. 1The atmosphere's two broad sea-level windows, visible light and radio; some infrared survives to a high, dry summit; ultraviolet and X-rays are absorbed high up and need airborne or orbiting observatories.

That single fact dictates where every telescope lives. Radio dishes work at ground level, day and night, straight through cloud. A big steerable dish is a reflecting telescope scaled up, and because the wavelength is so long its surface can even be open mesh. Infrared telescopes climb high, dry mountains, above as much water vapour as possible, and their detectors are cooled so the instrument's own warmth does not glow in the very band it is observing. Ultraviolet and X-ray telescopes must orbit above the atmosphere entirely. X-rays then raise a further problem, since they pass straight into an ordinary mirror, so they are focused by skimming off nested metal surfaces at grazing incidence.

A few kilometres of altitude is nothing against interstellar distances. Observatories climb instead to get above water vapour and the shimmering turbulence of warm air, which is also why the sharpest visible-light images come from orbit.

BandWhere it worksWhy
radioground levelthe atmosphere is transparent to it; cloud and daylight do not matter
infraredhigh, dry summits, or spacewater vapour absorbs it, and warm equipment glows in it
visibleground, better in orbita clear window, but turbulence blurs fine detail
ultravioletorbitabsorbed high in the atmosphere, much of it by ozone
X-rayorbit, grazing-incidence mirrorsthe atmosphere absorbs it, and it penetrates ordinary mirrors

How sharp: the Rayleigh criterion

Every telescope has a sharpness limit it cannot exceed, and diffraction sets it. The aperture is a hole that light waves must squeeze through, so a point star never images to a point. It becomes a small bright disc ringed by faint fringes, the aperture's diffraction pattern. Two stars close together in angle mean two overlapping patterns, and if the overlap is too great no instrument behind the telescope can untangle them.

The Rayleigh criterion: two point sources are just resolved when the central peak of one diffraction pattern sits on the first minimum of the otheranglebrightnessangular separation θjust resolved: one peak on the other's first dark ring
FIG. 2The Rayleigh criterion. Two point sources are just resolved when the central peak of one diffraction pattern falls on the first dark minimum of the other; any closer and the two blobs merge into one.

The Rayleigh criterion makes the judgement precise. Two sources are just resolved when the centre of one pattern sits on the first minimum of the other, which happens at an angular separation of about

θλD\theta \approx \frac{\lambda}{D}ON THE AQA DATA SHEET

with θ in radians, λ the wavelength observed and D the aperture diameter. Smaller θ means finer detail, so the resolving power improves with a bigger aperture and a shorter wavelength. The equation explains a real difficulty for radio astronomy:

WORKED EXAMPLE

The giant that cannot out-see your eye

The Lovell radio telescope has a 76 m dish and observes the hydrogen line at λ = 0.21 m. Your dark-adapted pupil is about 5.0 mm across, working at λ ≈ 5.0 × 10−7 m. Compare their resolving powers.

Lovell: θ ≈ λ/D = 0.21 / 76 = 2.8 × 10−3 rad.

The eye: θ ≈ 5.0 × 10−7 / (5.0 × 10−3) = 1.0 × 10−4 rad.

The unaided eye resolves detail nearly thirty times finer than a 76-metre national instrument, because its wavelength is nearly half a million times shorter. Radio astronomers claw the sharpness back by linking dishes many kilometres apart, which makes D effectively enormous.

GUIDED PRACTICE

A tale of two dishes

One radio dish is 64 m across, another 32 m, observing the same 0.21 m line. Compare their resolving powers and say which sees finer detail.

Show the working

θ for the 64 m dish is 0.21/64 = 3.3 × 10−3 rad; for the 32 m dish, 0.21/32 = 6.6 × 10−3 rad.

Doubling D halves θ, so the 64 m dish resolves detail twice as fine. Smaller θ is better, because the number is the angle at which two sources stop being separable.

How deep: collecting power, and what detects the light

Most objects astronomers care about are desperately faint, so a telescope's second figure of merit is its collecting power, the rate at which the aperture gathers light. That is proportional to the aperture's area, and hence to the square of its diameter:

collecting powerD2\text{collecting power} \propto D^{2}NOT ON THE AQA DATA SHEET: LEARN IT
Collecting power grows as the square of the diameter: doubling the aperture gathers four times the lightD2D×1×4collecting power ∝ diameter squaredlight gathered
FIG. 3Double the diameter, four times the light. Collecting power follows area, so observatories chase every extra metre of mirror.

An 8.2 m mirror against your 5.0 mm pupil is a diameter ratio of 1640, so it gathers 16402 ≈ 2.7 × 106 times the light. Two and a half million times deeper into the dark, before any camera tricks. Both figures of merit point the same way. Diameter, not magnification, is what a telescope is for.

The detector matters as much as the mirror. For a century that meant the eye, then photographic plates. Now it means the charge-coupled device, the CCD at the heart of every astronomical camera.

You are asked for a basic account of how one works, so learn this chain. A CCD is a silicon chip divided into millions of picture elements, or pixels. Light falling on a pixel frees electrons inside the silicon by the photoelectric effect, and the number freed is proportional to the intensity of the light that struck it. Those electrons stay trapped in a potential well at the pixel for as long as the exposure lasts, so charge builds up in a pattern matching the image. When the shutter closes, the charge is shifted pixel by pixel to the edge of the chip, measured, and turned into a number for each pixel.

Comparisons with the eye usually open on quantum efficiency, the fraction of photons arriving at the detector that are actually registered. A CCD detects roughly 80% of them. Your eye manages about 1%.

CCDthe eye
quantum efficiencyabout 80%about 1%
exposurecan accumulate light for hoursrefreshes many times a second
resolutionset by the pixel spacing, a few micrometresset by the retina, coarser in the dark
recordpermanent, shareable, measurablenone
wavelengthsvisible and into the infrared and ultravioletvisible only
convenienceneeds power, cooling and a computeralways with you

Convenience is the eye's one remaining advantage, and astronomy traded it away without regret. A detector that stores light for hours, records it permanently and measures it in numbers is a different kind of instrument. That is why a modest amateur telescope with a CCD now reaches objects the great Victorian refractors never saw.

INDEPENDENT PRACTICE

Could Hubble read a number plate on Mars?

The Hubble telescope's mirror is 2.4 m across, observing at 5.0 × 10−7 m. Mars at its closest is about 5.6 × 1010 m away. Estimate the smallest detail Hubble can resolve on the Martian surface.

Show the working

θ ≈ λ/D = 5.0 × 10−7 / 2.4 = 2.1 × 10−7 rad.

At distance d the smallest separable detail is s ≈ θd = 2.1 × 10−7 × 5.6 × 10101.2 × 104 m.

About twelve kilometres: whole craters, not rovers. Small angle times huge distance is the working pattern for every question of this shape.

ASSESSMENT FOCUS

  • θ ≈ λ/D answers live in radians. If a question supplies degrees or arcseconds, convert before comparing.
  • To compare resolving powers, compute θ for both instruments and then say explicitly that the smaller angle resolves the finer detail. The comparison sentence completes the answer.
  • Collecting power comparisons square the diameter ratio. Show the ratio first, then square it. Examiners are looking for D2, not D.
  • Siting questions want the atmosphere named and blamed, either the band the air absorbs or the turbulence that blurs. Altitude for its own sake earns nothing.
  • A CCD question can ask for the mechanism, so do not stop at the numbers. Photons free electrons in the silicon, the electrons are held in a potential well at each pixel, the count is proportional to intensity, and the pattern is read out at the end of the exposure.
  • Quantum efficiency has a one-sentence definition, the percentage of incident photons the detector registers. Start the CCD-versus-eye comparison there, then take resolution and convenience in turn.

CHECK YOURSELF

What diameter would a radio telescope observing at 0.21 m need to match the resolving power of a 10 cm optical telescope working at 5.0 × 10−7 m? Comment on the answer.

Show a hint

Find the optical telescope's θ first, then ask what D gives the radio dish the same θ.

Show the answer

Optical, θ ≈ λ/D = 5.0 × 10−7 / 0.10 = 5.0 × 10−6 rad.

Radio, D = λ/θ = 0.21 / (5.0 × 10−6) = 4.2 × 104 m.

A single dish 42 km across is not buildable, and that is the point of the comment. Matching even a small optical telescope at radio wavelengths takes networks of dishes spread across the countryside, acting together as one giant aperture.

θ ≈ λ over D, in radians, and smaller is sharper.

The light gathered grows as the diameter squared: double the mirror and it collects four times the light.

A CCD registers about 80 per cent of the photons that reach it. Your eye manages one.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

20 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the telescopes across the spectrum questions page.

6 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

CHECK YOUR PROGRESS

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  • Compare radio, infrared, ultraviolet and X-ray telescopes with optical ones in structure, siting and use.
  • Use the Rayleigh criterion θ ≈ λ/D, in radians, to compare resolving powers.
  • Compare collecting powers through diameter squared, and the CCD with the eye.

Open the full revision checklist to track your progress across the whole unit.