Spectrographs, photometry and astrometry

Builds on: Sections 5.4 to 5.6
Reading time: 35 min
Exercises: 4
Olympiad problems: 1

The yellow light of a sodium lamp, spread out by a good spectroscope, splits into two lines at 589.0 nm and 589.6 nm. The same pair of lines appears dark in the spectrum of the Sun, where Fraunhofer labelled it D. The lines are only 0.6 nm apart, so an instrument must tell apart wavelengths that differ by about one part in a thousand to show them as two. In this section we see how a spectrograph spreads light by wavelength and what limits the detail it shows, how astronomers measure brightness through standard filters, which kinds of light pass through the air at all, and how precisely the position of a star can be measured on an image.

5.8.1Spreading light with a grating

Before you readMake a guess first

A grating has 600 lines in each millimetre. At about what angle does it send green light of wavelength 550 nm in its first order?

A spectrograph spreads the light of a star into its colours and records the result, the spectrum, on a camera. Most astronomical spectrographs use a diffraction grating, a surface ruled with thousands of fine, evenly spaced lines. Light leaving neighbouring lines at an angle $θ$ travels distances that differ by $d \sin θ$ , where $d$ is the spacing of the lines. The waves reinforce each other when that difference is a whole number $m$ of wavelengths, so each wavelength leaves at its own angle, and each whole number $m$ , called the order, gives a complete spectrum.

The finest detail a grating can show depends on how many lines take part. The bright peak of one wavelength fades to its first dark point when the paths from the first and the last of the $N$ lit lines differ by one extra wavelength, as the two edges of a lens do for the first dark ring of Section 5.4. A second wavelength $λ + Δ λ$ has its peak exactly there when $m N Δ λ = λ$ , and by the Rayleigh criterion that is when the two can just be told apart. The ratio $λ / Δ λ$ is called the resolving power $R$ of the spectrograph.

ResultGrating equation and resolving power(5.12)

\begin{aligned} d \sin θ & = m λ \\ R = \frac{λ}{Δ λ} & = m N \\ Δ v & = \frac{c}{R} \end{aligned}

$d$	the spacing of the lines of the grating
$θ$	the angle at which light of wavelength $λ$ leaves the grating, for light arriving straight on
$m$	the order, a whole number
$R$	the resolving power, the wavelength divided by the smallest difference in wavelength that can be told apart
$N$	the number of lines lit by the light
$Δ v$	the smallest difference in radial velocity that shifts a line by $Δ λ$ , by the Doppler effect of Section S4

In practice the width of the slit and the pixels of the camera can limit

R

before the grating does. Spectrographs for finding planets round other stars reach

R

of about 100 000.

Quick checkThe sodium lines

The two sodium lines are at 588.995 nm and 589.592 nm. What resolving power is needed to tell them apart?

Example 5.10How small a velocity a spectrograph sees

A spectrograph has a resolving power of 50 000. What is the smallest change in the radial velocity of a star that it can detect as a shift of one resolution element?

Solution

By (5.12),

Δ v = \frac{3.00 \times 10^{5} {km s}^{- 1}}{50 000} = 6.0 {km s}^{- 1} .

AnswerA shift of one resolution element is 6.0 km/s. By measuring the centres of thousands of lines at once, each far more precisely than its width as the last part of this section explains, the best spectrographs find shifts of about a metre per second.

Example 5.11Splitting the sodium linesfinish it

A grating with 600 lines in each millimetre is used in the first order. How wide a strip of it must be lit to split the sodium lines, which need a resolving power of 987?

The start

In the first order $m = 1$ , so by (5.12) at least $N = 987$ lines must be lit.

Your stepNow find the width of grating that holds that many lines.

5.8.2Photometry and filters

A photometer measures the brightness of a star, today usually with a CCD and before that with a photomultiplier tube that counted photons one by one. To compare measurements made at different observatories, astronomers look through standard filters that pass only one band of wavelengths. The most widely used set, the Johnson system, has the U (ultraviolet), B (blue) and V (visual) bands, centred near 365 nm, 445 nm and 551 nm, with R and I added in the red and near infrared. A magnitude measured through the V filter is written $V$ or $m_{V}$ , and the difference between the magnitudes of a star in two bands, its colour, tells its temperature, as Chapter 4 explains.

5.8.3Light that reaches the ground

Before you readMake a guess first

Which of these kinds of light from space reaches a telescope at sea level?

The air lets through only some wavelengths, in a few ranges called atmospheric windows. The optical window runs from about 300 nm, where ozone starts to absorb ultraviolet light, to about 1 µm, and a few narrower windows continue into the near infrared. The radio window runs from about a centimetre to about ten metres, and reaches down to about a millimetre at high, dry sites such as the Atacama plateau in Chile, where the ALMA dishes stand. Water vapour absorbs most of the infrared, and gamma rays, X-rays and most ultraviolet light never reach the ground at all.

Telescopes for those wavelengths are put in orbit above the air, such as X-ray telescopes and the James Webb Space Telescope for the infrared. Even inside a window, the light of towns scattered by the air brightens the sky, and this light pollution, together with the moving air of Section 5.4, is why observatories are built on remote mountains.

5.8.4Measuring positions

Astrometry is the precise measurement of the positions of stars on the sky, and of how they change. The parallax of a star 100 parsecs away is only 0.010″, a hundred times smaller than the 1″ blur of the seeing, and yet parallaxes far smaller still are measured routinely, today mostly by the Gaia satellite in space, where there is no seeing. The centre of a star's image can be found far more precisely than its width, which is measured as the full width at half maximum, the width across the image where it is half as bright as at its centre. For a round image limited by photon noise, an image $w$ wide recorded with a signal-to-noise ratio $S / N$ has its centre located to about $w / (2.35 S / N)$ .

ResultPrecision of a position(5.13)

σ_{pos} \approx \frac{w}{2.35 S / N}

$σ_{pos}$	the uncertainty in the position of the centre of a star's image
$w$	the full width at half maximum of the image
$S / N$	the signal-to-noise ratio of the star, from (5.10)

The factor 2.35 belongs to a round image whose brightness falls off as a bell curve, and other shapes change it a little. The same idea lets a spectrograph measure the centre of a line far more precisely than its resolution.

Quick checkA bright star on a 2″ night

On a night of 2.0″ seeing, a star is recorded with a signal-to-noise ratio of 100. How precisely can its position be measured?

arcseconds

Common errorsSection 5.8

The resolving power $R = m N$ counts only the lines lit by the light. A grating with many lines in each millimetre but a narrow beam may have a small $N$ .
The smallest difference in wavelength a spectrograph shows is $λ / R$ , which grows with the wavelength. Taking $Δ λ$ as the same across the whole spectrum gives the wrong answer at the red end.
A position or the centre of a spectral line can be measured much more precisely than the width of the image or of the line, but only if the signal-to-noise ratio is high.

ExercisesSection 5.84 questions

5.8.1
What resolving power is needed to detect a change of 1 km/s in the radial velocity of a star as a shift of one resolution element?
5.8.2
A grating 25 mm wide has 600 lines in each millimetre and is fully lit. What is its resolving power in the first order?
5.8.3
A star's image is 1.0″ wide, and it is recorded with a signal-to-noise ratio of 100. Roughly how precisely can its position be measured?
arcseconds
5.8.4
Which of these can be observed with a telescope on the ground?

Olympiad problemsPast IOAA problems, restated1 problem

2023A spectrum during microlensing5 points

Adapted from IOAA 2023 (Katowice), theory problem 3. We have restated it in our own words and kept the original data. The solution was last checked on 20 October 2026.

A faint star in the bulge of the Milky Way, normally of magnitude 20.4 in the I band, brightened to magnitude 15.2 for a while, because the gravity of a star passing in front of it bent more of its light towards us. During the brightening, astronomers took a high-resolution spectrum of it with the UVES spectrograph on the Very Large Telescope, whose mirror is 8.2 m across. How large a telescope would be needed to take a spectrum of the same quality, with the same instrument and exposure time, of the star at its normal brightness? The sky background is negligible.

I solved it without the solution

AnswerA telescope about 90 m across would be needed, more than twice the size of the largest telescope now being built.