Plate scale and field of view

Builds on: Section 5.1
Reading time: 30 min
Exercises: 4
Olympiad problems: 0

With a camera whose sensor is 22.5 mm wide at the focus of the telescope of IOAA 2026, focal length 750 mm, Orion's belt does not fit on the picture. Alnitak and Mintaka are 2.74° apart on the sky, and their images land 35.9 mm apart in the focal plane. The same camera on a lens of focal length 50 mm covers about 26° along its sensor and takes in the whole of Orion. In this section we see how the focal length turns angles on the sky into millimetres on a sensor, how much sky each pixel and the whole sensor cover, and how long the shutter can stay open before the turning of the Earth smears each star into a short line.

5.2.1The plate scale

Before you readMake a guess first

Two stars are 1′ apart on the sky. How far apart are their images in the focal plane of a telescope of focal length 1 m?

Section S4 showed that two stars an angle $θ$ apart form images a distance $s = F θ$ apart in the focal plane, with $θ$ in radians. Astronomers usually measure small angles in arcseconds, and a radian is 206 265″, so an angle of 206 265″ on the sky becomes a distance $F$ in the focal plane. The angle on the sky that corresponds to a unit of distance in the focal plane is called the plate scale, from the days when the focal plane held a glass photographic plate.

ResultPlate scale(5.3)

\begin{aligned} plate scale & = \frac{206 265^{″}}{F} \\ p & = \frac{206 265^{″} \times ℓ}{F} \end{aligned}

$F$	the focal length of the telescope or lens
$p$	the angle on the sky covered by one pixel of a camera, called the pixel scale
$ℓ$	the width of one pixel, in the same unit as $F$

With

F

in millimetres the plate scale is in arcseconds per millimetre. Only the focal length counts, so a large telescope and a small one of the same focal length have the same plate scale.

Quick checkThe plate scale of the IOAA 2026 telescope

What is the plate scale, in arcseconds per millimetre, of a telescope of focal length 750 mm?

″ per mm

Example 5.2Stars on a camera

A camera with pixels 10 µm wide sits at the focus of a telescope of focal length 1.00 m. Two stars are 20″ apart. How many pixels apart are their images?

Solution

With $ℓ = 0.010$ mm and $F = 1000$ mm, (5.3) gives

p = \frac{206 265^{″} \times 0.010}{1000} = {2.06}^{″} per pixel,

so the two images are $20 / 2.06 = 9.7$ pixels apart.

AnswerThe images are about 10 pixels apart. This is part (4) of a problem from IOAA 2016, which Section 5.3 restates in full.

5.2.2The field of view of a camera

The field of view of a camera is the angle of sky that its sensor covers. A sensor of width $w$ at the focus covers an angle $w / F$ in radians, or $206 265^{″} \times w / F$ , as long as the angle is small. For a wide-angle lens the exact angle is $2 \arctan (w / 2 F)$ , which differs from $w / F$ by less than 1% for fields up to about 15°.

Example 5.3Orion's belt and the sensor

Will the whole of Orion's belt, 2.74° long, fit across a sensor 22.5 mm wide at the focus of the telescope of IOAA 2026, focal length 750 mm?

Solution

The sensor covers

\frac{22.5}{750} = 0.0300 rad = 1.72 °,

and the belt is 2.74° long, which would need $750 \times 2.74 \times π / 180 = 35.9$ mm.

AnswerThe belt does not fit. The sensor covers only 1.72°, and the belt would need 35.9 mm of the focal plane.

5.2.3Star trails

A camera on a fixed tripod stays still while the sky turns, so every star drifts across the sensor. A star on the celestial equator moves through 360° in one sidereal day of 23 h 56 min 4 s (Section 2.1), which is 15.04″ every second. A star at declination $δ$ moves on a smaller circle, of radius $\cos δ$ times that of the equator, so it moves only ${15.04}^{″} \cos δ$ each second. If a star moves by less than about one pixel during the exposure, its image stays a point.

ResultLongest exposure without trailing(5.4)

t_{max} = \frac{p}{{15.04}^{″} s^{- 1} \times \cos δ}

$t_{max}$	the longest exposure, in seconds, for which a star moves by less than one pixel
$p$	the pixel scale, in arcseconds per pixel
$δ$	the declination of the star

A telescope on a mount that turns with the sky, as Chapter O2 describes, can expose for minutes or hours instead.

Example 5.4A star near the polefinish it

With a pixel scale of 4.44″ per pixel, how long can a fixed camera expose before a star at declination 60° moves by one pixel?

The start

At declination 60° a star moves ${15.04}^{″} \cos 60 °$ each second.

Your stepNow find the longest exposure.

The olympiad problem on astrophotography from 2020, restated in Section 1.1, asks for exactly this exposure, for a camera with a 150 mm lens and pixels 3.23 µm apart.

Common errorsSection 5.2

The rule $s = F θ$ needs $θ$ in radians. With $θ$ in arcseconds it must be divided by 206 265 first.
The plate scale is an angle per unit length and the pixel scale an angle per pixel. They differ by the width of a pixel, which is usually given in micrometres and must be turned into millimetres.
The stars near the celestial pole move more slowly across the sky than those on the equator. Forgetting the factor $\cos δ$ gives an exposure that is too short for those stars.

ExercisesSection 5.24 questions

5.2.1
A camera with pixels 3.76 µm wide sits at the focus of the telescope of IOAA 2026, focal length 750 mm. What is its pixel scale?
″ per pixel
5.2.2
Jupiter is 45″ across. How many pixels does its image cover from side to side with a pixel scale of 1.03″ per pixel?
pixels
5.2.3
A camera lens of focal length 150 mm is used with a sensor 22.5 mm wide. What field of view does it give, in degrees?
degrees
5.2.4
A camera on a tripod photographs two stars with the same exposure, one on the celestial equator and one at declination 70°. Which leaves the longer trail?