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Pixel size and field of view

In images observed close to the optical axis of a well-designed telescope an angular displacement on the sky is simply proportional to a linear displacement in position in the focal plane. The constant of proportionality is usually called the plate scale (a name which betrays its origin in photographic techniques) and is traditionally quoted in units of seconds of arc / mm. That is:


\begin{displaymath}
p = \Delta '' / \Delta {\rm mm}
\end{displaymath} (1)

where $p$ is the plate scale in seconds of arc / mm, $\Delta ''$ is a displacement on the sky in seconds of arc and $\Delta {\rm mm}$ is the corresponding displacement in the focal plane in mm. If you know the plate scale and the size of either a single pixel in the grid or the linear size of the CCD then it is trivial to use Equation [*] to work out either the angle on the sky subtended by a single pixel or the field of view of the CCD respectively. For example, the sample data used in Part II of the cookbook were obtained with the Jacobus Kapteyn Telescope (JKT) on La Palma. The CCD detector used has pixels which are

24x24 micron in size. The plate scale of the JKT is 13.8 seconds of arc / mm. Thus, each pixel subtends an angle of

0.331x0.331 seconds of arc on the sky.

The manual for the instrument or telescope that you are using will usually quote a value for the plate scale. However, if necessary it can be calculated from other parameters for the telescope. By simple geometry the plate scale is the reciprocal of the effective focal length of the system:


\begin{displaymath}
p^{\prime} = 1 / f
\end{displaymath} (2)

where $f$ is the effective focal length of the system and $p^{\prime}$ is the plate scale in units of `radians / whatever units $f$ is in'. Thus, for $f$ in metres and applying the factor for converting radians to seconds of arc:


\begin{displaymath}
p = 206.26 / f
\end{displaymath} (3)

$f$ is itself related to the diameter of the primary mirror, $D$, and the focal ratio, $F$:


\begin{displaymath}
f = F . D
\end{displaymath} (4)

At larger distances from the optical axis there is no longer a simple linear relation between angular displacement on the sky and displacement in position in the focal surface. That is, $p$ varies as a function of position in the focal surface. This effect is usually not important in instruments containing a single chip because of the small size of individual CCDs. However it may be important if a grid of chips is used.



next up previous 63
Next: Instrumental Effects in CCD Detectors
Up: Overview of CCD Detectors
Previous: Advantages and disadvantages of CCDs

The 2-D CCD Data Reduction Cookbook
Starlink Cookbook 5
A.C. Davenhall, G.J. Privett & M.B. Taylor
16th August 2001
E-mail:starlink@jiscmail.ac.uk

Copyright © 2001 Council for the Central Laboratory of the Research Councils