{ Practical astronomy | Optics and imaging | Exposure }

Exposure

The question we address here is how long we should expose our image to record an object of certain brightness. The algorithm depends on whether the object is resolved or not, and also whether it is the optics (diffraction) or the pixel size that limits the resolution.

Point sources

Three comparisons of Airy disc radius Δα_A (blue curve) and pixel size Δα_P (pixelated red curve). Left is the case of a point source, but where the pixels resolve the Airy disc, centre is an unresolved point source, right is a resolved disc-shaped source.

Consider a star as the light source of interest. Its brightness is described by its magnitude V. The brightest stars are "first magnitude" and the faintest stars visible to the naked eye are "sixth magnitude". This magnitude scale goes back to Ptolemy (+137, 137 AD), but Hipparchus, around −130 (around 130 BC), used a similar scale in his star catalogue (Miles 2007). The system was put on a modern mathematical footing by Pogson (1856). It is convenient for us to convert this logarithmic measure to the linear measure introduced by radio astronomers during the 20th century. For a visual magnitude V the flux in Jansky is (UKIRT 1998):

S = 3540 · 10^−0.4·V Jy

The lens or telescope will not show a point source as a point of light, but as a small Airy disc. The flux S is spread out over an area of typical diameter Δα_A; the brightness distribution is similar to a two-dimensional Gauß curve. Behind the lens in the focal plane the surface brightness then is something like

B_I(α) = B₀ · exp[−0.5·(α/σ)²]

B₀ = S / (2 π σ²)
σ = Δα_A / [2·√(2·ln(2))]

Another question is how this is then pixelated by the detector. If the pixel size – a square with side Δα_P – exceeds the diameter Δα_A, then we have simply all the flux dropping into a single pixel; the measured surface brightness is the flux divided by the area of the pixel. But if Δα_A is larger, then the Airy disc is resolved by the pixels and we are interested in B₀ as the brightest pixel. Comparing the two cases, the brightest pixel is

B = ln(2) · S / [π · (Δα_A/2)²]           if      Δα_A > Δα_P
B = S / Δα_P²                                   if      Δα_A < Δα_P

The two variants are written down such that the denominator at the end makes sense as an "area", the effective area of the Airy disc or the area of the square pixel.

Extended sources

Now consider an extended source like a planet, the Sun or the Moon. We know the apparent radius α and the magnitude V, or the flux S, or the surface brightness B. We can convert magnitude to flux as above.

If the size of the source is less than the size of the Airy disc, then the object is not resolved and we treat it like a point source above. If the size of the source is larger than the Airy disc but less than the size of a pixel, then all the flux falls into a single pixel, just like above for point sources when the Airy disc is smaller than the pixel. In these cases, if given the surface brightness B, we need to convert to the flux S as B times source area, e.g. S = B π α²

If, finally, the source is really resolved both by the optics and the pixels, then the surface brightness seen by the pixels is simply the given surface brightness of the object. If necessary, we can calculate B as the flux divided by the area on the sky that the object occupies. Assuming a circular source of radius α

B = S / (π α²)                                 if      2 α > Δα_A      and      2 α > Δα_P

As before, this is written such that the denominator is an area, here that of the source itself.

Pixel brightness

Although the discussion above of point sources and extended sources makes sense, it muddles the question of how bright the pixels on our detector are. Here is how a software engineer might tackle that question.

• If the magnitude V and source radius α (possibly zero for a point source) are given, then convert V to flux S.
• Else if the surface brightness B and (non-zero) radius α are given, then convert B to flux S by multiplying with the source area.
  S = B (π α²)
• With S and α given or calculated:
  • If the source is unresolved by the optics (2α < Δα_A):
    • If the Airy disc is also unresolved by the pixels (Δα_A < Δα_P):
      The pixel brightness is the flux divided by the pixel area
      B = S / Δα_P²
    • Else (the Airy disc is resolved by the pixels):
      The pixel brightness is the maximum of the Gauß curve
      B = ln(2) · S / [π · (Δα_A/2)²]
  • Else (the source is resolved by the optics):
    • If the source is also resolved by the pixels (Δα_P < 2α):
      The pixel brightness is the flux divided by the source area
      B = S / (π α²)
      (If the source B was given to begin with and then converted to S, this is the same value B.)
    • Else (the source is not resolved by the pixels):
      The pixel brightness is the flux divided by the pixel area
      B = S / Δα_P²

Saturation exposure

The exposure time at which an object of given surface brightness saturates the detector pixels depends only on the filter transmission, f ratio and ISO setting. In a simple experiment we can determine the coefficient. To do this image an object of known surface brightness, say the Full Moon, such that it just about saturates. The result should be something like

t = 300,000 (s MJy/sr) · (f/D)² / (ISO · B · T)

As an example, take the Full Moon in an f/13.3 telescope with ISO set to 800; assume the camera has a light pollution filter, which takes out roughly half the light:

V = −12.6
S = 388,000,000 Jy
α = 900" = 0.0043 rad
B = 6,680,000 MJy/sr

t = 300,000 s · 13.3² / (800 · 6,680,000 · 0.5) = 0.0198 s = (1/50) s

As another example, use the same telescope and camera, add an ND5 solar filter at the front, and reduce the ISO to 200

V = −26.7
S = 170,000,000,000,000 Jy
α = 900" = 0.0043 rad
B = 2,900,000,000,000 MJy/sr

t = 300,000 s · 13.3² / (200 · 2,900,000,000,000 · 0.5 · 0.00001)
   = 300,000 s · 13.3² / (200 · 29,000,000 · 0.5)
   = 0.018 s = (1/50) s

The result is an estimate how long an exposure is a little too long (when it saturates the detector). Try half the calculated exposure, check the result and then adjust further. The result here is only a ballpark figure. Much depends on the altitude above the horizon, how clear or murky the atmosphere is on the day, etc.

Brightness of various objects