Optics
From chi and h
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Perhaps the central piece of equipment for astrophotography is the "lens". I put this in quotes, because the term gives an overly simple impression of what is involved. A simple photographic lens is made of several separate pieces of transparent glass, in general with different refractive indices. Some telephoto "lenses" use mirrors instead of transparent glass. Astronomical telescopes also come in these two varieties. First, there are refractors with an objective at the front, made from several pieces of glass. Second, there are reflectors with a large primary mirror at the back of the telescope and a small secondary mirror near the front.
Whether you use the lens fixed to a compact digital camera or replace the lens of a dSLR with a telescope of either kind, the optics can always be characterised by two quantities, the aperture and the focal length. A third quantity, the f ratio, is simply the ratio of the two principal quantities.
There are two other ways of combining a camera with a telescope. Both involve the use of an eyepiece in the telescope. You use afocal projection when you fix the camera including its own lens behind the telescope eyepiece. You use eyepiece projection when you remove the camera lens, but keep the eyepiece.
Aperture
The aperture of a photo lens or telescope is simply the diameter of the front lens or primary mirror. (In photography, the term "aperture" is often used as another term for the f ratio.) It is usually assumed that the aperture is clear, even though in a mirror assembly the primary is partially obstructed by the secondary. The aperture is important for two properties of the optics.
First, if you double the aperture you quadruple the light-collecting area. A larger aperture collects more light from the object, giving brighter images, detecting fainter objects, permitting shorter exposures, permitting larger image scales, etc.
Second, if you double the aperture you halve the separation of objects that you can distinguish in the image. Image resolution is limited by quantum mechanics; in optics this effect is called diffraction.
Diffraction
One of the principles of quantum mechanics is Heisenberg's uncertainty principle. It states that one cannot simultaneously measure with arbitrarily high precision both the location and movement of a particle. In our case, the particle is a photon of light. We do not set out to measure the position of the photon, only its direction, i.e. where on the sky is the object that sent us the photon. However, we do restrict the photon's position as being within the aperture. This restricts the precision with which we can measure its direction of movement. The larger the aperture the smaller the spot on the sky to which we can pin down the source.
The number used to quantify the diffraction of a circular aperture is [1]
- Δα = 1.22 λ / D
where λ is the wavelength observed and D the aperture. The wavelength is usually set to that of green light, 550 nm. We should rewrite this so that you can insert your aperture in mm and read out the resolution in arc second:
- Δα/" = 140 mm / D
Hence, a 140 mm aperture can resolve two equally bright stars 1" apart. 70 mm can resolve such a double star only if the stars are 2" apart. The following table gives some examples for optics I use:
Diffraction-limited resolution. D Δα optics 1 mm 110.0" f = 4.1 mm, f/3.3, Panasonic Lumix DMC-TZ8 set to wide angle 5 mm 27.0" f = 18 mm, f/3.5, dSLR wide angle lens 10 mm 14.0" f = 55 mm, f/5.6, dSLR tele lens 18 mm 7.8" f = 50 mm, f/2.8, standard lens for 35 mm film format 34 mm 4.0" f = 135 mm, f/4, SLR tele lens 10 mm 14.0" f = 49 mm, f/4.9, Panasonic Lumix DMC-TZ8 using 12x zoom 63 mm 2.2" f = 400 mm, f/6.3, SLR tele lens 200 mm 0.7" 200 mm, f/10, Schmidt-Cassegrain telescope
The rows in this table are sorted not by aperture, but by likely size of the field of view, i.e. by what sort of a picture one is trying to take. Observe how the smaller build of the compact camera - here a Panasonic travel zoom - compared to the dSLR causes degraded image resolution. The compact is about four times smaller, and the resolution element due to diffraction about four times larger. This may not matter; it depends on the pixel resolution we will try to use with the compact camera.
Focal length
For a thin lens, the focal length is the distance from the lens to the focal plane, where light from objects infinitely far away is concentrated to form an image. An example of a thin lens is a refracting telescope with a few pieces of glass at the front and a long empty tube behind. The tube ends a little before the focal plane where the image of the stars is formed. Similarly, the tube of a Newtonian reflector is about as long as the focal length of the telescope. However, a number of modern designs - e.g. zoom photo lenses, Schmidt-Cassegrain and Maksutov telescopes - are geometrically shorter than their focal length.
Image scale
The focal length is what determines the scale of the image in the focal plane. Consider an angular size or separation α on the sky; say the apparent diameter of a planet or the separation of two components of a double star. A lens or telescope of focal length f projects this to a linear size d in the focal plane given by
- d = 2 f tan(α/2)
For small angles, we can approximate the tangent function and solve for the image scale d/α. Furthermore, to use degrees, arc minutes, or arc seconds for the angle, mm for the focal length, and micron (μm) for the image size:
- d/α / (μm/°) = 17.5 f/mm
- d/α / (μm/') = 0.3 f/mm
- d/α / (μm/") = 0.005 f/mm
The table gives examples for optics I use.
Image scale. f d/α optics 4 mm 72 μm/° f = 4.1 mm, f/3.3, Panasonic Lumix DMC-TZ8 set to wide angle 18 mm 314 μm/° f = 18 mm, f/3.5, dSLR wide angle lens 50 mm 15 μm/' f = 50 mm, f/2.8, standard lens for 35 mm film format 55 mm 16 μm/' f = 55 mm, f/5.6, dSLR tele lens 135 mm 39 μm/' f = 135 mm, f/4, SLR tele lens 49 mm 15 μm/' f = 49 mm, f/4.9, Panasonic Lumix DMC-TZ8 using 12x zoom 400 mm 115 μm/' f = 400 mm, f/6.3, SLR tele lens 800 mm 230 μm/' same lens with 2x adapter 2000 mm 9.7 μm/" 200 mm, f/10, Schmidt-Cassegrain telescope 3900 mm 20.0 μm/" Panasonic Lumix DMC-TZ8, 12x zoom, telescope at 80x magnification 6150 mm 30.0 μm/" telescope with eyepiece projection
f ratio
The f ratio is a confusing concept. It is the ratio between the aperture and the focal length. The term "aperture" is also used for this, but we should reserve that term for the linear diameter of the lens. The f ratio being a ratio, different people will probably calculate it differently, some use f/D, others D/f. Very often, another letter "f" is added to confuse us even further.
Saying "the lens is f/2.8" means that the lens has
- D = f/2.8
- f = 2.8 D
- f/D = 2.8
- D/f = 1/2.8
So far, that's fine. However, when you actually use the term "f ratio" then you might say "the f ratio is 2.8" or "the f ratio is 1/2.8". Best to be clear and say "the f ratio is f/D = 2.8".
Although the f ratio does not tell us anything that aperture and focal length can't tell us, it is a useful quantity. Consider this: A given extended object like the Moon has a certain surface brightness. For a given lens, this will result in a certain brightness in the image. Now you decide you want a larger image and change to a lens with three times longer focal length. If the longer lens has the same f ratio, then the image brightness will be the same.
Afocal projection
Afocal projection is a very intuitive - although perhaps not ideal - way of combining any camera with any telescope. You stick an eyepiece in the telescope and focus visually by looking through the telescope. You focus the camera with its lens for infinity, just as you relax your eye when focussing the telescope. Hold the camera behind the eyepiece to replace your eye, and you can take a picture.
How bright will the image be and how large will the object be in the image? To answer this you need to calculate the aperture and focal length of the combined optics, telescope and eyepiece and camera lens. Fortunately, that is simple. The aperture is that of the telescope. The focal length is that of the camera multiplied by the magnification of the telescope. A telescope's magnification is the focal length of the telescope divided by the focal length of the eyepiece.
- D = D1
- f = f3 f1 / f2
where the index 1 is for the telescope, 2 for the eyepiece, and 3 for the camera. It is best to bring the camera close in on the eyepiece. That will give you a larger illuminated area, though it will not change the image scale or image brightness.
Another thing should be considered in afokal projection. This is the telescope's exit pupil. If the telescope has aperture D1 and magnification f1/f2, then the bundle of light that exits the eyepiece has a diameter of
- D2 = D1 f2 / f1
i.e. the telescope's aperture divided by the magnification. In order that we do not lose any of the precious light in the camera, its aperture should be somewhat larger than the exit pupil, certainly not a lot smaller. You can of course calculate the camera's aperture from its focal length and f ratio.
Consider an example. With a 200 mm, f/10, Schmidt-Cassegrain telescope I have the choice between 80x and 225x magnification. The exit pupils are 2.5 mm and 0.9 mm, resp. The aperture of a compact travel zoom may range from about 1 mm on the wide angle setting to about 10 mm at 12x zoom. Using wide angle is not a good idea, because the aperture may be less than the exit pupil.
Eyepiece projection
Some of the objects we might want to image are very small. Therefore, you remove the lens from your dSLR or webcam and use a telescope instead. 1000 or 2000 mm focal length may be nice for the Moon, but for a planet, this is not enough. You can use an eyepiece to re-project the primary image created by the telescope onto the detector in the camera. A lens - here the eyepiece - will project at a scale 1:1 when the primary image is twice its focal length in front. The secondary image will then be twice the focal length behind. To accomplish magnification you move the primary image closer to the eyepiece and the camera further away.
If the distance to the camera is x then the combined focal length of telescope and eyepiece is (index 1 is for the telescope, 2 for the eyepiece):
- f = f1 [(x/f2) - 1]
- D = D1
For example, I find that to image planets with my webcam I need 6000 mm focal length, but my telescope is only 2000 mm. To get that threefold boost I need to put the detector x = 4 f2 behind the eyepiece. Using a 9 mm eyepiece, that is about 35 mm.
There are a few practical difficulties that make the maths here all but meaningless. In particular, an eyepiece is not a thin lens and it is hard to know from where to measure the focal length or the separation x. You will have to find the best arrangement by trial and error. It is good to know, at least, that a shorter eyepiece and a larger distance to the camera each increase the image scale.
You will normally use a high magnification eyepiece. With an eyepiece for moderate magnification (longer f2) the optical assembly becomes unmanageably long. With a similar value of x and longer eyepiece focal length, you will find that eyepiece projection reduces the focal length rather than increase it.
Reference
- R. Willingale (2000). Lecture notes - 2nd year optics. University of Leicester, Department of Physics and Astronomy. http://www.star.le.ac.uk/~rw/courses/lect220.ps
