For a discussion of digiscoping technique, see the Technique Primer.
For an Excel spreadsheet to calculate optical parameters for any digiscoping setup, click here.
A lens is a material that transmits light, with a special property: that an image of a distant object is formed a certain distance behind the lens. That distance is the focal length. The focal length is the primary characteristic of a lens. Several lenses can be combined to form a compound lens with effective focal length f. The size, or field of view, of the image is proportional to the focal length. The other important parameter is the diameter, or aperture , of the lens. The brightness of the image depends on the aperture (as the square of the radius).
A refracting telescope is comprised of two effective lenses, characterized by their focal lengths. The optical performance of the telescope is defined by the focal length of the objective (front) lens, the aperture of the objective, and the focal length of the eyepiece assembly. The ratio of objective to eyepiece focal length is the magnification. The aperture divided by magnification equals the exit pupil. In general, telescopes with small objective focal lengths have larger fields of view, and telescopes with small magnification have large exit pupils. Both qualities are desirable.
For a given magnification, a large aperture scope is brighter than a small aperture scope. Image brightness increases as the square of the objective radius (aperture/2). For a given lens aperture, a scope is brighter at lower magnification. Image brightness is inversely proportional to magnification.
A commonly discussed photographic parameter is the aperture setting or f/#. For a single (simple or compound) lens, f/# is the ratio of effective focal length to effective diameter. For a telescope, f/# is not defined, because for a telescope the focal length is not defined, i.e., the telescope is afocal. Remember, a telescope requires both an objective and an eyepiece. If a telescope is described as having a given focal length (and f/#), that is the focal length of the objective only. It's not a telescope without the eyepiece, and once you add the eyepiece, the focal length becomes undefined.
Camera lenses can be defined by a focal length (a zoom lens is a lens of variable focal length), so f/# is defined. The camera's field of view decreases with increasing focal length (tighter zoom). Camera lenses usually have variable apertures defined by the diameter of the shutter; wide apertures correspond to small values of f/# for a given focal length. For a given focal length, images get brighter with smaller f/#. Keep in mind the difference between aperture, a diameter, and aperture setting, f/#, a ratio (of focal length to aperture). Camera lenses are not telescopes! Among other things, magnification is not defined for a camera lens--you need a telescope for that.
A camera is also defined by the size of the image frame. One common frame size is 35mm format, (actually, 36x24 mm). CCD detectors used in common digital cameras are smaller than 35 mm film frames (a common format is 7.1 x 5.3 mm); therefore, one can produce a full-frame digital image using a lens of shorter focal length. Digital cameras usually have lenses whose focal lengths range between about 8 and 25 mm. In 35 mm format the same fields of view require focal lenghts of approximately 35 to 120 mm.
In a camera, as focal length increases, field of view decreases. Likewise, in a telescope, as magnification increases field of view decreases. But since the size and brightness of the object is unchanged, and the area of the detector is fixed, the image must also darken; the image is being spread over a larger area.
When digiscoping, the telescope passes a magnified field of view to the camera that records the image. Camera effective focal length is now the camera lens focal length times magnification. In "native" camera units, this implies a focal length of 749 mm for Leica Televid at 32x and the Coolpix 990 at full zoom (f=23.4x32). The equivalent lens in 35 mm format would be 3640 mm.
The effective aperture of the camera can be no larger than the exit pupil of the telescope. This quantity is determined by the aperture and magnification of the scope. If the exit pupil is very small compared to the aperture of the shutter, then exposure brightness and depth of field is determined by the telescope, not the camera. However, if the exit pupil is large compared to the range of shutter diameters, the camera controls do indeed influence exposure and depth of field.
The camera shutter diameter is equal to the camera focal length (f) divided by the camera aperture setting (f/#). The exit pupil is equal to the scope aperture (A) divided by magnification. For the Leica Televid 77 at 32x, exit pupil is 2.4 mm. For the Nikon Coolpix 990 at full zoom, f=23.4 mm. The aperture setting corresponding to the 32x exit pupil is f/9.75 (23.4/2.4).
Aperture settings below f/9.75 have no influence on the exposure at full camera zoom, as the shutter diameter is smaller than the scope's exit pupil. (But for ease of alignment set the shutter wide.) However, at full wide, where f=8 mm, the equivalent f/# is 3.3. Aperture settings above f/3.3 do influence the exposure. For the Leica at 20x the exit pupil is 3.85 mm. At full zoom (f=23.4 mm), aperture settings above f/6.1 influence the exposure; at full wide (f=8 mm), settings above f/2.1 do.
Depth of field depends on a parameter known as the blur spot: the maximum acceptable imaged size of a point object. In film photography the blur spot size is often taken as the emulsion grain size. For digital photography, we may use the size of a pixel, 0.00345 mm for the Nikon Coolpix 990 (and many other 3.3 Mpix cameras). We divide this quantity by the camera focal length to obtain the angular blur spot (strictly valid only when the camera focus is set to infinite distance). This quantity is independent of the distance between object and image plane.
Using the angular blur spot, depth of field is a function of distance to object, focal length, and most conveniently, effective aperture diameter. Depth of field will be no smaller than that imposed by using the exit pupil as the limiting aperture. For the case where the depth of field is small compared to distance to subject, the depth of field is twice the square of the distance to subject times the angular blur spot divided by the scope aperture. For the Leica at 32x with subject 25 m distant, the depth of field is about 154 mm.
Resolution is the minimum distance between two separable points on the object; smaller values are better. In a perfect telescope resolution is limited by the effective aperture and the wave nature of light (i.e., diffraction effects). At the camera, effective aperture is the exit pupil, since this is the largest effective aperture available when digiscoping. In object space this number must be multiplied by the magnification, i.e., the scope aperture determines the digiscoped resolution. (Camera resolution, equal to the pixel angular size, is about 10x better than the scope's.) The Rayleigh criterion for resolution, in object degrees and for green light, is the factor 0.039 divided by aperture in mm. For the Leica 77, the resolution is about 0.0005 degrees, the approximate angular width of a human hair from a distance of 12 meters. Increasing magnification cannot improve resolution because the exit pupil will shrink correspondingly.
All of the above requires that the camera be able to see the full field of view available in the scope. The eye relief of a telescope is the maximum distance at which a subsequent optical element can see the full field of view. In other words, the next optical element (whether eye or camera) must be within the eye relief of the scope if vignetting is to be avoided. Eye relief is generally larger for lower magnification eyepieces, and is generally larger for fixed-magnification eyepieces than for zoom eyepieces.
When digiscoping, the trick is to couple the camera and scope such that the camera input aperture is within the eye relief. This generally means minimizing the separation.
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