I'm a computer engineering student working on my senior project that involves some optics. Illadvised though it may be, I'm contemplating building a telescope based on a simple achromatic refractor design for use in an automated astrophotography rig. I'm trying to work out how to build a compact telescope such as commercial manufacturers create, and still get an image that fills the sensor of a digital camera. The entire problem can be distilled to a simple example:

Suppose we're trying to get an image of the Orion Nebula, at a distance of 1.334e3 light years away. If we want to fill the entire sensor, what should our focal length be if the distance from the straightening lens to the sensor is 44mm?

From my EM course I recall that $\frac{1}{F} = \frac{1}{D} + \frac{1}{I}$ where $F$ = Focal Length, $D$ = Object Distance, and $I$ = Image Distance.

For large distances I can rewrite this as $\frac{1}{F} = lim_{D\longrightarrow{\infty}} \frac{1}{D} + \frac{1}{I} \longrightarrow \frac{1}{F} = \frac{1}{I}$

This doesn't fit with my intuition. I'm an amature photographer, and I don't know much about focal length except that if I want to take an image of something at a great distance I use a lens with focal lengths in the 100s of mm, and that's images of things within a mile or less from view, not light years. What I'm missing in all of this

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    $\begingroup$ I learned photography before auto focus was a common feature in cameras, and one of the things we learned back then was that if the subject is more than about 15 feet away, you can just focus at "infinity" (denoted $\infty$ on the focus ring) and you'll get a clear image. $\endgroup$
    – The Photon
    Mar 23, 2022 at 3:51
  • $\begingroup$ A telescope generates magnification but its output is still a (virtual) object at infinity. You thus want to place your camera entrance aperture at the eye-relief point of the 'scope and focus the camera at infinity. $\endgroup$ Mar 23, 2022 at 12:38

2 Answers 2


The question lacks some information. Specifically, the angular size of the Orion nebula, and the linear diameter of the camera sensor.

The Orion nebula is about $1$ arc degree in size (Google). This means that the angular size of the image will also be about $1$ arc degree in size, as seen from the lens.

If the sensor is $44$ mm from the lens, then the focal length of the lens must be the same $44$ mm. An object at infinity will always form a real image at the focal distance.

Combining these two parameters, the size of a $1^o$ image at $44$ mm is $S$:$$S=44 \times 1 \times \frac{\pi}{ 180}=0.768 \text{ mm}$$ where the $\frac{\pi}{180} $ changes degrees to radians.

So, the linear size of the object is irrelevant (as long as it is essentially an infinite distance away). Angular size of the object, focal length of the lens and sensor size are all related: pick two, and the third is determined

  • $\begingroup$ Thank you this makes it very clear to me. $\endgroup$
    – richbai90
    Mar 23, 2022 at 18:41

The focal length in this context determines the size of image. Therefore, when you photograph an object at a large distance you would select a larger focal length so that the object appears larger in the image.

There is always a range of distances in which objects would appear in focus. This range becomes larger for larger object distances. Therefore, it not necessary to match the object distance to the exact position of the object.

When the object is far enough away the image is effectively formed in the focal plane, which is one focal length from the lens. Thanks to the range over which objects are in focus, it does not matter if the object is not at infinity. So what is true for an object a mile away is also true for an object lightyears away.


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