I'm interested in taking a picture of the Sun / Moon using a telephoto lens. For example like this.

It would be very helpful for me to know what the approximate image size of the sun / moon would look like at xxx mm in FF / Crop sensor.

Is there a formula or a similar image that can give me an idea of it?

  • 4
    \$\begingroup\$ Just a safety note, be sure to use a VERY powerful ND filter if you want to shoot the sun directly or you will damage your camera. Additionally, as John Kemp pointed out, do not use NDs if you are looking through it optically as the IR is still damaging. A solar filter is the correct way to go. \$\endgroup\$
    – AJ Henderson
    Sep 27, 2013 at 14:39
  • \$\begingroup\$ Yes, I do have a 9-stop ND filter that I will use. \$\endgroup\$
    – Viv
    Sep 27, 2013 at 14:44
  • 1
    \$\begingroup\$ Buy solar filters from an astronomy dealer as many camera shops are out of their depth and expertise. They must be properly certified (e.G. Ce mark), looked after, and used according to the instructions. Many websites give good advice - e.G canon learn.usa.canon.com/resources/articles/2017/solar-eclipse/… \$\endgroup\$
    – John Kemp
    Sep 2, 2018 at 18:31

2 Answers 2


The size of the sun or moon in mm in the sensor plane will be approximately

f / 110

where f is your focal length. A typical APS-C sensor is 16mm tall (or 15mm for Canon), hence a 1760mm lens would be required to fill the frame (vertically). 800mm would get you about half the frame, 400mm one quarter etc.

A "full frame" sensor is 24mm tall, so you'd need 2640mm to fill the frame, better get stacking those TCs!

  • 2
    \$\begingroup\$ to illustrate, here's a shot I just did of the moon using a telescope with a focal length of 1250mm and a Nikon D200 (not perfectly sharp, it's very hard to accurately focus the moon in the camera viewfinder due to the glare). hornet.demon.nl/photos/DSC_3148.png Cropped by about 10% around the edges to center the image. That's 1250mm at 1/50 of a second, ISO 800. \$\endgroup\$
    – jwenting
    Sep 28, 2013 at 3:05

My answer will be the same as @MattGrum's, but I would like to add a teeny bit of explanation.

Say the diameter of the sun/moon is h, the distance is d and your lens' focal length is f. Assume that the image has come to a focus at a distance l behind the lens and has a diameter of i on the sensor.

If you quickly sketch out things assuming a thin lens you will find (by equal triangles) that

h / d = i / l

For large d the lens basically focuses at it's focal distance f so we can rewrite things as:

h / d = i / f

Now the tangent of the angle subtended by the object is given by

tan(theta) = h / d

When d >> h we can actually write

theta = h / d when theta is expressed in radians.

So we can say

i = f x theta

Now, from observation we see that theta = 0.5 degrees for the objects in question, which after proper conversion to radians leaves us with

i = f x 0.0087

i = f / 115


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