# Image size of the Sun / Moon from a telephoto lens

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?

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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. – AJ Henderson Sep 27 '13 at 14:39
Yes, I do have a 9-stop ND filter that I will use. – Vivek Sep 27 '13 at 14:44

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!

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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. – jwenting Sep 28 '13 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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