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I am trying to decide what lens should I use for eclipse and I am wondering if my 70-300 IS USM would be enough or should I get (rent) a lens with higher magnification. Since my solar filters are going to be only in a week, I was wondering, how can I determine how big the sun would be if I shoot at 300mm? What about at other focal lengths?

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  • \$\begingroup\$ check this out learn.usa.canon.com/resources/articles/2017/solar-eclipse/… \$\endgroup\$ Aug 3, 2017 at 2:16
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    \$\begingroup\$ I took the liberty of slightly changing your question from a request for people to post images of the sun @ 300mm, and made it a more generic question about determining the size. As originally written, a request for images wouldn't yield as good answers as a request to explain how to determine the size. The original form resembled a question on typical internet discussion forums, as opposed to the Stack Exchange Q&A format. I hope I didn't substantially change the meaning of the question in your eyes; if so, you can of course revert my edit. \$\endgroup\$
    – scottbb
    Aug 3, 2017 at 2:27
  • \$\begingroup\$ Since the Sun and moon are almost exactly the same size as seen from the surface of the Earth: Please see here \$\endgroup\$
    – Michael C
    Aug 3, 2017 at 7:06
  • \$\begingroup\$ Related: What focal length lens do I need for photographing the moon? \$\endgroup\$
    – Michael C
    Aug 3, 2017 at 7:06
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    \$\begingroup\$ Possible duplicate of Image size of the Sun / Moon from a telephoto lens \$\endgroup\$
    – scottbb
    Aug 3, 2017 at 12:51

3 Answers 3

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As Alan Marcus mentions, the sun (as well as the moon) are very close to the same size in the sky, at 0.5° of arc.

Instead of computing angle of view, we can use a simpler rule of thumb. Using the pinhole projection formula (which non-wide angle lenses follow), an object of extent (e.g., width) \$w\$ at a distance \$d_o\$ from the lens is projected onto a the film or image sensor with a size of

$$ s = \frac{d_i\cdot w}{d_o} $$

For the sun, this becomes \$s = f\cdot\frac{{\mathrm{1.4\,\times10^6 \,km}}}{\mathrm{1\,AU}} \approx f\,/\,108 \$

This handy rule says that the actual size of the sun on your film or sensor is the lens's focal length divided by 108. This is regardless of the size of the camera format. Don't convert lenses to "35 mm equivalent" or worry about crop factor, etc.

Example: a typical Canon APS-C sensor might be 22.2 mm wide by 14.8 mm tall. Using the short dimension of the sensor (the height), d = 14.8 mm. A Nikon or Canon full-frame sensor is 36 mm wide by 24 mm tall. Therefore, in the table below, for a given focal length \$f\$, the sensor size in sun diameters is given:

focal length \$s\$ (size of sun on sensor, mm) APS-C height, in sun diameters Full-frame height, in sun diameters
300 mm 2.78 mm 5.3 sun diameters 8.6 sun diameters
400 mm 3.70 mm 4.0 sun diameters 6.5 sun diameters
500 mm 4.63 mm 3.2 sun diameters 5.2 sun diameters
600 mm 5.56 mm 2.7 sun diameters 4.3 sun diameters

With this approach, you can determine exactly how big the sun will be in your photo, depending on what focal length you use.

A very good resource is Fred Espenak's Mr. Eclipse pages, specifically How to Photograph a Solar Eclipse. That page specifically covers the size of the sun for different focal lengths, both on full frame and crop sensor cameras, as well as suggestions for focal lengths depending on what part(s) of the eclipse you are interested in shooting (solar disk, prominences, coronae, etc.).

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    \$\begingroup\$ Technically, Canon APS-C sensors vary slightly in size from 22.2x14.7mm (1100D/Rebel T3) to 22.7x15.1mm (10D and 300D/Digital Rebel). \$\endgroup\$
    – Michael C
    Aug 3, 2017 at 6:57
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With your camera and existing 70-300 lens, shoot pictures of the moon. The moon and the Sun appear in the sky as a disk, they image exactly the same as to size. Both are 1/2 degree of arc disks.

Addendum: Figuring out the size of the Sun's image is easy math. You can trace out a triangle from the edges of the Sun to the camera lens. The image forming rays from the lens trace out a similar triangle inside the camera. Ratio to the rescue: Sun's diameter is 864,576 miles. Sun’s mean distance from Earth is 92,955,807 miles. The ratio diameter to distance is 92955807 ÷ 864576 = 0.00934.

The image forming rays of the camera lens trace out a triangle with this exact ratio. Thus if a 300mm lens is mounted, we multiply 300 X 0.00934 = 2.8. That tells us that if a 300mm lens is used, the image of the Sun at film or digital sensor will be a circle 2.8mm in diameter.

We view the images of our miniature cameras by making an enlargement, be it print on paper or image on a computer monitor. To make an 8X10 print from a full frame (FX), the magnification requires a minimum of 8.5X. If this is the magnification, the final image of the Sun will be 2.8 X 8.5 = 23.8mm (about 1 inch). If the camera is a compact (DX or APS-C), the magnification will be about 1.5X more - thus 23.8 X 1.5 = 35.7mm (Sun’s image diameter or about 1.4 inches).

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I've found the following sentence in a 1999 NASA bulletin:

For any particular focal length, the diameter of the Sun's image is approximately equal to the focal length divided by 109.

Using this equation it gives 300/109 which is approximately 2.75mm.

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  • \$\begingroup\$ To solve we multiply or divide the focal length by a calculated factor. Dividing by 109 solves as does multiply by 1/109 = 0.0092 the reciprocal of 109. \$\endgroup\$ Aug 3, 2017 at 15:31
  • \$\begingroup\$ Sure. I usually divide by 100 (which is quite easy) and round down. \$\endgroup\$
    – asalamon74
    Aug 4, 2017 at 7:05

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