How to choose the right focal length for your astrophotos

To perfectly frame objects in your field of view, you’ll need to know your telescope’s focal length — and a bit of math.

My first image of the Lagoon Nebula (M8) was effectively zoomed in so far that the object looked nothing like a lagoon. It was taken with a Nikon D3100 on my Celestron 8-inch SCT, with an exposure time of 30 seconds at ISO 3200. Credit: Molly Wakeling

On the first night out with my brand-new Celestron NexStar 8SE back in July 2015, I knew little about the relative sizes of astronomical objects. I was bowled over by my first view of Saturn — an impressive sight through an 8-inch Schmidt-Cassegrain telescope (SCT) — and I was eager to see more.

I knew the view through the scope would not be like images I had seen online, but when I slewed to the Lagoon Nebula (M8), all I could see were stars, with no hint of nebulosity. On my third night out, with a newly acquired T-adapter to connect my DSLR to the telescope, I took a 30-second exposure, and could finally see the colorful nebula! But it filled the entire frame and didn’t look anything like pictures I’d seen. “Why?” I wondered.

Focal length and field of view

One important number associated with every telescope is its focal length, or the distance the light travels after passing through the objective to your eyepiece or camera. This figure is commonly given in millimeters (mm). For example, my 8-inch SCT has a focal length of 2,032 mm. Imagine it like an enormous 2,000mm camera lens!

This number can also be expressed via focal ratio, or f-number. Unlike camera lenses, telescopes have no iris to reduce the effective aperture, so the focal ratio is simply the focal length divided by the aperture. For my 8-inch, whose aperture in metric units is 203 mm, the focal ratio is f/10. Another example is my 106mm f/5 refractor, whose focal length is 106 × 5 = 530 mm.

People at stargazing events often ask me what my magnification is. But when framing a particular object in your eyepiece or image, the more useful value is the field of view (FOV), or how large an area of sky can be seen with your telescope and a given camera or eyepiece.

The quick and dirty calculation for a camera is FOV = 3,436 × (sensor dimension) ÷ (focal length), where the sensor dimension is the length or width of your camera sensor in millimeters, and the focal length is also in millimeters. The result is the FOV in arcminutes; divide by 60 to get degrees. For a Nikon DSLR with an APS-C (crop) sensor, with dimensions of 23.6 mm by 15.7 mm, the FOV of my 8-inch SCT is then 39.9′ by 26.5′, or 0.67° by 0.44°. For an eyepiece, FOV = (eyepiece apparent FOV) × (eyepiece focal length) ÷ (telescope focal length).

There are also online calculators; my favorite is https://astronomy.tools/calculators/field_of_view.

Choosing your target

Now that you have calculated your FOV, you can find out which targets will be too big, too small, or fit just right. You can find the sizes of objects in apps like SkySafari or online; the Lagoon Nebula, for instance, is 90′ by 40′. With an FOV of 39.9′ by 26.5′, no wonder I couldn’t see much of it in my image — the FOV is far too small! It would be a much better target for my 530mm-focal-length refractor.

On the other hand, there are many galaxies and planetary nebulae that appear tiny in my refractor but are perfect for my Schmidt-Cassegrain; for example, the peculiar galaxy Arp 273 is a mere 1.4′ by 0.3′ in size, but shows up nicely in my Celestron 9.25 Edge HD with a focal length of 2,345 mm.

Now that you are more aware of the effects of different focal lengths, you can plan out images for your combination of telescope and camera or eyepiece. A great resource for finding targets of a certain size (among many other criteria) is the Imm Compendium, an impressive and detailed spreadsheet from astrophotographer Gary Imm, which can be found at https://tinyurl.com/imm-compendium.

Happy hunting!