I know it's a beginner question, but I spent the last 3 days reading this forum and other resources on the Internet and I reached a breaking point.

So decided to ask the question after all, as I couldn't figure the answer based on the other posts.


I have a compact camera with a tiny sensor (crop factor 5.6) and max focal length of 1365 mm. That camera allows me to get satisfactory close to my subjects (wildlife). I want to upgrade to a camera with interchangeable lenses that will have a crop factor of 1.6 (among other advantages over a compact super-zoom)

Am I correct that I will need a lens with a focal length of 390 mm, to get as close to my subject as before?

How I got to this conclusion was (1365 / 5.6) * 1.6 = 390

Is this correct or am I oversimplifying and confusing things? Is there any other parameter that can tell me "how close" I can get to my subject?

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    \$\begingroup\$ Consider micro-4/3. They have some very nice zooms, with a crop factor of 2, plus teleconverters; with a moderate weight and bulk. And they have some crazy image stabilisation, which is very good for that long lenses. \$\endgroup\$
    – Davidmh
    Commented Aug 9, 2019 at 14:20
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    \$\begingroup\$ Is the value 1365 an actual focal length or "full frame equivalent"? \$\endgroup\$
    – Agent_L
    Commented Aug 9, 2019 at 14:49
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    \$\begingroup\$ @Agent_L yes, 35mm-equivalent, this is the thing I was missing. So my "calculations" were all wrong. \$\endgroup\$
    – Tommy
    Commented Aug 9, 2019 at 19:23
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    \$\begingroup\$ @Agent_L If it's the actual focal length it wouldn't be a "point-and-shoot", it would be an unwieldy behemoth of a telescope! \$\endgroup\$
    – J...
    Commented Aug 10, 2019 at 10:55

6 Answers 6


In photography, what is interesting is mostly the angle of view (AOV). The AOV is the angle that a lens offers on a sensor - it can be specified horizontally, diagonally, or vertically.

AOV [°] = 2 * arctan ( sensor_height|width|diagonale [mm] / (2 * focal_length [mm]) )

The formula to get from a specified focal length (FL) on a non-full-frame sensor to the full-frame-equivalent focal length is:

equivalent_FL [mm] = true_FL [mm] * crop_factor

The crop factor can be determined by comparing the diagonals:

crop_factor = full_frame_diag [mm] / your_sensor_diag [mm]

This means:

  • With the same focal length, a larger sensor (but same aspect ratio) will give a greater AOV
  • With the same sensor dimensions, a smaller focal length will give a greater AOV
  • AOV is different in vertical, horizontal, and diagonal (except in a quadratic sensor, where vert&hor would be the same) axes

Or, in practical terms:

  • A 10mm lens on your 5.6-crop-factor sensor will give you an AOV that is equivalent to that of a 56mm lens on a full frame sensor.
  • The same 10mm lens on a 1.6-crop-factor sensor will give you an AOV that is equivalent to that of a 16mm lens on a full frame sensor.
  • A 1600mm lens on a full frame sensor will give the same focal length as a 1000mm lens on APS-C (1.6 crop) or a ~ 285mm lens on your point and shoot.
  • A 16mm lens on a full frame sensor will give the same focal length as a 10mm lens on APS-C or a ~ 2.85mm lens on your point and shoot.
  • All other factors left aside, smaller sensors favor smaller AOVs / higher reach, while larger sensors favor wider AOVs.
    • Among the ignored factors are:
    • Pixel density (a 20mm² sensor with 20MP has half as large pixels than a 40mm² sensor with 20MP) which influences noise (smaller pixels typically are worse at collecting light and thus contain more noise)
    • Aperture (f/4 on a 5.6-crop-factor is something like f/24 on full frame)
    • Physical limitations (e.g. negative-valued focal lengths (-1mm) are not possible)

Why do we then use focal lengths (in mm) on lenses? Because AOV is not a function of the lens, but of the sensor-lens combination. A lens will keep its focal length forever, but based on the sensor it is mounted on, its AOV will vary. (Of course, the image circle that a lens can provide will limit its abilities at some point, so mounting a smartphone 3mm lens on a medium format sensor wouldn't do much good ;-) )

Oh, and why compare it to full frame? Because we needed some metric to compare it to - we could also use IMAX or Super35 or 1 / (⅔ * π) [inches] if we want to.

Now to actually answering the question:

Your formula was:

(1365 / 5.6) * 1.6 = 390

Which would mean:

effective_FL / crop_factor_PnS = real_FL_PnS
real_FL_PnS * crop_factor_APS-C = ??

What you calculate is therefore the effective focal length of the lens of your point and shoot camera on the sensor of your new camera.

Your 1365mm are already full-frame equivalent, so you can calculate the APS-C related true focal length with this value already.

This means:

1365 / 1.6 = 853.125 [mm]

So you would need a lens with that focal length to get the same narrow AOV with a 1.6-crop-factor sensor.

Note that the difference in AOV between 100-200mm is bigger than that between 500-600mm!

Note that - as twalberg already stated - 400mm+ lenses are usually very expensive and mostly limited to primes (and/or the use of tele-converters, which might deactivate your camera's AF if your lens is not fast enough). This is because they are typically a niche market built for professionals who need/want every last bit of image quality, and most 15000€ lenses on 5000€ bodies offer better image quality in the worst of circumstances than any 500€ camera ever can. Does that mean it will make you a better photographer or that you need that setup? No!

I have no stake in this, but if you want a modular system with that kind of reach, I think that µ4/3 might be a better choice if you are on a budget - it offers a 2x crop and 100-400mm lenses are not quite as costly as an 800mm prime from Canon ;-)


Assuming that the quoted 1365mm focal length is in 35mm full frame equivalent terms (because otherwise, it would be huge), then the actual focal length of the lens assembly is around 1365 / 5.6 ~= 244mm. To accomplish an equivalent 1365mm focal length with a 1.6 crop factor, you would need about an 854mm actual focal length lens. I'm not aware of anything quite that long that isn't prohibitively expensive (Canon's EF 800mm f/5.6L IS USM is only $13000 - but maybe money's not an issue for you), but you might be able to afford a 400mm lens with a 2X tele-extender, which would get you to 800mm actual focal length, or about 1280mm equivalent.

  • 2
    \$\begingroup\$ A note of caution re: 400mm + 2x tele: most entry-level to mid-level consumer camera bodies can't autofocus a lens with a maximum aperture smaller than ƒ/5.6 or ƒ/6.3 or so. It's only the higher end prosumer and professional cameras that can autofocus with max ƒ/8 lenses. So that means that in order to be able to autofocus with the 2x teleconverter in place, the 400mm lens needs have a max aperture no smaller than ƒ/2.8. \$\endgroup\$
    – scottbb
    Commented Aug 8, 2019 at 13:37
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    \$\begingroup\$ @Tommy Yes, a lot of point 'n shoots (and even bridge cameras) have superzoom lenses that far out-reach entry and prosumer model DSLRs/mirrorless. However, there's more to e.g. wildlife photography than just reach. In particular, image quality with a superzoom will never be as good as with a good prime lens (or even most standard non-super zooms - those restricted to around a 4X range). As long as the quality of the photos you're capturing is satisfactory to you, then upgrading might not really be necessary. \$\endgroup\$
    – twalberg
    Commented Aug 8, 2019 at 14:23
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    \$\begingroup\$ @Tommy How many good pictures have you taken at 1365mm? That focal length is not easy to aim. Also, a $20 ghillie suit will get you closer to your subject that a $1000 lens. \$\endgroup\$
    – xenoid
    Commented Aug 8, 2019 at 17:30
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    \$\begingroup\$ @Tommy There is no law that states that you cannot own two cameras for two purposes or that you have to sell your old camera ;-) \$\endgroup\$
    – flolilo
    Commented Aug 8, 2019 at 17:47
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    \$\begingroup\$ For ”cheap” long lenses old mirror lenses like the 1000mm mto-11 could be an idea if manual focus is acceptable. \$\endgroup\$
    – lijat
    Commented Aug 8, 2019 at 20:45

Based on others' calculation that you would need approximately an 850mm lens - which is not going to really be in any normal person's budget, nor even really portable - see the humorous article The Question of 18-300mm Lenses, Part Deux to see how big [& expensive, $16,000] the Nikon 800mm is.

The article compares it to the 18-300mm lens which personally I love for the same reasons as the author - but that's not going to be enough to get close to what you're actually looking for.

Neither Nikon nor Canon[1], as far as I'm aware make a 150-600, but others do - see this comparison between the Nikon 200-500mm, Tamron 150-600mm & Sigma 150-600mm
Now, neither of these are going to be as sharp as a Nikkor 600 or 800 prime, but they're a mere $1,000 rather than more than $10,000.

You won't have quite the reach you had with your compact, but weighed against that you will have a massive sensor in comparison, probably with a lot higher megapixel count, so at the end of the day you could probably shoot wider & still crop to the same field of view, whilst having an overall higher-resolution image than you had before.

BTW, I shot this with one of the 18-300s mentioned above. They're not perfect, but they're really not bad at all ;)

enter image description here

Click for full [actually half] size

[1]From comments - Canon does make a 200-400mm f/4 lens with a built-in 1.4X extender, which makes it a 280-560mm f/5.6 lens with the extender engaged. Put it on a 1.6X crop body and it gives the equivalent FoV of a 448-896mm lens on FF.
It's $11,000.

  • 1
    \$\begingroup\$ I love that article - "But the results were impressive. Check out the creamy bokeh. Awesome." Gets me every time :D \$\endgroup\$
    – flolilo
    Commented Aug 8, 2019 at 14:34
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    \$\begingroup\$ @flolilo - I know. It cracks me up too. Heck, I'd love one, if I had a full-time sherpa ;) I do want one of those 150-600s though, too. Next time I'm feeling flush... \$\endgroup\$
    – Tetsujin
    Commented Aug 8, 2019 at 14:36
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    \$\begingroup\$ Canon does make a 200-400mm f/4 lens with a built-in 1.4X extender, which makes it a 280-560mm f/5.6 lens with the extender engaged. Put it on a 1.6X crop body and it gives the equivalent FoV of a 448-896mm lens on FF. \$\endgroup\$
    – Michael C
    Commented Aug 8, 2019 at 16:58
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    \$\begingroup\$ Or why not the Coolpix 1000? [not a serious question] ;) \$\endgroup\$
    – Tetsujin
    Commented Aug 8, 2019 at 17:40
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    \$\begingroup\$ The Sigma 120-300mm f/2.8 DG OS HSM Sports at $3,600 is a steal. \$\endgroup\$
    – Michael C
    Commented Aug 8, 2019 at 18:24

Camera math provides lots of equivalent ways to solve this optical problem. You know that your heart’s desire is a compact digital crop factor 1.6 (might be 1.5 depending on model). One approach is to find the inverse value of the crop factor and multiply.

For a camera crop factor 1.6, the math is 1/1.6 = 0.625. For this format you would search for a 1365 x 0.625 = 853mm. For a camera with a 1.5 crop factor, the formula is 1/1.5=0.66. Thus you need to search for a 1365 x 0.66 = 900mm.

Another approach: The 1365mm value is the actual focal length of a lens mounted on the venerable 35mm film camera circa 1935 to present. For this format, a 50mm is labeled as “normal” meaning not wide-angle and not telephoto. Mount a 1365mm on this format and the results are a formidable telephoto. This lash-up images distant objects as if they were near. Say a bird is 100 meters (328 feet) distant. With a 1365mm mounted the bird is magnified 1365 ÷ 50 = 27x. The bird images as if it were 100 ÷ 27 = 3.7 meters (12 feet) away.

What focal length equivalent would I need if I use a compact digital? The compact digital sports a 30mm “normal” lens. This is derived from the diagonal measure of the frame. The frame size is 16mm height by 24mm length. The corner to corner measure is 30mm. The diagonal measure of any format, in the terminology of photography is understood to be the “normal” focal length.

The answer is 30 x 27 = 810mm.

  • 1
    \$\begingroup\$ The different answers are induced because we typically round values. The 35mm frame actually measures 43.3mm diagonal. We round to 50mm by tradition. Thus a better magnification value = 1365 ÷ 43.3 = 31.5x. The compact digital diagonal is actually 28.8mm. Thus 31.5 x 28.8 = 907mm. \$\endgroup\$ Commented Aug 8, 2019 at 16:19
  • \$\begingroup\$ Except most APS-C cameras are not actually 24mm x 16mm. Even the 1.5X number used by all but Canon is rounded. The typical Nikon APS-C DX sensor is about 23.6x15.7 mm which figures to 1.52X. Canon's 1.6X APS-C sensors run about 22.3x15.9 mm, or 1.58X. \$\endgroup\$
    – Michael C
    Commented Aug 8, 2019 at 16:55
  • \$\begingroup\$ @ Michael C -- Trying hard to cut to the chase and give clear and concise answers, but I am often guilty of tossing out minutiae. \$\endgroup\$ Commented Aug 8, 2019 at 18:03

I have a compact camera with a tiny sensor (crop factor 5.6) and max focal length of 1365 mm.

No, you don't. If it had a focal length of 1365mm, the description "compact" would never be used for it and it would be more like a telescope than a camera. The actual focal length is printed on the front of the lens. It is likely to be something like 244mm. Which is not to be sneezed at, but with a small sensor like that doable with a tolerable amount of glass likely distributed across a whole lot of elements.

The 1365mm is the "equivalent" focal length after applying the crop of the small sensor. So the bad news is to get the same equivalent focal length, you need to get the same equivalent focal length.

In contrast to what is printed on the lens (and what lenses in isolation are sold with as specification), compact cameras usually advertise the 35mm equivalent focal length since it sounds more impressive.

So with a crop factor of 1.6, for same reach you'll want to get a camera advertised as having 1365mm of (equivalent) focal length while printed on the front of its lens will be something like 835mm focal length (and the detailed specs may mention this when describing the lens). If you buy camera body and lens separately, the lens will be advertised with 835mm focal length.

Yes, this is somewhat silly, but those are the current conventions. Of course, the obvious conclusion is "no way". Now you can get a bit more apparent reach by sacrificing resolution and cropping the image yourself. However, you will waste much of the sensor, and much of the glass, and you are going to pay for all of the waste nevertheless.

For that kind of reach, if needed on a consistent basis, the small sensor camera provides a focused solution that you cannot improve upon a lot with reasonable effort. A change makes mostly sense when you find that you use the full extent of the range rather rarely.

  • 1
    \$\begingroup\$ compact cameras usually advertise the 35mm equivalent focal length since it sounds more impressive. - It's also sort of a necessity: If I do not know how big the sensor is, the real focal length tells me nothing of interest. Does a 20-100 Nikon P981 offer the same view as a 20-100 Canon G673 or a Panasonic FZ321? Who knows! That's also why we do it this way - you cannot promote a lens with its AOV (which really is what 35mm-equivalent is about), since AOV changes with sensor size. \$\endgroup\$
    – flolilo
    Commented Aug 10, 2019 at 12:58

If you want a 1365mm lens for your sub frame digital body all you need to do is divide the 35mm equivalent of 1365mm by the crop factor of 1.6. Which is 853mm. I hope you have lots and lots of cash on hand. You will need a 1000mm f3.5 or faster lens with a 1.4x teleconveter. You may also need to hire someone to carry it for you.


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