I read another SE post that shows how focal lengths can be converted to optical magnification, but it isn't clear to me how it would work the other way around. My binoculars are adjustable from x10 to x30 times zoom, but what would that be in an equivalent lens focal length? Working with the equation you have some reference focal lengths to work off of, but working with magnification it is not clear to me what information to use. Would I use the focal length of the human eye as the minimum focal length, the optical zoom as the magnification number, and the maximum focal length would be what I'm solve for?

  • \$\begingroup\$ just as an observation often when shooting birds my 500mm on an aps-c body appears to "see" the same as my 10x binoculars I often leave my binoculars at home because of this \$\endgroup\$
    – craig
    Commented Mar 15, 2017 at 20:42
  • \$\begingroup\$ @craig This is a coincidence. Looking on B&H, the top five results for 10× binoculars all have different angles of view — 5.6°, 6.2°, 6.3°, 7.0°, and 6.0°. On APS-C, a 500mm lens has a diagonal field of view of about 3.3°. You may be comparing apparent size in the viewfinder or on the LCD screen, which is yet another thing entirely. \$\endgroup\$
    – mattdm
    Commented Mar 15, 2017 at 22:47
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    \$\begingroup\$ Magnification can be defined as either apparent magnification or FoV. The magnification for those binoculars on B&H could theoretically have identical magnification but different sized exit pupils and thus different FoVs, just the same as a 500mm lens yields various FOVs with different sized sensors. \$\endgroup\$
    – Michael C
    Commented Mar 16, 2017 at 1:26
  • \$\begingroup\$ Different viewfinders in cameras also have varying magnifications, so even two cameras with the same sensor size and same lens could give the same angle of view at two different apparent sizes as viewed through the viewfinder. \$\endgroup\$
    – Michael C
    Commented Mar 16, 2017 at 1:29
  • \$\begingroup\$ related: Panasonic Lumix FZ1000 - How does it compare to binocular magnification? \$\endgroup\$
    – Michael C
    Commented Mar 16, 2017 at 1:42

5 Answers 5


Binocular zoom and lens focal length just aren't equivalent and can't be converted in any meaningful way.

A camera lens is an image-forming system. It has a focal length. Binoculars and telescopes (with eyepiece in place) are "afocal" systems — there's no focal plane onto which the image converges. Individual elements (or even groups) within the binoculars have a focal length, but they are arranged together so that they produce magnification, but their "effective focal length" is infinite.

In photography, we often use focal length as a stand-in for field of view, because — for a given film or sensor size — there's a direct correlation. And the phrase "equivalent field of view" is used when comparing different sensor sizes, with 35mm-film-equivalent being the most common (see crop factor).

Unfortunately for your wish here, there's no way to tell this from the two numbers normally given for binoculars — the objective size (diameter of the front element, basically) and magnification. The angle of view is given as a separate specification, and from that you could simply work back to what camera lens would have a similar field of view. But you don't get that from the 10× or 30× alone, nor from something like 10×50. You'll need to look in the specs for a particular model of binoculars to get this information.

For example, these Pentax 8×40 binoculars have (not surprisingly) 8× magnification and a 40mm objective lens. They also have a 6.3° angle of view. On a "full-frame" DSLR, this is the diagonal angle of view of approximately a 400mm lens. But these Olympus 8×40 binoculars have the same magnification and objective size, yet an 8.2° angle of view — similar to a 300mm on a full-frame DSLR.

(Note that some binoculars will give an "apparent" field of view, which is the actual FoV multipled by the magnification. This is useful when comparing binoculars of different magnification, but isn't what you're looking for the purpose here. Look deeper in the specifications — it may be labeled actual field of view.)

Note that if you're looking through the viewfinder and comparing to binoculars side by side, that's yet another thing — see What does "viewfinder magnification" mean? and What is it called when an object appears to be the same size with the eye and through the camera viewfinder? for some details. But, assuming your goal is really to take pictures and not just replace your binoculars with a camera, this is a curiosity rather than something to really worry about.

  • \$\begingroup\$ I see, my binoculars are 10-30x25 with 3.5 degrees view at 10x and 2 degrees at 30x. I thought maybe I could just convert the FOV to mm length, but then again another set of binoculars is 10-100x32 with the same FOV because the objective lens is bigger... which leads me to assume that the FOV alone can't be used to convert to lens focal length. Can my question be answered provided this new information? \$\endgroup\$
    – Ryan
    Commented Mar 15, 2017 at 23:34
  • \$\begingroup\$ We're getting pretty far afield from photography, but as I understand it, the field of view is related primarily to the design of the eyepiece, and is a trade off with eye relief (how close you need to press your eyes to the thing). \$\endgroup\$
    – mattdm
    Commented Mar 15, 2017 at 23:45
  • \$\begingroup\$ But that said, 3.5° is roughly like a 700mm lens on a full frame camera, zooming in to 2° being roughly what you'd get at 1250mm. (Why this is not a 3× difference, I'm not sure; possibly simply a matter of imprecision.) \$\endgroup\$
    – mattdm
    Commented Mar 15, 2017 at 23:47
  • \$\begingroup\$ And I should stress that I'm not really converting to lens focal length, here — just giving an idea of what lenses will give a view across the diagonal of the image roughly like that you see in the binoculars. \$\endgroup\$
    – mattdm
    Commented Mar 15, 2017 at 23:50

Matt approaches the question from a Field of View (FoV) approach and, accepting that assumption, everything he says is correct. This answer approaches the same question from the different approach of apparent magnification when looking through the viewfinder.

You can only meaningfully compare binocular "X" magnification to camera lens focal lengths in terms of apparent magnification if other variables such as the sensor/focusing screen/viewfinder coverage/eyepiece magnification are defined and held constant.

In such a case you can compare apparent magnification obtained by a particular binocular to the apparent magnification seen via the viewfinder of a particular camera with a particular focal length lens. Note that the same apparent magnification may not necessarily give the same FoV because the exit pupil sizes may be different, sometimes significantly so.

If you use the common "rule of thumb" from the film era that a 50mm lens gave a 1X magnification, then your 10-30X binoculars would give an apparent magnification of about a 500-1500mm lens!

But as we will see in the details below, the 50mm rule of thumb wasn't that accurate for film SLRs back in the day, much less for the current varieties of digital cameras on the market. In the 1980s your 10-30X binoculars would have given about the same apparent magnification when looking through the viewfinder on a typical SLR at the time as somewhere between a 550-1650mm lens to 600-1800mm lens. Today, just for the Canon EOS line, it could be anywhere from 500-1500mm (7D Mark II) to 620-1860mm (1200D) to 700-2100mm (5D Mark IV).

First let's discuss what apparent magnification is: Most simply put it is the size objects look to our eyes when viewed through a lens system compared to how big they look when viewed without that lens system. If I look through a camera viewfinder with my right eye and leave my left eye open I will see an object in front of me with both eyes. If the apparent sizes are the same for both eyes, we would say the lens system (consisting of the total combination of elements in the camera lens as well as the mirror, viewscreen/focusing screen, prism, and eyepiece elements in the viewfinder) to be a magnification of 1X. If the object looks twice as large with my right eye, we would say the magnification is 2X. If the object looks half as large as seen with the right eye via the viewfinder then we would say the magnification is 0.5X.

Now let's discuss the viewfinders in typical SLR cameras. How large something appears when viewed in a camera's eyepiece depends on two factors:

  • The focal length of the lens. This affects the size of objects as they are projected on the camera's focusing screen (sometimes also called a viewscreen) as well as projected on the camera's imaging medium. Since the mirrors in every SLR I have ever seen are flat, they provide no magnification as they flip the image up onto the focusing screen. The same is true of the pentaprism or pentamirror in the viewfinder. Since all of the reflecting surfaces are flat they provide no magnification.
  • The magnification of the eyepiece. The lenses in a camera's eyepiece are very much like the lenses in a telescope or binocular eyepiece. They provide a magnification, usually a fractional one (that is they make things smaller), and project collimated light through the exit pupil. Our eyes then focus on this collimated light to view the image through the eyepiece. The size of the cylinder (or rectangle) of collimated light projected by the eyepiece is called the exit pupil size.

Many, if not most, 35mm SLR cameras during the second half of the 20th century had viewfinders that provided magnification similar to each other. With a 55-60mm lens attached the apparent magnification was about 1X.¹ That means what we saw through the viewfinder with our right eye was approximately the same size as what we saw with our unaided left eye looking directly at the same scene.

In the digital age that standardization has been severely altered. Cameras have a wide variety of sensor sizes. Viewfinder sizes vary more from camera to camera. In the manual focus only portion (which was most) of the film era even lower priced cameras needed large, bright viewfinders to enable their users to focus them properly. With the advent of autofocus large bright viewfinders have become more of a luxury than a necessity and are seen mostly on the more expensive models. Differences in sensor sizes affect how much magnification is needed for the viewfinder to display approximately the same field of view as the FoV the imaging sensor will capture.

Here are a couple of examples:

  • Compare the EOS REbel XTi/400D to the EOS 7D. Both have the same sized sensor: an APS-C sensor measuring approximately 22.2x14.8mm. The viewfinder on the entry level Rebel shows 95% of the sensor coverage at a magnification of 0.80X. The viewfinder on the 7D shows 100% of the sensor coverage at a magnification of 1.0X. Thus when a user looks in the two viewfinders from the same eye relief (distance behind the exit pupil), the image in the viewfinder of the Rebel XTi looks about 3/4 as large (0.76) as the image seen in the viewfinder of the 7D. If the same focal length lens is mounted on both cameras, the apparent magnification of items in the scene would be about 4/3 more (1.316) with the 7D than with the Rebel XTi.

  • Compare the EOS 1Ds Mark II and the EOS 1Ds Mark III. They both had identically sized 36x24mm FF sensors. Both viewfinders provided 100% coverage. The 1Ds Mark III had a larger viewfinder with a 0.76X magnification compared to the 1Ds Mark II with a 0.70X magnification. With the same lens mounted on each they both provided the same FoV. But the larger viewfinder made the same FoV appear 8.6% larger. This can be compared to viewing a 25" television next to a 23" television with identical resolutions. Both are showing the same information, but each item in the picture is 8.6% larger on the 25" screen.

Now let's compare modern digital cameras to the old SLR standards.¹ Remember that viewfinder magnification specs are measured with a 50mm lens focused at infinity. Canon's three most recent 1-Series FF cameras have viewfinder magnifications of 0.76X. That translates to about a 65mm lens need to get apparent 1X magnification through the viewfinder. All of Canon's other FF digital cameras (the 5-series and the 6D) have viewfinder magnifications of 0.71X which translates to roughly a 70mm lens needed for 1X apparent magnification.

Canon's recent APS-C models range from 0.8X for the Rebel T5/1200D to 1.0X for the 7D Mark II. That leaves quite a bit of range for apparent 1X magnification: Anywhere from 50mm for the 7D2 to 62mm for the 1200D. See how the rule of thumb is disappearing with the various differences between camera models? We need anywhere from a 50mm to a 70mm lens to get the same apparent magnification looking through the viewfinders of various EOS DSLRs.

Keep in mind that the FoV with an APS-C camera will be much smaller than the FoV of a FF camera if both have the same magnification. In fact, the total apparent width of the view in the 1.0X 7D2 viewfinder is not as wide as the 0.76X viewfinder of the 1D X or even of the 0.71X 5D3. That is because the APS-C sensor/mirror/viewscreen size is only about 0.63X as wide as the FF sensor/mirror/viewscreen. So now the waters are even muddier!

By now it should be apparent that only when you have a specific camera model in mind can you compare the "X" magnification of a pair of binoculars with the apparent magnification observed when looking through the viewfinder of a camera with a lens of a particular focal length.

¹ The following cameras listed with their viewfinder magnifications with a 50mm lens focused at infinity: Canon F1 - 0.8X, Nikon F - 0.8X, Canon AE-1 - 0.86X, Minolta X-570 - 0.9X, Pentax K2 - 0.88X, Pentax ME-F - 0.87X. A 0.9X viewfinder would give 1X apparent magnification at roughly 55mm, a 0.8X viewfinder would do so at roughly 62mm.


Keeping it simple. YES. Your 30X magnification binoculars are 30X lifesize. In photography, lifesize is 50mm on a full frame camera (35mm or full frame digital). That is the standard. So, 50mm x 30 = 1500mm.

Now, if you are using a sub frame (APS-C) digital, you must figure the crop factor. So, 1500mm / 1.6 = 937mm.

To get 30x magnification on your sub frame digital you need a 937mm lens. To get 30x magnification on your 35mm or full frame digital you need a 1500mm lens.

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    \$\begingroup\$ This is a fairly concise and close answer. With telescopes or binoculars, the magnification is found by dividing the focal length of the instrument by the focal length of the eyepiece. E.g. a 400mm scope with a 25mm eyepiece is 16x magnification (400 ÷ 25 = 16). With a camera, the magnification is the focal length of the lens divided by the diagonal size of the sensor. For a full-frame camera that's about 43mm (not 50mm but close). Assuming a full-frame sensor camera then multiply 16x magnification by 43mm to get 688mm. So a 688mm lens would provide a similar magnification. \$\endgroup\$ Commented Aug 12, 2019 at 14:43
  • \$\begingroup\$ @TimCampbell Only if the apparent size of the image in the viewfinder is the same size as the camera sensor (i.e. a VF with 1.0X magnification). If the viewfinder has a fractional magnification (0.70X, 0.86X, etc.) that must be applied to the computation made using the focal length and sensor/focusing screen dimensions. \$\endgroup\$
    – Michael C
    Commented Mar 18, 2021 at 3:53

I am an astronomer....I never really think much about magnification when using my 'scope. Usually, less is more, as you mainly and clarity, brightness and wide field of view ideally. My large scope is very clear and very bright. You can't magnify stars anyway.......but there is light magnification. Camera lenses are limited for both.....they don't have a lot of light magnification if any, and the more you magnify the image the more distortions you tend to get optically with the exception of very expensive glass. Digital zooms are very limited. Fine lenses can give some magnification with good quality, but you really pay for it. If I need magnification, I just hook up a 2100mm Schmidt to my camera.....that gives about 40x magnification with a very bright image...the only drawback is the bokeh has mirror remnants, but that never bothers me. The 40x of the Schmidt is totally different than a camera.....you can't compare them as there are so many variable.....depth of field image resolution and brightness, etc.

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    \$\begingroup\$ This seems to address why you might not want to over-magnify a subject, but I'm not seeing a relationship to the question regarding relating zoom factor to a focal length. If you have ideas on how this might be done, you can edit your answer to include that information. \$\endgroup\$ Commented Aug 12, 2019 at 14:31

The 50mm = mag. 1 is not format related. The sun and the moon have a visual angle of 1/2°. When imaged with a 50mm lens, the image size of the projected image of these bodies is approx. 0.5mm in diameter. Such a circle size is at the threshold of being recognized as a disk when viewed by the average observer with the unaided eye. This is the basis of this rule of thumb.

Astronomers continue to use 50mm as magnification 1. I am talking about "prime focus" astrophotography.

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    \$\begingroup\$ Astronomers typically place a photographic plate or film or digital sensor at the focus of their telescopes. Universally, they accept 50mm has magnification 1. Visually, they place an eyepiece lens at this position. Now the magnification is the focal length of the objective lens divided by focal length of the eyepiece lens. Check astronomy sites and you can verify. \$\endgroup\$ Commented Mar 15, 2017 at 23:50
  • \$\begingroup\$ @ mattdm, you are correct, I described afocal imaging. Astronomers measure an image of the moon or planet on the plate. 50mm translates to magnification 1. \$\endgroup\$ Commented Mar 16, 2017 at 0:22
  • \$\begingroup\$ @ mattdm -- I misspoke -- I was describing the film / plate in the prime position. I stand on 50mm = 1X. We image on film or chip and then view that image as a paper print or on a display screen. The 1X mag. on film or chip is then enlarged for viewing. We must take this additional magnification into account. The 1X M is a contact print (same size as projected by the lens. \$\endgroup\$ Commented Mar 16, 2017 at 6:20
  • \$\begingroup\$ I think you have a very narrow view of what constitutes an "astronomer," not to mention what sort of imaging apparatus she (or he) may use. \$\endgroup\$ Commented Mar 16, 2017 at 11:33
  • \$\begingroup\$ @mattdm -The moon is 3474 km diameter and 384400 km distance. So my notion is, computing your 0.5 mm size is 0.0000005 km diameter / 3474 km diameter is 0.00000000014x magnification. Or if computing the distances behind and in front of the 50 mm lens in classic fashion, this is 0.00005 km focal length / 384400 km distance, which computes magnification 0.00000000013x in a 50 mm lens. Neither is imaginably close to 1x, and not imaginably related to 1:1 macro. We might argue the 50mm "normal lens" is chosen to vaguely match the human eye view, but that only applies to 35 mm film. \$\endgroup\$
    – WayneF
    Commented May 6, 2017 at 21:15

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