Focal Length to magnification conversion formula

What is the formula for converting focal length to magnification?

For standard 35mm.

 

Example: I have a Canon 10D and a pair of 7x binoculars, What focal

length lense will give me the same magnification as the binos? (Feel

free to disregard the 1.6 crop unless it really changes things)

 

I'm guessing that my 50mm lens (80mm on my 10D) would be approximate

to 1x magnification meaning that (all math aside and only as an

example) would be 1.6x magnification on my 10D.

 

I'm guessing that the focal length X = magnification X math involved

is more complex and that a 50mm lens doesn't really equal 1x

magnification.

It depends on your final image size and the distance from which it is viewed. You are trying to determine angular magnification as seen by the viewer, such as that stated on binoculars.

The magnification in cameras is a bit useless. For binoculars, it's X times life size. Digital camera makers call 1x the wides focal length, so it bears no real resemblence to your binoculars.

 

In 35mm frame size, you can probably regard 50mm as 1x. So on you 10D, 35mm would be about 1X.

I'm not worried about final image size or anything like that.

 

How do I figure out what focal length I need to see the same level of magnification as I see through my binos?

Sorry, I mean through the viewfinder of my camera.

Magnification in photography is usually used to mean the ratio of

the image size to the subject size. Magnification

is related to both focal length and distance, because

images of objects get smaller as you get farther away

from the objects. To determine

magnification, you put a ruler up to your subject, and put a ruler

up to its image, and divide the second number by the

first.

<p>

The word "Magnification" in binocular terminology refers

to <em>angular</em> magnification. It's the angle subtended

by the image of the object, divided by the angle subtended by

the actual object. To determine meaningful angular magnification

in photography, you've got to take into account final print

size and viewing distance, as has already been mentioned.

<p>

It's possible to contrive a definition of "Magnification" which

is something like "Magnification relative to a 50mm lens". That's

the definition by which a 300mm lens would give 6x magnification.

That definition isn't used much -- it's usually easier and

clearer just to state the focal length outright.

My Nikon full-frame cameras (F4 and F5) will give unity magnification (subject at infinity) for a lens focal length of 70mm. For other focal lengths the magnification is proportional: a 210mm would have a "magnification" of 3X, a 35mm would have 1/2X, and so on.

The answer is 50mm for the 10D, and probably most Canon bodies (film or digital). The combination of viewfinder magnification and focal length most closely matches what you see through your unaided eye at that focal length (ie, if you had your other eye open with an unobstructed view, it would look just about normal). As you correctly point out, the digital "crop factor" plays no part in this at all.

 

This came up here the other day, too.

David, you may already have the answer you were wanting, but I wanted to through this in. Many years ago when I had a Pentax Spotmatic with a 55f1.8 on it. I noticed that if I rotated it to the vertical and I looked through it's viewfinder with one eye and the same object at the same time with other, I could see the two views were the same size. So by that, I've always figured that 55mm times power of the binoculars(usually 7), so 55 X 7= 385mm....Hope this is of some value to you....Jim

Using my 50mm Summicron or Elmar as a frame of reference, and assuming that lens provides 1x power, when I use my 200 Telyt I am getting 4x power. When I hook on my 500 Takumar I am gettint 10x power. Now, if I start with my 35mm Summaron then I have another set of proportions to contend with.

<p>Using the simultaneous eye and viewfinder method as above, my D750 results in 70mm being the same size as the naked eye view.<br>

However, I think eyepiece magnification is a critical factor in this method.</p>

 

<p>Addendum:<br>

I just measured (approximate) field of view on D7200 and D750 with a 200-500mm lens and compared that to the FOV of 8x and 9x binoculars.<br>

My results show that dividing focal length by 50 on FX and 33 for DX is a good rule of thumb for FOV.<br>

The result from the eyeball test above are consistent as the D750 has an eyepiece magnification of 0.7, so that 70mm x 0.7 = 49. Close enough for government work. :)</p>