How would I be able to take the multiplication value of a teleconverter (1.7, 1.4, 2, etc) and find out how many stops of light are lost when using it just based on that number?


Although very little (single digit percentages) light is lost to a teleconverter, the ratio of entrance pupil to focal length grows proportional to the magnification of the telecon. For example, if a 100mm f/2 lens has a 2x teleconverter attached to it, the lens will effectively become a 200mm f/4. This is because the lens will always have a 50mm entrance pupil (remember, the name f/number literally means the entrance pupil diameter is focal length/number)

Because the amount of light gathered by the aperture series is based on doubling (f/2.8 gathers half as much as f/2) the series progresses not linearly but rather along a log-base-2 line. The equation for the increase in f-number as a result of teleconverter is, approximately:

E = 2 x log2(M)


  • E is the (negative) shift in Ev
  • M is the magnification of the teleconverter
  • log2 is the operation logarithm base 2

So if you combine 2 each 2x teleconverter and 1 each 1.4x teleconverters (4.8x total) you would "lose" about 4.5 Ev or stops of light. If you attached that teleconverter to an f/2 lens, it would become a f/9.5 lens.

This equation could be combined with the expression of aperture in Ev to predict new effective aperture based on old effective aperture.


It's not really that you are losing light, per se. I mean, you might be a little, as the extra glass is going to soak up some - but I think what you're looking for is this: Why does using a tele-converter cause my aperture to be smaller?

The f-number is the ratio of the lens's focal length to the diameter of the entrance pupil. (More reading at the Wiki)

So, let's say that you are using Canon's 135 f/2L. At f/2, the diameter of the entrance pupil (aperture) will be: 2 = 135/D, so D = 67.5mm.

Now, let's say you throw a 1.4 teleconverter onto the lens. The lens specs say that you effectively lose 1 stop, but here's the math. 1.4*135 = our lens is now a 189mm lens. However, the entrance pupil did not get any bigger, it's still 67.5mm.

So, our aperture calculation now reads: f = 189/67.5 = 2.8.

And, as you know, f/2.8 is one stop down from f/2.

So, it's not that you lost light per se, it's that the physical size of the entrance pupil doesn't change while the focal length does, causing the effective aperture to be smaller when using the teleconverter.

Rinse and repeat the calculation using the f-number = focal length / entrance pupil diameter for any size teleconverter you like to prove how many stops it will soak up.

  • But is there a formula to calculate it without knowing the "pupil"? Is there a way to just take f/1.8 and find out how much is lost from a 1.4x teleconverter?
    – Joe Scotto
    Jun 11 '18 at 16:35
  • The entrance pupil is always focal length divided by f-stop. So a 50mm f/2 has a 25mm pupil. If you do not know the focal length of the lens then you cannot calculate the entrance pupil. Jun 11 '18 at 17:14
  • @PhotoScientist So for example a 50mm f/1.8 with a 1.4x teleconverter would act as if it was a: 70mm f/2.52, correct?
    – Joe Scotto
    Jun 11 '18 at 17:56
  • @JoeScotto That is correct. I'm not aware of any 50mm lens which will accept a teleconverter but that's not really the point of the question. Note that if, instead of calculating a new f/number, you are trying to calculate the difference in stops, see my answer. Jun 11 '18 at 18:02
  • I think that’s right. I’m trying to find what f stop it would be after the teleconverter hits it. Thanks
    – Joe Scotto
    Jun 11 '18 at 18:03

The easiest way to figure how much a teleconverter (TC) increases the f-number without doing any complex math is to do this:

  • Take the linear magnifying power of the teleconverter and compare it to how many stops away from "1" it is on the f-number scale.¹ That's how many stops you lose.
  • For a focal reducer (FR), use the reciprocal of the magnifying power (1/M) and compare it to how many stops away from "1" it is on the f-number scale.¹ That's how many stops you gain.
  • If you are using multiple teleconverters and/or focal reducers, figure the loss/gain for each one separately and then add the number of stops each TC loses and subtract the number of stops each FR gains to get the number of total stops difference.
  • Take the f-number of the bare lens before adding the teleconverter, start there on the f-number scale,¹ and go up or down the number of stops that equals the difference you calculated. Go up for stops lost using a TC and down for stops gained using an FR.

A couple of examples:

You have a 135mm f/2 lens and add a 1.4X TC. what is the new maximum aperture of the lens?

We can look at one of the charts below and see that f/1.4 is exactly one stop slower than f/1. This means that we will lose one f-number. We look at the same chart and start at f/2 (the max aperture of our bare lens) and count up one full f-stop. We see that now our maximum f-number is f/2.8

You have a 50mm f/1.8 lens and want to add a 2X TC and a 0.71x focal reducer (just for kicks since this would make absolutely no sense at all from a cost/benefit or image quality standpoint). What is the max f-number of the full combination?

For the 2X TC, we see that 2 steps removed from 1 on the f-number scale is "2", so we lose two stops when using a 2X TC.

For the 0.71x FR, we use the reciprocal of 0.71 which is 1.41 (1 ÷ 0.71 = 1.41. Hmmm.... where have we seen that before?). We see that 1.4 is one stop removed from 1 on the f-number scale, so we gain one stop using a 0.71x FR.

When we add the two stops we lose and subtract the one stop we gain, that leaves a net effect of a loss of one stop.

Looking at the one-third stop scale, we see that f/1.8 is one-third stop less than f/2. By counting up three spaces on the one-third stop scale (because three-thirds = one), we see that the f-number for our f/1.8 lens + 2X TC + 0.71 FR will be f/2.5.

(1) A working knowledge of the f-number scale, at least in whole stops, is something every photographer should have committed to memory. There are just too many places in photography where intuitively knowing the progression of the powers of the square root of two, which is the same progression as the f-number scale, comes in handy. If you haven't yet learned the f-number/powers of √2 scale well enough, you can carry a "cheat sheet" with the whole, one-half, and one-third stop scales printed on it. Many photographic equipment stores once sold handy laminated cards with such scales printed on them, usually for a fairly low cost. Some stores would give customers a free set with each new camera purchase. With so many free printable card templates now available online, finding such cards available from brick and mortar retailers is becoming much less common.

enter image description here
This sheet, posted at phototraces.com shows whole stop f-numbers in the left column, one-half stop f-numbers in the middle column, and one-third stop f-numbers in the right hand column, as well as explanations of how different apertures affect images to the right of the columns and a visual representation of what an aperture diaphragm might look like at different f-number settings to the left of the columns.

enter image description here
This one is more basic but also includes the AV (aperture value) scale, which is simply a numerical scale that shows what power of the √2 is used for each f-number. On each scale, the full step f-numbers are shaded in green.

You can find other versions of the same thing here and here. A few third party sellers still offer sets of pocket sized field reference cards via amazon. Without being able to see the cards, though, it's impossible to say if they include f-numbers all the way down to f/1 that would enable using the method outlined above.

A more detailed explanation:

It's all simply based on how much area over which the same amount of light is spread out. When you increase magnification by a factor of two with the same size entrance pupil, you spread the same light out over four times the area. Thus the field density of the same amount of light spread out over four times the area is one-fourth as bright as before. That's two 'stops' in photography, where each 'stop' is double or half of the next or previous one, respectively.

F-number is a measure that, among other things, attempts to approximate, before losses due to reflection and absorption of light as it passes through the lens, the light strength per unit area falling on the film/sensor plane based on the ratio of the diameter of the entrance pupil to the focal length of the lens. In other words, it is a measure of how much light energy per mm2 is falling on the film/sensor plane if the brightness of the scene is constant.

Since f-number is based on the diameter of the entrance pupil divided into the focal length of the lens, but the amount of light allowed through the opening is based on the area of the opening, each increase or decrease of the area of the entrance pupil by a factor of two results in an f-number that is based on the powers of the square root of two (√2).

If you double the area of the entrance pupil, you increase the area of the entrance pupil by a factor of 2. To do this you only increase the diameter of the entrance pupil by √2, or approximately 1.414. To halve the area of the entrance pupil you decrease the diameter of the entrance pupil by a factor of 1/√2, or approximately 0.71.

What this means is that the basic scale we use for f-numbers, the so-called "whole-stop" scale, is based on powers of the √2 (taken here to three significant digits past the decimal for values in the series that are not exact integers):

√20 = 1, √21 = 1.414, √22 = 2, √23 = 2.828, √24 = 4, √25 = 5.657, √26 = 8, √27 = 11.314, √28 = 16, √29 = 22.627, √210 = 32, √211 = 45.255, √212 = 64, √213 = 90.510, etc.

We usually round off the √2 to 1.4 and use the multiples of 1.4, also rounded off to whole numbers past 8, to represent the powers of the √2:


Notice that what we refer to as f/5.6 (f/5.657...) is really closer to being f/5.7 than f/5.6!
What we call f/11 (f/11.314...) is really exactly double what we call f/5.6, but 11 is not exactly 5.6 x 2!
What we call f/22 (f/22.627) is really closer to being f/23!
What we call f/90 (f/90.510) is really closer to being f/91!

We use the 'rounder' numbers for the odd-numbered powers of √2 because they are easier to remember as approximate multiples of 1.4, rather than memorizing the actual exact powers of the √2, a number with never-ending digits past the decimal point, or even the exact multiples of 1.4. When the f-number scale was established during the early days of photography it didn't matter at all because the mechanics of the cameras used at that time were not anywhere close to being precise enough in terms of aperture size and shutter timing for it to make any difference. Most of the cameras/lenses we use today for artistic photography are still not that precise, but now they do more or less target f/22.627 when we select f/22 with our aperture control, just as they more or less target 1/1,024 (1/210) seconds shutter time when we dial in a 1/1,000 shutter "speed".

  • This is a great way to handle the problem proposed in the field. I doubt most people will be doing log-base-2 while out photographing wildlife :) Your answer is pretty close to one of my favorite photo teaching tools: the APEX system or its modern variant Ev system. I wonder if anyone has explained telecons in the scope of APEX/Ev? Jun 12 '18 at 13:56

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