How do I calculate f/stop change for teleconverter that increases lens magnification?

I'm trying to calculate the f/stop change of attaching a x2 teleconverter to a 80mm f/2.8 lens. In this instance, the teleconverter attaches to the front of the lens, and has a larger opening than the original 80mm lens.

The original lens would have a lens diameter 80mm/2.8 which is about 30mm.

The new lens seems to be about 40mm, and it's a x2 teleconverter, so the new f/stop should be 160mm / 40mm = 4.

All the documentation I found suggested that this is wrong and since it's a x2 teleconverter the f/stop should also be doubled.

Where's my mistake here?

Edit: turns out you can't trust numbers on off-brand teleconverters, it's actually a 1.4 when measured. still the question stands.

• A picture of the equipment wouldn't go amiss either. Jul 15 '21 at 17:29
• it goes at the front of the lens, yes. Also I'm sure it's a teleconverter. here is the specific product. alibaba.com/product-detail/… Jul 15 '21 at 21:10
• That's not really a teleconverter. That's a magnifying lens. Jul 16 '21 at 6:09
• Magnifiers that go on the front of lenses are rated according to the increase in area, not on the increase in linear size. A magnifier that has a linear magnification ratio of 1.4:1 does double the size of the subject in terms of the area it covers on the sensor or film, and is therefore referred to as a 2X magnifier. Jul 16 '21 at 6:12
• The amount of light that traverses the lens is based on the diameter of the front entrance pupil diameter. Using a flashlight that outputs a parallel beam, point it so the beam enters the lens from the rear. Using a white paper target mounted over the front of the lens, you can see an illuminated circle. This will be a good representation of working diameter of the exit pupil. Measure the diameter of this circle, divide this value into focal length to calculate the revised focal ratio (f-number). The difference, working f-number vs. click-stop f-number should yield a suitable tweak factor. Jul 21 '21 at 16:04

With a rear-mount converter, it's always the factor of 2 (unless there's a gross mismatch between lens and converter), but with your front-mount one, it's not that simple.

The relevant question is where the light path is effectively limited.

Or, to put it another way: Does all the light collected in your converter's front lens reach the sensor, or is some part of it blocked by the front opening of your base lens? If part is blocked, then the large converter front lens doesn't help and is just a waste of material.

As a quick check, you can detach the combo from the camera body and look into it from the rear side, a few centimeters behind the lens, roughly where you'd expect the sensor. Do that with aperture full open. If you can fully see the circular edge of the converter's front lens, then the 40mm-based calculation is indeed valid (all the light collected on that 40mm circle reaches the sensor). If you don't, that means that the 30mm opening of the base lens still is the limiting factor, and the 30mm calculation will probably give better results.

And, if you want to do some experiments, compare the exposure times for full-open shots with and without the converter (of course, in a constant-lighting situation). If there's a factor of 4 between the two times, the classical calculation applies, if it's a factor of about 2, your 40mm-based calculation is correct.

Having said all that, I bet that you won't be able to see the converter front lens edge, and that the exposure-time experiment will result in a factor of 4.

• It is a front mount indeed. Surprisingly, i can just see the converter's edge at filming distance. As for exposure time experiments, this is a film camera and i'm not sure I want to waste all that film. Jul 15 '21 at 21:18

A teleconverter attaching to the front of your lens and not vignetting will retain the aperture number (that is quite different from a tele extender put between lens and camera body). A 2× teleconverter would thus turn an 80mm/2.8 lens into a 160mm/2.8 lens. Now here is the rub: almost everything marketed as a "2x teleconverter" for the front of the lens is quite far from actually being a 2x teleconverter. Typical factors for realistic converters with reasonable quality are 1.4, 1.5, 1.7. Once fantasy numbers come into play, they may try to justify themselves as indicating the area size factor rather than linear size. That's ridiculous but not uncommon. So you are more likely in the 120mm/2.8 area (assuming that the glass and coatings are good enough not to cause significant light loss).

Now here is another rub with cheap teleconverters: they tend to lose more resolution via their optical quality than they gain by enlarging the image.

Another problem with those converters is not really much of their fault: image stabilisation will undercompensate since it doesn't expect the image to move as much as it does with a tele converter in front. So it is a good idea to use a tripod. Or tell your camera your converter's magnification factor in its menus, but that is rarely available.

• Is there a way to measure the actual focal change? Jul 15 '21 at 21:19
• @buzzgo Sure: just take a picture of some reasonably far object (infinity focus) with and without the converter and measure some dimension in pixels in both images. The ratio is your magnification factor. Jul 15 '21 at 21:55
• thanks @user98068. This is a film TLR so I took a pictures of the ground glass with my phone. it is indeed 1.4 so the resulting lens is a 110mm. Jul 15 '21 at 22:20

I'm not sure I understand the description/question.

If your added magnification is at the objective end it is a diopter or telescope type addition... it does not convert or change the lens' optical formula in any way. It increases the magnification of the scene prior to entering the lens... And because the increased magnification is between the subject and the aperture, it also equally magnifies/increases the size of the entrance pupil (effective size of the aperture opening as seen by the subject). Therefore the F# does not change. You can think of this as using the camera to photograph through a window that has wavy glass... the glass affects what the camera sees, but it doesn't change anything about the camera/lens.

If the added magnification is between the lens and the camera it is a teleconverter... telephoto lens element(s) result in an optical focal length greater than the lens' physical length. It is called a teleconverter because it converts a standard (prime) lens' optical formula/design into a telephoto formula/design. I.e. if the lens was an 80/2.8 prime it is converted into a 160/5.6 telephoto lens if a 2x TC is added.

This is because telephoto elements (converters) function by magnifying the image circle leaving the lens. This makes the image circle larger, which reduces the exposure due to the inverse square law. If you make the image circle 2x larger in area that is 1/4 the light density (2 stops less). And because the image circle is now larger than the sensor, it has created a crop factor of 2x as well (DoF/CoC/EFL).

• Tele converters in front of the lens are not diopters. They are essentially telescopes often consisting of a convex and a concave group spaced apart and are a quite common accessory for compact cameras without interchangeable lens. Jul 15 '21 at 22:02
• @user98068 They're effectively magnifying lenses designed to be able to still allow the camera to focus at infinity. But the rest of the answer still holds. Any magnification that occurs in front of the aperture diaphragm determines the size of the entrance pupil does not affect the f-number because the size of the entrance pupil is also magnified by the same amount. Any increased magnification that occurs behind the most restrictive aperture will increase the f-number because the increase in magnification does not also increase the size of the entrance pupil. Jul 16 '21 at 6:17
• "magnifying lens" is the common term for a diopter. it does not magnify in itself when held close to the eye but allows to reduce object distance. What the question is talking about is what generally the manufactorers of compact cameras refer to as "tele converter" and that's generally the marketing term for third-party producers. Wikipedia calls it "teleside converter", a term i never heard anywhere else. They do "convert" the scene even when focus-neutral (not every of them is perfectly so) since the minimum focusing distance is increased. Jul 16 '21 at 7:33
• @user98068, I've added "or telescope type addition" to include that information... I've never used nor paid attention to those kind of adaptors (though I do own compact cameras). Jul 16 '21 at 12:31

The lens will become 160mm/f5.6

80mm x 2 = 160

2.8 + 2 stops of light = 5.6

• but why? does a teleconverter with a wider lens not increase the lens diameter in our calculation, since it takes in more light? Jul 15 '21 at 11:08
• @buzzgo, those calculations are true independently of lens. Do not forget that inside teleconverter there is glass... Jul 15 '21 at 11:10
• This happens because a 2x teleconverter doubles the effective focal length without changing the effective aperture diameter -- and f stop is defined as focal length divided by aperture. So a 50 mm lens with 25 mm aperture is f/2, but if you extend the focal length to 100 mm, that 25 mm apterure is now f/4. Jul 15 '21 at 11:49
• @ZeissIkon This is not a tele extender between lens and body but a tele converter in front of the lens. It changes the apparent entrance pupil size according to the extension factor and thus retains the aperture number. Jul 15 '21 at 22:04

The job of the camera lens is to project an image of the outside world onto the surface of film or digital sensor. The image size of object (magnification) is determined by the actual size of the object intertwined with distance from the camera and the focal length of the camera lens used. If you increase the focal length of the camera lens, the projection distance is also increased. This results and image that displays greater magnification. As an example, if you increase the distance screen to projector of a slide or movie projector and re-focus, the image projected on the screen is enlarged.

Peter Barlow, English Mathematician / Optician, invented an achromatic (without color error) supplemental lens that increased the magnification of telescopes in 1833. The Barlow lens design is the one used in modern teleconverters.

Such supplemental lenses increase the versatility of our camera lens. Commonly they double or nearly double the focal length. A 2X teleconverter doubles the focal length granting a 2X focal length increase which results in 2X grater magnification.

This increased magnification comes with a price. Along with the increased image size comes a reduction in the intensity of projected image. To calculate the impact of this magnification gain on image brightness, we square the magnification gain. Thus for a 2X teleconverter the math is 2 X 2 = 4. We find the reciprocal of this reduction factor by annexing 1/ before the number. Thus, a reduction factor of 4 tells us that the amount of light reaching film or image chip is ¼ or 25% of the former.

Now the f-number system we use is based on an incremental change of 2. In other words, each f-number change doubles or halves the exposing energy. Thus, we divide the magnification increase granted by the teleconverter by 2 to find out how many f-stops reduction results. In this case, a 2X doubling of the magnification results in a reduction factor of 2 X 2 = 4. This value, divided by 2 = 2. This tells us that the functioning f-number is 2 f-stops so we open up 2 f-stops. Go left on the below f-number set.

The f-number set: 1 – 1.4 – 2 – 2.8 – 4 – 5.6 – 8 – 11 – 16 – 22

Thus if the f-number is f/8 and we add a 2X teleconverter, the working f-number changes two f-stope to f/4. Also note – the f-number set is its neighbor multiplied going right by the square root of 2 = 1.4.

Let me that that understanding the resulting reduction factor holds for figuring out exposure when adding filters (filter factor). This value is a multiplier used to manipulate exposure time. Thus if the factor is 4, we can multiply the exposure time by this factor to calculate a compensating exposure time.

Suppose the exposure without filter or teleconverter is 1/400 of a second at f/8. We mount a filter of teleconverter with a factor of 4.

The revised exposure time is 4/1 X 1/400 = 4/400 = 1/100 the revised shutter time @ f/8 Or 1/400 second @ f/4

• Unfortunately wrong since it ignores "In this instance, the teleconverter attaches to the end of the lens, and has a larger opening than the original 80mm lens." and discusses an extender placed between lens and body rather than in front of the lens. Jul 15 '21 at 21:58

First, you have to understand what the f-number actually is. It's the ratio of the focal length of the lens to the diameter of the aperture opening (not the lens's filter diameter or the aperture opening in the TC):

f-number = focal_length / aperture_diameter

And it describes the proportional amount of light we get from a given lens based on the aperture opening diameter and focal length. We use f-numbers instead of the actual width measurement to describe apertures so we don't have to remember that on a 200mm lens, a 50mm wide aperture opening gives the same amount of light as a 12.5mm opening on a 50mm lens. F-numbers normalize across all focal lengths.

And solving for the aperture diameter gets:

aperture_diameter = focal_length / f-number

Which is why we express the aperture diameter setting or the max. aperture of a lens as f/# (e.g., f/4).

When you use a teleconverter, however, you're multiplying the focal length, but the diameter of the aperture opening in the lens hasn't changed. So you are getting proportionally less light, since the lens is now longer. The teleconverter's opening is bigger to keep from obstructing light to the lens, but the physical limit on how wide the aperture blades in the lens can open up remains the same.

So, if you have a 2x teleconverter, then your f-number also doubles (increases by two stops); if you use a 1.4x teleconverter, your f-number multiplies by 1.4x (increases by one stop).

Remember stops are doubling/halving of the light. But the amount of light you get from the aperture is directly proportional to the area of the opening, not its diameter (which is square-root-of-two proportional to the area, 'cause of the whole circle_area = πr2 thing). What doubles in the full-stop f-number scale is the square of the numbers:

• 1.42 ≈ 2
• 22 = 4
• 2.82 ≈ 8
• 42 = 16
• 5.62 ≈ 32
• 82 = 64
• 112 ≈ 128
• 162 = 256
• It's actually entrance pupil size, not the actual size of the physical aperture, that determines f-number. f-number = focal length / entrance pupil diameter Magnifiers placed on the front of a lens (which do not restrict any light that reaches the front of the main lens) increase the size of the e.p. by the same factor as they increase the size of the image projected by the lens. Thus, there is no reduction in f-number. Magnifiers placed behind the physical aperture do not increase the size of the entrance pupil, thus do reduce the f-number. Jul 16 '21 at 21:01