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

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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).

Answer 4

The lens will become 160mm/f5.6

80mm x 2 = 160

2.8 + 2 stops of light = 5.6

P.S. And this about aperture is equivalent when you add teleconverter

Answer 5

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

Answer 6

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