BetterPhoto Q&A
Category: How to Work with Photos and Film Equipment

Teleconverters

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0 January 14, 2004

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0 January 13, 2004

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0 January 13, 2004

For

1.4x = -1 f/stop

2x = -2 f/stops

3x = ... -3 f/stops?

... and what about those lovely 1.5x and 1.7x teleconverters?

Although I can "approximate", and with TTL meters it doesn't matter since the meter will read the available light through the lens/teleconverter setup ...

... has anyone seen a formula where

teleconverter strength number x C = -change in f/stops?

Thanks

Peter Blaise
peterblaise@yahoo.com
http://www.peterblaisephotography.com

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0 August 24, 2004

You can't say TC strength x C = change in f/stops because that would mean TC increases with f/stop change linearly. And this is definitely not the case because what an f/stop means is a halving or doubling of the amount of light...that is, it goes up or down exponentially. So in order for f/stops to go up one by one as you increase the TC strength, the TC strength must increase exponentially by some factor C as you said.

From that pattern you started above, you can see that the number of f/stops goes up faster than the teleconversion factor. Specifically, the f/stops go up by 1 f/stop each time. Meanwhile, the teleconverter sizes go up by 1.4 each time. (Or actually the square root of 2 to be exact.)

Anyway, since it's NOT a linear progression, to derive an equation from that you have 1.4^x = y or in other terms, y (the f/stops) equals 1.4 raised to the teleconverter strength power.

Now keep in mind the f/stop answer you get is not actually in f/stops. It's a factor of change in the f/stop.

1.414^TC=f/stop factor

So if you have a 1.7x teleconverter on a lens with a Max Aperture of 2.8, the new aperture is:

1.4^1.7 = 1.77 change factor. Then multiply that times 2.8 to get the new max aperture of roughly 5 stops. About in between 4 and 5.6 stops.

Now say you want to go the other way and find out what TC magnification to use at a certain aperture.

For example, say you can't autofocus past f/8, and you'r Max. aperture is f/2.

The equation is 1.4^TC = (8/2) so simplify and you have TC log1.4 = log4 so TC = log4 / log1.4 which conveniently equals 4. So the largest TC you can autofocus with using that lens and camera is a 4x TeleConverter. In the same fashion, the -3 f/stops change you guessed above doesn't equal 3, but rather log3 / log1.4 = 3.17.

Whew...and you thought you'd never use logarithms again.

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0 August 24, 2004

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0 October 10, 2004

When you use M42 screw mount lenses with an adaptor on your bayonet mount camera, there is no linkage of the aperture control to the camera. So you have to use stop-down metering:
(1) focus with the aperture wide open,
(2) set the lens aperture to the desired setting - it will "stop down" the opening and the viewfinder will darken,
(3) now use the camera TTL meter to set the shutter speed,
(4) trip the shutter.

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0 October 11, 2004

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0 October 11, 2004

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0 October 29, 2004

With a dedicated TTL flash you don't do anything different.

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0 October 30, 2004

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0 October 30, 2004

With a screw-mount lens and adaptor, you cannot use shutter priority or full progam modes because the camera cannot control the aperture. When you set the aperture wanted, the lens stops down and the viewfinder goes dark making it difficult to focus. So you have to focus at full open aperture, then stop down, take the photo, then open the aperture again to focus the next shot.

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0 October 30, 2004

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0 October 30, 2004

First of all, TTL Through The lens metering obviates the need for all this below, but ...

Okay, there are 2 formulae for teleconverter power to f/stop changes, and they are both simple:

--------------------------

FORMULA ONE: To find the resulting f/stop of a lens with a teleconverter, multiply the f/stop setting of the lens by the teleconverter marking! Doh!

Example, an f/~1.4 lens * with an ~1.4x teleconverter would equal ~1.4 x ~1.4 = ~1.96 ... that is, f/~1.96, which is effectively equal to f/2, so an f/1.4 lens with a 1.4x teleconverter makes a setup with an f/2 maximum aperture.

The numbers work cleaner when you use 2^.5 or the square root of 2 (2 to the power of 1/2) which is 1.4142135623731 ... but nobody labels their lens 50mm f/1.4142135623731! ;-)

So, that is ONE rule - what is the resulting f/stop.

--------------------------

Now, how many f/stops away does a teleconverter push a lens? If an 1.4x teleconverter moves a lens 1 f/stop darker, and a 2x teleconverter moves a lens 2 f/stops darker ... then, as the original question went, where does a lens move to when using an inbetween teleconverter or greater, such as one of those odd 1.5x or 1.7x?

There are two ways to resolve the question. One is to COUNT after multiplying, let's use that never
before accurately labeled f/1.4142135623731 lens:

f/1.4142135623731 x 1.7 = 2.40416305603427 ... which is 1.531070 f/stops darker. How do I know?
I measured! Okay, I cheated, I used the f/Calc program from http://www.tangentsoft.net/ - FREE!

FORMULA TWO: Alternatively, we could take a LOG base 1.4142135623731 (or LOG base square root of 2, or LOG base 2^0.5) of the teleconverter. Here, in Microsoft Excel is the formula:

=LOG(TC,2^0.5)

... where TC = the teleconverter power as labeled. So, for the 1.7x teleconverter in our example,

Log(1.7,14142135623731) = 1.53106949272595

Wow, that f/Calc program is quite accurate! And for the 3x teleconverter:

Log(3,1.4142135623731) = 3.16992500144231

... which equals the 3.17 that Steven found (see below and previous post).

Now, it should be quite easy for you all to build a spreadsheet of the powers of your own teleconverters versus the f/stop results of your own lenses! Enjoy.

* I write f/~1.4 as the math works better with f/stops NOT rounded off. Here's the sequence where each lower f/stop is the previous f/stop times the square root of two. We don't always write them this way, but here's how the computers in our cameras see our f/stops:

f/ 1
f/ 1.414213562
f/ 2
f/ 2.828427125
f/ 4
f/ 5.656854249
f/ 8
f/ 11.3137085
f/ 16
f/ 22.627417
f/ 32

===================================

PS - I found problems with Steven C. formulae as written, though I'm sure there are many ways around the topic and many resolutions that produce accurate results.

1.414^TC=f/stop factor

fails for me. This is what I get:

1.414^1.41 = 1.63252691943815
1.414^1.70 = 1.80250092522166
1.414^2.00 = 2.00000000000000
1.414^3.00 = 2.82842712474619

BUT, a 1.4x teleconverter as in the first line does NOT produce a 1.63252691943815 f/stop change.
The 2x teleconverter is okay, but this formula is way off!

Then, the log3 / log1.4 example says a formula would read:

log TC / log square root of 2 = f/stop change

And that works for me, though it's more typing in Excel, where the formula for f/stop factor by this method is:

=(LOG(TC))/(LOG(2^0.5))

... where TC = the teleconverter markings. The results agree with my example FORMULA TWO above, quoted again here for comparison:

=LOG(TC,2^0.5)

THANKS, STEVEN!

Click!

Love and hugs,

Peter Blaise Monahon
Konica Minolta Olympus Contax Yashica Pentax Canon Nikon Nikkor Vivitar Tamron Samyang Cosina
Fujifilm Ilford Kodak Agfa Adobe Hewlett Packard et cetera Photographer (have I left anyone out?)
peterblaise@yahoo.com
http://www.peterblaisephotography.com/

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0 November 03, 2004