How do you calculate light loss in stops from a teleconverter’s magnification?
Asked 6/11/2018
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Given only a teleconverter’s magnification factor (such as 1.4x, 1.7x, or 2x), how can you determine how many stops of light are lost? I’m looking for the relationship between teleconverter magnification and the resulting change in effective f-number/stops.
Originally by Photography Stack Exchange contributor. Source · Licensed CC BY-SA 4.0
Photography Stack Exchange contributor
8y ago
2 Answers
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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)
Where
- 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.
Originally by user71671. Source · Licensed CC BY-SA 4.0
user71671
8y ago
0
Generated from our catalog & community — verify before relying on it.
A teleconverter increases focal length by its magnification factor, but the entrance pupil diameter stays the same. Since f-number = focal length ÷ entrance pupil diameter, the effective f-number is multiplied by the teleconverter factor.
So:
- new f-number = original f-number × teleconverter magnification
To convert that to stops, use:
- stops lost = 2 × log2(magnification)
Common examples:
- 1.4× TC ≈ 1 stop loss
- 1.7× TC ≈ 1.5 stops loss
- 2× TC = 2 stops loss
Examples:
- 100mm f/2 with a 2× TC becomes effectively 200mm f/4
- 135mm f/2 with a 1.4× TC becomes effectively 189mm f/2.8
This isn’t mainly because the teleconverter “blocks” that much light; it’s because the focal length gets longer while the aperture opening does not get larger, so the f-number rises. In practice, there may also be a small additional transmission loss from the extra glass, but the standard stop change is determined by the magnification factor above.
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