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

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

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