Helen B

10-23-2006, 01:38 PM

Would some example numbers help? A slightly simplified example:

Think of a 100 mm (10 D) lens set at f/8 and focussed at infinity. The rear nodal point is 100 mm from the film plane, and the diameter of the effective aperture (entrance pupil) is 12.5 mm.

Now add a +3 diopter lens, and suppose that it is 50 mm in diameter. Being simple, it has an exit pupil of 50 mm, which is larger than the entrance pupil of the main lens.

Because of the diopter, the lens is now focussed on an object 1/3 m away, ie 333 mm. The combined lens has a focal length of 13 D or 77 mm. The rear node is still 100 mm from the film plane and the object is 333 mm from the front node, so the magnification is 0.3x.

If the entrance pupil diameter was unchanged you might say that f/8 had changed to f/6.2 because the focal length of the lens was now 77 mm – in fact it would be marked as f/6.2 if it was marked for infinity. But the rear node is 100 mm from the film, not 77 mm, so the effective aperture is still 12.5/100 mm, ie f/8. Magic.

In fact the entrance pupil may be larger, because of the magnifying effect of the diopter. It could be about 5% larger, so 12.5 mm would change to 13.1 mm and the aperture would be 13.1/100 = f/7.6. Offsetting that would be the extra air-glass reflection losses, so you could be back where you started from.

If you wanted the same magnification with only the 100 mm lens without a diopter the rear node would be 130 mm from the film plane and the effective aperture would be 12.5/130 = f/10.4 when the lens was set at f/8. If the lens was set at f/6.2, the effective aperture would be f/8.

If you wanted the same magnification with a 77 mm lens set at f/8 with no diopter, the diameter of the entrance pupil would be 77/8 = 9.62 mm and the rear node to film distance would be 100 mm (as above), giving an effective aperture of 9.62/100 = f/10.4

Anupam’s statement that "effective aperture and effective focal length being equal, tubes and diopters need the same exposure for the same magnification" appears to be in agreement with the above simplified calculations. The point is that, in this situation, with a diopter the marked aperture is close to the effective aperture; but with an extension tube with no diopter the marked aperture is not the effective aperture. More or less.

Best,

Helen

Think of a 100 mm (10 D) lens set at f/8 and focussed at infinity. The rear nodal point is 100 mm from the film plane, and the diameter of the effective aperture (entrance pupil) is 12.5 mm.

Now add a +3 diopter lens, and suppose that it is 50 mm in diameter. Being simple, it has an exit pupil of 50 mm, which is larger than the entrance pupil of the main lens.

Because of the diopter, the lens is now focussed on an object 1/3 m away, ie 333 mm. The combined lens has a focal length of 13 D or 77 mm. The rear node is still 100 mm from the film plane and the object is 333 mm from the front node, so the magnification is 0.3x.

If the entrance pupil diameter was unchanged you might say that f/8 had changed to f/6.2 because the focal length of the lens was now 77 mm – in fact it would be marked as f/6.2 if it was marked for infinity. But the rear node is 100 mm from the film, not 77 mm, so the effective aperture is still 12.5/100 mm, ie f/8. Magic.

In fact the entrance pupil may be larger, because of the magnifying effect of the diopter. It could be about 5% larger, so 12.5 mm would change to 13.1 mm and the aperture would be 13.1/100 = f/7.6. Offsetting that would be the extra air-glass reflection losses, so you could be back where you started from.

If you wanted the same magnification with only the 100 mm lens without a diopter the rear node would be 130 mm from the film plane and the effective aperture would be 12.5/130 = f/10.4 when the lens was set at f/8. If the lens was set at f/6.2, the effective aperture would be f/8.

If you wanted the same magnification with a 77 mm lens set at f/8 with no diopter, the diameter of the entrance pupil would be 77/8 = 9.62 mm and the rear node to film distance would be 100 mm (as above), giving an effective aperture of 9.62/100 = f/10.4

Anupam’s statement that "effective aperture and effective focal length being equal, tubes and diopters need the same exposure for the same magnification" appears to be in agreement with the above simplified calculations. The point is that, in this situation, with a diopter the marked aperture is close to the effective aperture; but with an extension tube with no diopter the marked aperture is not the effective aperture. More or less.

Best,

Helen