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Author Topic: Lens magnification factor - a question  (Read 3132 times)

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« on: February 24, 2007, 03:24 »
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Hi all,

given that the sensor size of the XTi is 22.2 x 14.8 mm, a lens with a 1:2 magnification would allow me to fill the photo with a object of the size 32 x 22 mm, right? With a 1:3 magnification the object would need to be about 39 x 24 mm to fill the photo, or?

If that is about right then I think I won't need a macro for my work right away (see my other threads) but could do my isolations also with a very good standard zoom, which would also be the much better walkaround lens.

Many thanks,
Michael
« Last Edit: February 24, 2007, 03:32 by Daneel »


« Reply #1 on: February 24, 2007, 05:04 »
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not a hundred percent sure but i think for the 1:2 lens you need 44.4mmx29.6mm
for the 1:3 lens you would need an object 66.6x44.4mm to fill the frame.

How did you come up with your numbers by the way?


« Reply #2 on: February 24, 2007, 05:06 »
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Hi!

This is what I thought first as well. But then I thought: Ok, if I multiply both the length and the height by 2, won't this then correspond to 4 times the size?

I'm not sure either...maybe someone else can enlighten us :-)

All the best,
Michael

« Reply #3 on: February 24, 2007, 05:32 »
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Yes you have to multiply both the height and width by 2.


« Reply #4 on: February 24, 2007, 05:37 »
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How did you come up with your numbers by the way?

What I did was to calculate the area of the chip first (i.e., 22.2 * 14.8 = ~328 mm^2). Then I took these 328 mm^2 and multiplied them by 3 because of the 1:3 magnification factor which brought me to about 960 mm^2. Then I just tried to find fitting values for height and length which give me about 960 mm^2 if multiplied.

Are you sure that you have to double both height and length to get the object size for a magnification factor of 1:2?

All the best,
Michael


 

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