How do you compare 2 lenses with different focal lengths and minimum focusing distances? e.g. comparing the Nikon DX Micro 40 mm f/2.8 G and the Nikon DX Micro 85 mm f/3.5. The 40 mm's minimum focusing distance is 16.3 cm, the 85 mm's is 28.6. So which will give a bigger image?
Based on this article: http://petapixel.com/2013/06/15/a-mathematical-look-at-focal-length-and-crop-factor/ and something I learned in astronomy.
The gist is:
- The lens can be simplified to a single lens at a distance of the focal length from the sensor
- Similar triangles are used: sensor width / focal length = object width / focusing distance
Using similar triangles, sensor width/2 /focal length = half minimum focusing width / focusing distance
What is half minimum focusing width? An object twice this length will, at full magnification, fill the width of the image. Smaller is better. Does this check out? For the iPhone 5, it does seem to check out. I can still focus on 7 cm, 6 cm is a bit fuzzy, and 5 is right out. For the ZR200, it doesn't seem to work. The shortest thing I can focus on is 1.9 cm long.
So between the 2 Nikon lenses, the 85 mm will have a bigger magnification. Not that impressive, really. They can't even magnify as much as the iPhone 5, although the longer focal length means the camera will be positioned further away, which means the perspective will be less distorted. To compare, the ZR200's minimum focusing distance at 86 mm focal length is 30 cm, similar to the 85 mm Nikon lens. But if I want to maintain focus at 30 cm away, I can zoom the focal length all the way to 110 mm. So... it looks like, "Anything you can do I can do better"?
Anyway, the final magnification depends on the resolution of the sensor. For the Nikon lenses, they'll be used on the same sensor, so it doesn't matter. But if you want to compare different cameras, you could divide the HMFW by the number of pixels along the width to get the "resolution" (lower is better). The real resolution won't be this high, because of noise and diffraction.
Ok I noticed people normally use "length" for the longer side and "width" for the shorter", but here I'm using "width" and "height", and "width" is the longer side.
Update 26 Nov 13
Ok I figured the best way to compare magnification is to take the most closeup picture of a ruler you can. The smaller the width you can take, the higher the magnification. Examples:
Nikon AF-S 105mm f/2.8 VR Micro-NIKKOR macro lens
I had the opportunity to play with a "true" macro lens and here are my observations:
- It's HUGE and heavy. I thought of using it as an everyday lens, but its bulk combined with the long focal length makes it impractical. I guess its length is unavoidable, since focal length is the distance between some part of the lens and the sensor, so it has to be at least 105mm long - like a zoom lens.
- The camera (D7000) had difficulty focusing it. Not sure if it's a lens or camera problem. Not to say focusing was bad - when it worked, it was perfect. But it often could not find the focus, or stopped when it was obviously out of focus.
- You need to use a very narrow aperture. Don't even think of using the f/2.8. At first my pictures weren't as good as those from my compact camera and I didn't know what was wrong. The wide aperture gave a very narrow Depth Of Field (DOF). I ended up using f/25, and the flash. This takes away some of the advantage of less noise of SLRs, since you have to use a higher ISO.
It looks like my formula above doesn't work. I thought:
- On the camera side of the lens: half sensor width / focal length =
- On the subject side: half subject width / focusing distance