Re: Question about Canon EF 180mm f3.5L Macro USM AutoFocus Telephoto Lens


"Dave" <dave2681@xxxxxxxxxxx> wrote:
Thanks all! How would I figure the minimum distance from the front of the
lens to the subject that would be possible to obtain focus?

Subject distance is measured from the film plane. For this lens, presumably
the "min_focal_range: 18.9 inches" spec refers to that. Subtract the length
of the lens (7 inches) and the flange to film/sensor distance (44mm, 1.5
inches or so), and you get about 10 inches.

Note that at that distance, you will be imaging an area 24x36mm with the 5D
or 15x23mm with an APS-C camera.

FWIW, the English on the US site is a bit better than that in the specs you
quoted.

http://www.usa.canon.com/consumer/controller?act=ModelInfoAct&fcategoryid=155&modelid=7324

David J. Littleboy
Tokyo, Japan


"Dave" <dave2681@xxxxxxxxxxx> wrote in message
news:QPuWi.30341$aJ3.27338@xxxxxxxxxxx
Reviews often speak of this lens as an excellent telephoto lens.
I guess I'm rather ignorant about the lens specs:
* Lens Type: teleconverter
* mounting_type: Canon EF
* Camera Magnify Power: 1 / 1
* lens_system_special_functions: Macro
* maximum_aperture_range: F/3.5
* Minimum focal length: 180 millimeters
* min_focal_range: 18.9 inches
* Focus Type: auto-focus, manual-focus
* Real Angle Of View: 13.5
* Weight: 2.4 pounds
* Item Display Diameter: 3.3 inches
* Length: 7 inches

Zoom is often referred to as 6X, 10X etc. Can anyone tell me what X would
this be equivalent to?



.

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