Zeiss PhotoScope 85 T* FL: 1800 mm f/5.6!
Has anyone seen this?
Zeiss has a new spotting scope with an embedded 7 megapixel camera. The camera is a 6.4x crop sensor. But the zeiss web page is blatant false advertising.
http://www.zeiss.com/c1256bcf0020be5...25755c006de445
"As a digital camera, the PhotoScope provides a truly superior telephoto lens: photographically the equivalent of a 600mm @ f4.0 to 1800mm @ f5.6 zoom lens on a 35mm camera²"
² Nominally an f2.4 to f3.3 at its true (non-35mm equivalent) focal lengths of 93mm–280mm.
Specifications page:
http://www.zeiss.com/c1256bcf0020be5...257561004d4c12
Effective f/ratio: f4.0 @ 600mm, f5.6 @ 1800mm
MSRP: $6,466.99
Wow! I would have thought that Zeiss engineers would know that just because you crop the sensor and use effective focal lengths doesn't make the f/ratio at that effective focal length.
If anyone is confused, equivalent focal length ONLY applies to the equivalent field of view. It is no different if you crop the image in post processing. To claim effective f/ratio is blatant false advertising and a misunderstanding of basic optics. An 1800 mm f/5.6 lens would have an aperture of 321 mm and deliver markedly different depth-of-field than the Zeiss photoscope. On a normal camera without the loss from the beamsplitter in the photoscope, a true 1800 mm f/5.6 lens would deliver over 14 times the amount of light as the zeiss photoscope (280mm at f/3.3 = 84.8mm aperture; 321^2/84.8^2 = 14.3). If you tried to do low light work with the photoscope, you would find the small pixels and small lens would deliver about 1/40th the signal of a real 1800 mm f/5.6 lens on a full frame camera, or at the other end of the photoscope's zoom: 1/40th the signal of a real 600mm f/4 lens.
I'm stunned that a major optics manufacturer would make such a mistake. And then at the price they are charging for a tiny sensor, smaller than many P&S cameras, what a profit margin they must have.
Roger
