How to explain the blurring while using a USAF chart to determine the resolution

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Originally Posted by xpatUSA

The tube is 400mm, not 40mm, but I get the idea . . .

I still have no idea what "image distance" means, however :(

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From lens end to subject / subject to focal plane etc.

Wow - 400mmm extension that would most certainly do something to the F16.

John
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Specially for those that say they don't have difficulties with light http://hyperphysics.phy-astr.gsu.edu...pt/lenseq.html

@John,

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The only effect is that the F ratio is changed by the extension tube - the rad of the spot is 1.22 * wavelength * focal length / lens diameter. Image distance is generally taken as not to matter. So with the set up the diffraction spot will increase by 240/200. Not that it matters because a lens of this nature is really unlikely to be capable of producing a sensible one anyway.

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Let's use the corrected extension value. Your calculation will be 600/200. The same factor 3.
The 600 stands for the image distance.

George

Gentlemen, gentlemen,

According to Wikipedia, the angle in radians to the first minimum is:

theta = 1.22(lambda/aperture diameter)

(small angle approximation)

The aperture diameter d = f/N = 200/16 = 12.5mm

So, theta, for the usual green light value of lambda is 1.22(555e-9/12.5e-3) = 1.3225e-3 rad.

The "image distance" for a magnification of 1:2 is maybe 700mm, where "image distance" means whatever I mean it to mean today :)

The disk radius in the image plane is, therefore, tan(theta) x the "image distance" = tan(1.3225e-3)x700e-3 = 926um

and, therefore, the Airy disk diameter (to the first minimum) is . . . . er, quite a lot bigger than the OP's "10um object" :) However, we can say that this large blur circle will be reproduced very well by the sensor.

Do pardon my pedantry. I'm getting used to a new scientific calculator so please excuse any arithmetic errors. The formulae above are correct though, IMNSHO.

Quote:

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Originally Posted by xpatUSA

Gentlemen, gentlemen,

According to Wikipedia, the angle in radians to the first minimum is:

theta = 1.22(lambda/aperture diameter)

(small angle approximation)

The aperture diameter d = f/N = 200/16 = 12.5mm

So, theta, for the usual green light value of lambda is 1.22(555e-9/12.5e-3) = 1.3225e-3 rad.

The "image distance" for a magnification of 1:2 is maybe 700mm, where "image distance" means whatever I mean it to mean today :)

The disk radius in the image plane is, therefore, tan(theta) x the "image distance" = tan(1.3225e-3)x700e-3 = 926um

and, therefore, the Airy disk diameter (to the first minimum) is . . . . er, quite a lot bigger than the OP's "10um object" :) However, we can say that this large blur circle will be reproduced very well by the sensor.

Do pardon my pedantry. I'm getting used to a new scientific calculator so please excuse any arithmetic errors. The formulae above are correct though, IMNSHO.

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I think you still don't understand it.
George

Quote:

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Originally Posted by xpatUSA

I still have no idea what "image distance" means, however :(

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This might help explain it.

http://www.rags-int-inc.com/PhotoTec...agram_1024.gif

No way am I going to ray trace one to find out but I doubt if it is a 200mm focal length set up any more. Briefy

A telephoto lens is one who's length is significantly less than it's focal length ie Not a telescope. They do this by including lenses with a negative focal length behind / mixed in with lenses with a positive focal length. The focal distance setting will be close to reality if the lens is mounted at the correct flange distance from a sensor. Stick an extension on of say 400mm and it could be said that it is now a lens of 600mm focal length. Trouble is that when things are moved to focus with it fitted that figure is rather unlikely even more so than it obviously is.

If some one really must get a feel for the likely out come sums around Barlow lenses might help. Not usually viewed this way but they are negative focal length lenses that convert telescopes into telephoto lenses. The further they are into the back focal point of the front element the more effect they have but the image position shifts as well.

Ted's angular radius is correct and illustrates why F ratio sets spot size and also in a round about way why focal length sets image scale so bigger apertures have higher resolution - hence n meter telescopes. Apertures literally not aperture settings, not the goofy use of the word aperture used by photographers who should use the term F stops.

John
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