(17250)

Flying Rock Obsevatory in Fountain Hills, AZ

>1930s Zeiss 4HF12 binoculars (photo is at bottom of page):

> http://gator.naples.net/clubs/eas/calusa_nature_center.htm

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Subject: Exit Pupil Questions

From: "Rafael Chamón Cobos" <rchamon@___s>

To reply to Kenny's exit pupil questions I would recall the following, although I am not by no means an expert in Optics.

The resolving power of an instruments is limited, on one hand, by the geometric faults (spherical aberration, coma, astigmatism, etc.), and on the other hand, by the diffraction fenomena that take place at the edges of the lens mounts and stops. The smaller the diameter of the stop or lens is, the greater is the diffraction. It can be proven that the resolving power of a telescope depends only on the diameter of the objective lens. The formula is: resolving power in secondes of arc at the object = 120 / Diameter of the objective in millimeters.

(This teoretically, because the magnifying factor sets a limit to the resolving power, and this leads to the concept of "resolving magnifying factor", in which the physiological resolving power of the eye is involved. The resolving magnifying factor corresponds to the maximal magnifying factor one can expect from the telescope of a given objective diameter, and this is equal to the radius of the objective in mm. For example, if we consider a telescope of 70 mm, it is possible to achieve magnifying factors up to a limit of 35x without loosing definition. Over this value of magnifying factor one will see the image greater but blurred, this is, without details).

Responses

To the case (1):

Perhaps the lost of qualiyt perceived by Kenny when he masks down the objective lenses of a binocular is due to the fact that in this case the diffraction faults will increase. By stopping down the objective lenses of a 7x 42 bino to 25 mm the resolution will be reduced to 60% (=25/42). By stopping down the objective lenses of a 10x 70 bino to 30 mm the resolution will be reduced to 43% (=30/70).

To the case (2):

In case of reducing the exit pupil by a stop I think the result would be similar as in the case (1) The exit pupil is an irreal stop, located between the eyepiecethe and the eye, that corresponds to the image of the objective after passing the telescope. Therefore, stopping down the exit pupil will be equivalent to stopping down the objective, as in case (1).

To the case (3):

In this case the eye pupil stops down the beam of rays comming into the eye. This case would also be similar to the case (1), because the exit pupil is now determinated by the eye itself, and this new exit pupile is translated back to the objective by the whole optical system. This has the effect of stopping down the objective, as in case (1).

Kind regards

Rafael

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