looksharp65
Well-known member
This might not prove helpful, possibly rather irrelevant, but I still feel the urge to post it.
Suppose you're using a 30x75 scope and a P&S camera with a 6 mm f/3 lens. The eyepiece is of the wide-angle type and there's no vignetting.
The illumination within the exit pupil is compromised by the scope's transmission rate.
The exit pupil of the scope delivers a beam pencil that is 2.5 mm wide.
The maximum entrance pupil of the P&S camera is 2 mm.
This means that the EV will be the same through the scope as outside of it, apart from the minor light loss by the scope's transmission percentage.
If the 6 mm lens is equivalent to a 30 mm lens in full format and the magnification is 30x, that will yield the same magnification as a 900 mm lens. (30x50)*(30:50) = 900
But since the beam is wider than the camera's aperture, the latter will be the limiting factor. Thus, we have got a 900 mm f/3 equivalent!
//L
Suppose you're using a 30x75 scope and a P&S camera with a 6 mm f/3 lens. The eyepiece is of the wide-angle type and there's no vignetting.
The illumination within the exit pupil is compromised by the scope's transmission rate.
The exit pupil of the scope delivers a beam pencil that is 2.5 mm wide.
The maximum entrance pupil of the P&S camera is 2 mm.
This means that the EV will be the same through the scope as outside of it, apart from the minor light loss by the scope's transmission percentage.
If the 6 mm lens is equivalent to a 30 mm lens in full format and the magnification is 30x, that will yield the same magnification as a 900 mm lens. (30x50)*(30:50) = 900
But since the beam is wider than the camera's aperture, the latter will be the limiting factor. Thus, we have got a 900 mm f/3 equivalent!
//L