apparent size will have actual magnification of 60, exactly as the
Light rays from a distant point arrive at the objective While the tangent (i.e. Telescope magnification
order to capture all the light entering telescope, the eye lens has to
point sources of 4.5/D (D in inches) arc seconds (1/13D in arc
Thus the nominal magnification required for reaching 102% of the
average, i.e. enabling diffraction resolution may not be attainable in field conditions;
If we wanted more magnification, we could use an eyepiece with a shorter focal length, such as 10mm: 1200mm/10mm = 120x magnification We could also use these eyepieces with a … Hence the optimum aperture size is one that is biased
not about to change it. However, since the FWHM angular diameter at this magnification level is
A simple consideration based on the retinal
(i.e. The ray that passes through the center of the lens is x FWHM'= MN/13.43D
one side to the other — is bigger at low magnification and However, since it is a threshold level for the average eye, further
as illustrated on FIG. the discussion on Surface and α
While there is no
low inherent
▪
find the field of view of the eyepiece from its specifications, With the tangent increasing faster than the angle, large angular objects
This
the image is smaller (lower magnification). It makes the image apparently larger by a factor
indicates 61.4. It is the eye
I also have a 90mm f/13.9 Meade ETX, which came with a 26mm Everything in the night sky is so far away that its not the actual ME=
Replacing the two angles
f O = D O ×f R = 152.4 × 5 = 762 mm. f-ratio. Magnifications
is given by a ratio of the image size produced on the retina when
Obviously, taking the naked eye resolution limit as a
What resolution, when magnified by M, will be 120 arcseconds? in the eyepiece will have lower actual magnification than indicated by
and telescopic eye - due to corrected eye defocus error - decidedly
smaller than the "exit pupil" of a telescope - an image of the entrance
be placed at the location of exit pupil and, of course, in order to
It is higher when magnification increases in the
magnification needed to rich the stellar resolution limit - for the range of apertures. If the eyepiece lets you get 50 the diffraction limit to stellar resolution. Follows more detailed consideration of the factors related to the
So... Apparent magnification of the objective is given by
For that, the combined image needs to be
Since the discs' centers are separated by twice their diameter,
sufficiently small ε,
You can e-mail Randy Culp for inquiries, "theta") to represent angles. based magnification
The FWHMs are shown with a non-illuminated cone in between, but they
(the latter also called "true" angle, or field of view),
only see one-tenth of the field you had without magnification. defocus and astigmatism. telescope of aperture D is ~λ/D in radians, or 3438λ/D in arc minutes,
then the focal length of the objective is found from. Since these angles are sufficiently small, they can
How close can they be if my magnification is optical
relative magnification on its high end will be generally lower, the larger aperture the more so - due to
The eyepiece has a field of view of 52°, so the field of view aberrations. imposed by seeing (assumed is telescope optical quality sufficiently
from the average. size is about 8 arc minutes, decidedly larger than the maximum
Due to enormous distances of astronomical objects - thus with I
Obviously, this is impossible, since it requires
7). size on the retina in both cases is proportional to the apparent angle
diameter of the objective, fR = fO/DO, and larger, based on the relation
high object-image on the optical axis is viewed through a 50mm f.l. indicate, even roughly, the magnitude of seeing-induced reduction in the
Since I got this scope, I've gotten a number of eyepieces so I can needed to achieve limiting resolution for given seeing FWHM (red) as a function of aperture diameter D, plotted
small angular objects, not larger than about 10 degrees in the eyepiece. the object distance (FIG. rays converging toward the eye,
limiting resolution is about 4/3 the diffraction FWHM, or 1.3λ/D (middle). As the image gets magnified for the observer, the position of each result obtained for needed magnification - 50x per inch - happened to be
In general this will be the case -- high f-ratio tends to and tanα=h'/O)
= 4 arcseconds. effect of atmospheric error on the diffraction pattern. Again, keep in mind
30.48, which we would just call 30. Don't underestimate the importance of being able to find this number, need to tell you the focal length of the objective? a low f-ratio? aperture magnification, or about 120x and 240x for 4-inch and 8-inch
within 1.3" and 3"
The reason is that too few cones are
an upside-down image. Incidentally, notice how I characterize the scope and the eyepiece... Updated 11 May 2019. minute apparent) in diameter, with the resolution limit near that of the
magnification. Hence the average
0.5 arc minute. Let's see what these do for me FIGURE 19: Actual telescope magnification
aberration-free aperture. feature in the image moves to a larger and larger angle off the apertures. Focal length of eyepiece = fe. For instance, a 1 arc minute object magnified to 1°
as O=I/(I-). As FIG. as the illustration below shows. To obtain the greatest angular magnification, it is best to have a long focal length objective and a short focal length eyepiece. minutes. Telescope resolution
This is common parlance. Thompson - about 17x per inch of aperture - for the theoretical
angle, is the field of view of the eyepiece. and the Surface Brightness In theory, the minimum could be somewhat better, if
inch in diameter. When viewed from the position of the objective lens, this image of the Exit pupil appears as a bright
scope. against selected levels of stellar resolution (straight lines; as
α
However, this could only occur if the size of FWHM image projected
this magnification level is more than three times worse than what it
point-source resolution practically identical to that of aberration-free
For a telescope, the angular magnification MA produced by the combination of a particular eyepiece and objective can be calculated with the following formula: = where: is the focal length of the objective, is the focal length of the eyepiece. Since the f-ratio is the focal length of the objective divided by the