Stupid question about magnification

Ok, so magnificiation is focal length of the telescope divided by the width
of the eyepiece.

What about when there is no eyepiece? My refractor has some magnification
without one, but I have no clue what it is.

Cerdic wrote:
Ok, so magnificiation is focal length of the telescope divided by the width
of the eyepiece.




Magnification = Telescopes F.L. / Eyepiece F.L.

Sam Wormley wrote:
Cerdic wrote:

Ok, so magnificiation is focal length of the telescope divided by the
width of the eyepiece.



Typical homework problem for Cerdic:

The Astro-Physics refractor telescope that we use in class has an
aperture of 130 mm (5.1 inches). This telescope has a focal
length-to-aperture ratio of 8 (written as f/8). These f-numbers are
written on all camera lenses. This telescope has a focal length of 8
times 130 mm, which equals 1040 mm. The magnification that you get with
a 1040 mm focal length telescope is simply the telescope's focal length
divided by the eyepiece's focal length.

Magnification = Telescopes F.L. / Eyepiece F.L.

For example a 25 mm eyepiece would give a magnification of 1040mm/25mm =
40x. So if you were looking at the moon, it would appear 42 times
bigger. It would look 42 time closer! Below is a table of typical
eyepieces one might use with 1200 mm and 1040 mm focal length
telescopes. Calculate the magnification for the missing entries.

Eyepiece F.L. Magnification


160mm f/7.5 1200mm 130mm f/8 1040mm FoV Relief
__________________________________________________ ____________________

40mm 30x (2.33°) ______ (2.7°) 70° 20mm
30mm 40x (1.75°) 35x (2.0°) 70° 20mm

25mm 48x (43') 42x (50') 35° 38mm

20mm 60x (70') ______ (81') 70° 20mm
14mm 86x (48') ______ (57') 70° 20mm
10mm 120x (35') ______ (40') 70° 20mm
7mm 171x (24') ______ (28') 70° 20mm
5mm 249x (17') 210x (20') 70° 20mm
3.5mm 343x (12') 297x (14') 70° 20mm

What brand of tennis shoes you use?


Sam Wormley wrote:

Sam Wormley wrote:
Cerdic wrote:

Ok, so magnificiation is focal length of the telescope divided by the
width of the eyepiece.



Typical homework problem for Cerdic:

The Astro-Physics refractor telescope that we use in class has an
aperture of 130 mm (5.1 inches). This telescope has a focal
length-to-aperture ratio of 8 (written as f/8). These f-numbers are
written on all camera lenses. This telescope has a focal length of 8
times 130 mm, which equals 1040 mm. The magnification that you get with
a 1040 mm focal length telescope is simply the telescope's focal length
divided by the eyepiece's focal length.

Magnification = Telescopes F.L. / Eyepiece F.L.

For example a 25 mm eyepiece would give a magnification of 1040mm/25mm =
40x. So if you were looking at the moon, it would appear 42 times
bigger. It would look 42 time closer! Below is a table of typical
eyepieces one might use with 1200 mm and 1040 mm focal length
telescopes. Calculate the magnification for the missing entries.

Eyepiece F.L. Magnification

160mm f/7.5 1200mm 130mm f/8 1040mm FoV Relief
__________________________________________________ ____________________

40mm 30x (2.33°) ______ (2.7°) 70° 20mm
30mm 40x (1.75°) 35x (2.0°) 70° 20mm

25mm 48x (43') 42x (50') 35° 38mm

20mm 60x (70') ______ (81') 70° 20mm
14mm 86x (48') ______ (57') 70° 20mm
10mm 120x (35') ______ (40') 70° 20mm
7mm 171x (24') ______ (28') 70° 20mm
5mm 249x (17') 210x (20') 70° 20mm
3.5mm 343x (12') 297x (14') 70° 20mm

Oh Yea ? wrote:
What brand of tennis shoes you use?


Sam Wormley wrote:


Sam Wormley wrote:

Cerdic wrote:


Ok, so magnificiation is focal length of the telescope divided by the
width of the eyepiece.


Typical homework problem for Cerdic:

The Astro-Physics refractor telescope that we use in class has an
aperture of 130 mm (5.1 inches). This telescope has a focal
length-to-aperture ratio of 8 (written as f/8). These f-numbers are
written on all camera lenses. This telescope has a focal length of 8
times 130 mm, which equals 1040 mm. The magnification that you get with
a 1040 mm focal length telescope is simply the telescope's focal length
divided by the eyepiece's focal length.

Magnification = Telescopes F.L. / Eyepiece F.L.

For example a 25 mm eyepiece would give a magnification of 1040mm/25mm =
40x. So if you were looking at the moon, it would appear 42 times
bigger. It would look 42 time closer! Below is a table of typical
eyepieces one might use with 1200 mm and 1040 mm focal length
telescopes. Calculate the magnification for the missing entries.

Eyepiece F.L. Magnification

160mm f/7.5 1200mm 130mm f/8 1040mm FoV Relief
________________________________________________ ______________________

40mm 30x (2.33°) ______ (2.7°) 70° 20mm
30mm 40x (1.75°) 35x (2.0°) 70° 20mm

25mm 48x (43') 42x (50') 35° 38mm

20mm 60x (70') ______ (81') 70° 20mm
14mm 86x (48') ______ (57') 70° 20mm
10mm 120x (35') ______ (40') 70° 20mm
7mm 171x (24') ______ (28') 70° 20mm
5mm 249x (17') 210x (20') 70° 20mm
3.5mm 343x (12') 297x (14') 70° 20mm



Take out the 25mm Collins Electro Optics Eyepiece and assume the others
are a series from one manufacturer.

On 2006-01-18, Cerdic wrote:
Ok, so magnificiation is focal length of the telescope divided by the width
of the eyepiece.

What about when there is no eyepiece? My refractor has some magnification
without one, but I have no clue what it is.

The telescope makes a real image that you can see with the unaided eye -
an "aerial image". The magnification is the focal length of the telescope
divided by the distance from your eye to the aerial image.

--
The night is just the shadow of the Earth.

William Hamblen wrote:

The telescope makes a real image that you can see with the unaided eye -
an "aerial image". The magnification is the focal length of the telescope
divided by the distance from your eye to the aerial image.

I'll expand on this. Apparent size of the prime image does depend on
what distance
it is looked at from, but it is usually taken to be the least distance
of clear vision,
250mm or 10". It is the maximum apparent magnification an objective can
give:
f(mm)/250, or f"/10. An eyepiece does the trick of making it possible
to observe this image from the distance of its focal length. Thus,
eyepiece magnification is given by 250/f*
for the ep f.l. (f*) in mm. Telescope magnification is a product of the
two, objective and eyepiece magnification, M=(f/250)(250/f*)=f/f*.

Vlad

On 2006-01-18, nick wrote:

William Hamblen wrote:

The telescope makes a real image that you can see with the unaided eye -
an "aerial image". The magnification is the focal length of the telescope
divided by the distance from your eye to the aerial image.

I'll expand on this. Apparent size of the prime image does depend on
what distance
it is looked at from, but it is usually taken to be the least distance
of clear vision,
250mm or 10". It is the maximum apparent magnification an objective can
give:
f(mm)/250, or f"/10. An eyepiece does the trick of making it possible
to observe this image from the distance of its focal length. Thus,
eyepiece magnification is given by 250/f*
for the ep f.l. (f*) in mm. Telescope magnification is a product of the
two, objective and eyepiece magnification, M=(f/250)(250/f*)=f/f*.

I was a little vague in the matter of eye distance because it varies
with the eye. In my younger days I could focus on the end of my nose.
Nowadays close focus is at an arm's length. 250 mm is out of the question.

For readers of this thread who haven't tried it, looking at the aerial
image is a neat trick. The moon works well because it is bright.
The image seems to float in the air. You can compare the size of the
aerial image to the moon by glancing from one to the other. This shows
that you do get a magnified view.

BTW, the image is called aerial because it is not projected onto a ground
glass or other surface.

--
The night is just the shadow of the Earth.

"nick" wrote in message
ups.com...

William Hamblen wrote:

The telescope makes a real image that you can see with the unaided eye -
an "aerial image". The magnification is the focal length of the
telescope
divided by the distance from your eye to the aerial image.

I'll expand on this. Apparent size of the prime image does depend on
what distance
it is looked at from, but it is usually taken to be the least distance
of clear vision,
250mm or 10". It is the maximum apparent magnification an objective can
give:
f(mm)/250, or f"/10. An eyepiece does the trick of making it possible
to observe this image from the distance of its focal length. Thus,
eyepiece magnification is given by 250/f*
for the ep f.l. (f*) in mm. Telescope magnification is a product of the
two, objective and eyepiece magnification, M=(f/250)(250/f*)=f/f*.

It seems to me that for astronomical telescopes, the distance one has to use
for this formula is the closest distance at which the human eye is at
infinity focus.

Stephen Paul wrote:
"nick" wrote in message

...Telescope magnification is a product of the
two, objective and eyepiece magnification, M=(f/250)(250/f*)=f/f*.

It seems to me that for astronomical telescopes, the distance one has to use
for this formula is the closest distance at which the human eye is at
infinity focus.

Why? It expresses apparent sizes of object's image, not object itself.

Vlad