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Focal length eyepiece for telescope

by grouper
Tags: eyepiece, focal, length, telescope
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grouper
#1
Nov18-11, 04:27 PM
P: 52
1. The problem statement, all variables and given/known data

Suppose that you wish to construct a telescope that can resolve features 7.0 km across on the Moon, 384,000 km away. You have a 2.2 m-focal-length objective lens whose diameter is 10.5 cm. What focal-length eyepiece is needed if your eye can resolve objects 0.10 mm apart at a distance of 25 cm?

2. Relevant equations

M=f(objective)/f(eyepiece)

RP=resolving power=s=fθ=(1.22*λ*f)/D where s=distance between objects, f=focal length, D=diameter, and θ=angle between objects

or θ=(1.22*λ)/D

1/f=1/d(object)+1/d(image)

m=h(image)/h(object)=-d(image)/d(object)

3. The attempt at a solution

I'm sure this problem uses the resolution equations somehow but I'm not sure how to go through two lenses with that.

I tried to figure it out another way using the lens and magnification equations above. I used do=3.84e8 m and f=2.2 m to figure out that di≈2.2 m and then using the magnification equation of hi/ho=-di/do to get hi=-4.0104e-5 m.

This image can then be used as the object for the eyepiece and since we want hi=0.0001m (the distance apart for the final image is supposed to be 0.10 mm) and di=-.25m (25 cm from the person and negative because the image will be on the same side as the object if do is positive), we can again use hi/ho=-di/do to get do=-0.10026m. With 1/f=1/do+1/di then, I got f=-0.07156 m.

Not only did I go about this in a roundabout way that I'm sure the problem writer did not intend, my method also yielded the wrong answer. (I tried a positive value for f as well but that's not correct either.) I'm not concerned with correcting what I did wrong above because I'm sure there's a simpler and more direct way to get the correct answer but I'm not seeing it. Any help would be appreciated, thanks.
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ehild
#2
Nov19-11, 12:52 AM
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Quote Quote by grouper View Post
This image can then be used as the object for the eyepiece and since we want hi=0.0001m (the distance apart for the final image is supposed to be 0.10 mm) and di=-.25m (25 cm from the person and negative because the image will be on the same side as the object if do is positive), we can again use hi/ho=-di/do to get do=-0.10026m.

The object distance is positive for the eyepiece do=0.10026 m.

ehild
grouper
#3
Nov19-11, 08:43 AM
P: 52
Using do=0.10026 m and di=-0.25 m yields f=0.167m, which is also incorrect. Our AI hinted that all of homework problems in this assignment use Rayleigh's criterion though, so I think this is the wrong way to go about it. I'm not sure how to apply Rayleigh to this situation though.

ehild
#4
Nov19-11, 11:51 PM
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Focal length eyepiece for telescope

You can use the resolution formula for the objective lens to find out if those two points 7 km apart on the Moon are resolved in the first image. As the angle of view (7/3.84)10-5 is greater then the resolution angle of the objective lens for the visible light θ=(1.22*λ)/D, it is resolved and can be magnified further. (For the wavelength of 600 nm it is 7˙10-6).
The problem asks the focal length of the eyepiece, and the correct result is f=0.167 m or 16.7 cm.
You would need to use the resolution formula to find the proper diameter of the eyepiece lens. The angle of view of the image is about 6˙10-4 radian, so θ=(1.22*λ)/D must be smaller than that: D>(1.22*λ)/6˙10-4=1.22˙10-3 m, that is the diameter of the eyepiece should be greater than 1.22 mm, but it was not the question.

I wonder why your teacher said your result was wrong. There can be other approximate) equations for a telescope, but you used the basic equations which are true and give correct result.

ehild
grouper
#5
Nov20-11, 07:15 AM
P: 52
Ok, I understand how you did that and I agree the answer should be f=0.167 m either way. It's an online program though and it keeps telling me that that answer is incorrect. It's due tonight and I'm not sure what else to do with this problem.
ehild
#6
Nov20-11, 07:40 AM
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It might be also the number of significant digits or the unit. Try 0.17 m or 16.7 cm ...

ehild
grouper
#7
Nov20-11, 07:54 AM
P: 52
No, it's in meters and automatically only grades the first two significant figures. I'm not sure what's wrong. We both must be making an assumption that isn't correct since we both got the same answer. I have until tonight, I'll look around elsewhere on the web for some hints, but if you think of anything else don't hesitate to drop a note here. Thanks for all the help so far as well.
grouper
#8
Nov20-11, 02:43 PM
P: 52
Luckily we're allowed to guess as many times as we want so I just started plugging in numbers because I figured it had to be around 17 cm. I eventually got the right answer with 10 cm. Not sure how they got that though, I'll have to bring it up with my professor. This online thing we use hasn't always been 100% correct.
ehild
#9
Nov20-11, 03:43 PM
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http://en.wikipedia.org/wiki/Magnification

Angular magnification — For optical instruments with an eyepiece, the linear dimension of the image seen in the eyepiece (virtual image in infinite distance) cannot be given, thus size means the angle subtended by the object at the focal point (angular size). Strictly speaking, one should take the tangent of that angle (in practice, this makes a difference only if the angle is larger than a few degrees). Thus, angular magnification is given by:

[tex] \mathrm{MA}=\frac{\tan \varepsilon}{\tan \varepsilon_0}[/tex],

where ε0 is the angle subtended by the object at the front focal point of the objective and ε is the angle subtended by the image at the rear focal point of the eyepiece.


Telescope: The angular magnification is given by

[tex] M= {f_o \over f_e}[/tex]

where fo is the focal length of the objective lens and fe is the focal length of the eyepiece.

(It is assumed that the focal point of the objective is at the same place as the focal point of the eyepiece, and one looks the image from the rear focal point of the eyepiece. See picture http://en.wikipedia.org/wiki/File:Kepschem.png)

The angle subtended by the object is 7/3.84E5 and the angle subtended by the image at the rear focal point of the eyepiece is 0.01/25 =4E-4.
So the angular magnification fo/fe=22, that is fe=0.10 m.

Uhhhh...

I never look into a telescope from the focal point of the eyepiece. I do not understand why is the magnification defined this way.

ehild
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grouper
#10
Nov20-11, 04:02 PM
P: 52
Yeah, I don't know either. I mean, that explanation makes sense in the context of defining everything that way but it goes against what we learned in class about telescope design and use. Oh well. At least I got the point for that problem.
sophiecentaur
#11
Nov20-11, 04:07 PM
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Quote Quote by ehild View Post
http://en.wikipedia.org/wiki/Magnification

Uhhhh...

I never look into a telescope from the focal point of the eyepiece. I do not understand why is the magnification defined this way.
Do you not arrange for the image to appear at infinity? It's the least tiring place to put it.
ehild
#12
Nov20-11, 10:39 PM
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Quote Quote by sophiecentaur View Post
Do you not arrange for the image to appear at infinity? It's the least tiring place to put it.
Well, you are right that looking at infinity with relaxed eyes is the less trying. But I usually want the image at the distance of my clear vision when reading or looking into a microscope or telescope, wanting to see clear details, and I do not mind if my eyes are not relaxed. And I usually put my eyes very near to the eyepiece of a telescope, and my glasses are very near to my eyes. It is different with a magnifying glass, I put the object near to the focal point when I want maximum magnification, with the image at my far sight.

When a problem asks "What focal-length eyepiece is needed if your eye can resolve objects 0.10 mm apart at a distance of 25 cm?" I understand it that the image is at 25 cm distance from my eyes and it is the same distance of the image from the eyepiece lens as I usually put my eyes at the lens. I accept that the magnification of a telescope is defined this way, but I do not use the arrangement shown in the picture when I want to see something clearly.

ehild
sophiecentaur
#13
Nov21-11, 04:25 AM
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Quote Quote by ehild View Post

When a problem asks "What focal-length eyepiece is needed if your eye can resolve objects 0.10 mm apart at a distance of 25 cm?" I understand it that the image is at 25 cm distance from my eyes and it is the same distance of the image from the eyepiece lens as I usually put my eyes at the lens. I accept that the magnification of a telescope is defined this way, but I do not use the arrangement shown in the picture when I want to see something clearly.

ehild
Isn't that just a way of describing the angular resolution? I don't think it constitutes a suggestion that anyone would necessarily put the image just 25cm in front of their eye, given the option of a more relaxing distance (for hours of start gazing).

Would you have a better suggestion for specifying angular magnification? I can't think of one involving fewer variables.

There is a similar problem in specifying the magnification factor of a magnifying glass (loupe). They stamp a number on the side of a loupe but it only refers to a particular eye / lens setup - it's limited by just how good your powers of accommodation are.
ehild
#14
Nov21-11, 11:05 PM
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You are right, I understood at last how simple it is, and I draw a more clear picture, showing only those rays which go through the centre of the lenses and do not change direction.
The angular size of an object is the angle these rays enclose.

The focal distance of the objective is fo and that of the eyepiece is fe.

The real image of the distant object appears in the focal point of the objective lens and the eyepiece is put with its focal point at the same place, so the virtual image is at infinity.

The object of size So is at distance do from the objective lens. The size of the real image is Si.
The angular size of the object is the same as that of the image at distance fo from the objective lens.

[tex]\alpha=\frac{S_o}{d_o}=\frac{S_i}{fo}[/tex]

The angular size of the image seen from the centre of the eyepiece is

[tex]\beta=\frac{S_i}{fe}[/tex],

which is the same as the angle subtended by the virtual image at infinity,
so the angular magnification is

[tex]M_A= \frac{\beta}{\alpha}=\frac{\frac{S_i}{fe}}{\frac{S_i}{fo}}=\frac{fo}{fe }[/tex].

The angular size of the final image was given as the angle of an object of size S=0.1 mm at 0.25 m distance from the eyepiece:

[tex]\beta=\frac{S}{0.25}=4 \cdot 10^{-4}[/tex]

The angular size of the object on the Moon was

[tex]\alpha=\frac{7000}{3.84 \cdot 10^8}=1.823 \cdot 10^{-5}[/tex]

The angular magnification should be

[tex]M_A=\frac{\beta}{\alpha}=\frac{4 \cdot 10^{-4}}{1.823 \cdot 10^{-5}}=22=\frac{fo}{fe}[/tex]

that means fe=0.1 m.


ehild
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sophiecentaur
#15
Nov22-11, 03:44 AM
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That simplified diagram shows it all nicely.
Takes me back to Mr. Scales, my hero.
ehild
#16
Nov22-11, 04:22 AM
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Quote Quote by sophiecentaur View Post
That simplified diagram shows it all nicely.
Takes me back to Mr. Scales, my hero.
Thank you. Who is Mr Scales?


ehild
sophiecentaur
#17
Nov22-11, 10:51 AM
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My legendary A level Physics teacher. respect


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