# Optics, Telescope magnification, missing data?

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1. Jun 9, 2012

### Cruikshank

1. The problem statement, all variables and given/known data

From Optics by Frances Sears, 1949, Chapter 6, Problem 11: (I've been working all the problems in the book on my own.)
A crude telescope is constructed of to spectacle lenses of focal lengths 100cm and 20cm respectively. (a) Find its angular magnification. (b) Find the height of the image formed by the objective of a building 200ft high and distant one mile.

2. Relevant equations
Part a is straightforward. Angular magnification of a telescope = -f1/f2, where f1 is the focal length of the objective and f2 that of the eyepiece (ocular in Sears' language). So I got -5x, which matches the back of the book.

I don't know what the relevant formulas are for part b, that is part of the problem.
1/p+1/q = 1/f, of course, and lateral m = -q/p = y'/y. The book claims the answer is y'=1.49 inches. I have no idea where they got that.
Sears gives angular magnification as tan u' / tan u, where u' and u are the angles subtended with and without the instrument.

3. The attempt at a solution

tan u = 200 ft/5280ft, where u is the angle subtended without the telescope.
so tan u = .0379 rad.
u' is the angle subtended with the telescope in use, so with -5x angular magnification, I would think that would make tan u' = .1916 (approximating u=tan u and u' = tan u'.)

The real problem is, I have no idea where the final image is supposed to be. In some problems, the final image is supposed to be at infinity (not that they ever SAY so) but that would make no sense here. I tried to brute force it by finding the image of the first lens, use that as the object of the second lens, and find the final image, and with it the lateral magnification. Unfortunately, that requires me to know the spacing of the lenses, which is NOT given. Again, when looking at infinity I get the general idea that the focal points should overlap so the separation of the lenses should be the sum of the focal lengths, 120cm. However, that gives me a final image height of -40.3 ft, while Sears claims 1.49 inches is correct.

I am clearly missing a piece of information, or an assumption I am supposed to make. This is getting frustrating as I have tried Wikipedia and other optics texts and they all get a little bit vague and hand-wave when it comes to this point. I actually took up optics this summer because I got sick of it being the only topic in intro physics I have trouble teaching to my students. Microscopes and telescopes both are frustrating. This should be simple. I'm getting nearly every odd problem right in the text (it has odd answers in the back), but I finally get to the point I really wanted to know, and I'm STILL stuck. Help?

2. Jun 9, 2012

### cepheid

Staff Emeritus
Part b is concerned with the image of the building formed by just the objective lens. So you have a simple thin lens problem with a single converging lens. As a result, the two equations in red above are all you need. I'm not sure which symbol is supposed to represent what in your notation, but let's say that p is the object distance and q is the image distance. Then:

1/q = (1/f - 1/p)

q = (1/f - 1/p)-1

If you like (optional), you can do a little bit of algebra to find that:

q = pf/(p-f)

I'm way too lazy to deal with unit conversions, especially imperial to metric, so I'll just compute the above like this:

To get a result of 1.00062176 meters for the image distance. Then we just use the fact that:

y' = (q/p)y

and compute this like so: