Microscope magnification question

[academi4]

Homework Statement

A compound microscope has a length of 25 cm. The objective lens has a linear magnification of -20cm, and the eyepiece has an apparent magnification of 10, what are their focal lengths (fe and fo)?
How close to the objective lens should the object be placed for viewing the final image?
What is the total magnification?

Homework Equations

Mapparent = (25/f) +1
Mlinear = -i/o
Mtotal = -(i/o)*[(25/f)+1]
1/i + 1/o = 1/f
Mtotal = -25i/(fe * fo)

The Attempt at a Solution

I tried plugging in everything I know:

10 = (25/f)+1
solving for f
f= 25/9

-20 = -i/o
-i=-20o
i=20o

1/i+1/o = 1/f
1/(20o) + 1/o = 9/25
o=35/12

i=175/3

now I plug in i,o,and f to find Mtotal = -200

Now, my book says the object distance is approximately the focal length of the objective lens.

So, I guess o = focal of the objective lens.

fo = 35/12

Now, using:
Mtotal = -25i/(fe * fo)
solving for fe gives:
fe = 5/2

I think this is wrong though. Using this website:

http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/micros2.html#c1

If I put the length of the microscope = 25cm, fo = 12.5mm, fe = 25mm I get the correct answers for total magnification and linear magnification.

Apart from my book I cannot find anything about apparent magnification, so I doubt this equation in my book:

Mapparent = (25/f) +1

The website I linked to has an equation for Mangular = 25/fe

My book does not talk about angular magnification.

Any idea? Are M angular and M apparent the same?

Thanks

 

[tiny-tim]

Hi academi4!

From http://www.telescope-optics.net/telescope_magnification.htm" …

Apparent magnification of the objective is given by the ratio of the viewing angle of its object-image from the least distance of distinct vision (250mm average) to the viewing angle of the object observed directly.

So apparent magnification and angular magnification are the same: M_e = 25/f_e.

See the site for more details.