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.

 

[tomcue]

for providing the equations you used to solve this problem. It seems that you have made some errors in your calculations, which is why your final answer does not match the correct solution.

Firstly, let's clarify the difference between angular magnification and apparent magnification. Angular magnification refers to the ratio of the angle subtended by the image at the eye to the angle subtended by the object at the objective lens. It is given by Mangular = 25/fe. On the other hand, apparent magnification refers to the ratio of the size of the image as seen through the eyepiece to the actual size of the object. It is given by Mapparent = (25/f)+1.

Now, let's go through your calculations. You correctly solved for the focal length of the eyepiece (f=25/9). However, your calculation for the focal length of the objective lens is incorrect. The correct equation to use is 1/i+1/o = 1/f, where i is the image distance, o is the object distance, and f is the focal length. Plugging in the values you know (i=20o and f=25/9), you get o=45/8.

Next, to find the total magnification, you can use the equation Mtotal = -(i/o)*[(25/f)+1]. Plugging in the values you know (i=20o, o=45/8, and f=25/9), you get Mtotal = -200/9. This means that the total magnification is -200/9, which is equivalent to -22.22.

Finally, to find the object distance for viewing the final image, you can use the equation 1/i+1/o = 1/f. Plugging in the values you know (i=20o and f=25/9), you get o=45/8. This means that the object distance should be placed at 45/8 cm from the objective lens.

In summary, the focal length of the objective lens is 45/8 cm, the total magnification is -22.22, and the object distance should be placed at 45/8 cm from the objective lens for viewing the final image. I hope this helps clarify any confusion and helps you in solving future problems involving microscope magnification.