# Magnification problems

1. Sep 23, 2009

### estanton

1. The problem statement, all variables and given/known data
A microscope with a tube length of 200 mm achieves a total magnification of 800\times with a 40 \times objective and a 20\times eyepiece. The microscope is focused for viewing with a relaxed eye.
How far is the sample from the objective lens?

2. Relevant equations
1/s+1/s'=1/f

3. The attempt at a solution
I calculated the two focal lengths for the lenses to be 1/40=0.025m (objective) and 1/20=0.05 m (eyepiece).
Next I calculated the position of the virtual image projected by objective lens. Assuming the final image should be at -25cm the virtual image would be at 4.16cm before the objective lense.
Since there is 20cm between the two lenses this means that the image projected by the objective lens would be at 15.84cm behind the objective lens. Finally I calculated that the object should be at 2.97 cm before the objective lens. However Mastering Physics is telling me this answer is not correct.
Can anyone help steer me in the right direction? Thanks

2. Sep 24, 2009

### rl.bhat

Magnification Mo = L/fo
Magnification of eye-piece = Me = 25 cm/fe.

3. Sep 24, 2009

### estanton

that would make the focal distances 0.5cm and 1.25 cm for the objective and eyepiece lenses respectively? Using those foci I still do not get the correct answer (0.51 cm).