What is the approximate magnification of a compound microscope with objective and eyepiece focal lengths of 0.3 cm and 3.6 cm, respectively, and a separation between lenses of 20 cm?

Question

The answer is 460.

I tried finding the magnification for the eyepiece which was m=25/f=25/3.6=6.94.
For the objective lens I tried to solve for q by 1/20 + 1/q = 1/.3 and solved for magnification of objective lens to be m=-q/p.. but i didn't get the correct answer.. is the given value of 20cm=p? i think that's where i went wrong.

Answer 1

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Answer 2

A resonable approximation for magnification of a microscope is

M= L/f0 * 25/fe
See:http://hyperphysics.phy-astr.gsu.edu/hbasees/geoopt/micros.html
With that approximation, I get
M= 20/.3*25/3.6=463 and it has to be rounded to two sig figures.

Answer 3

Within a compound microscope, the objective and eyepiece have focal lengths of 0.9 cm and 2.5cm respectively. The distance between the eyepiece and objective is 23.2 cm. A real image formed by the objective is 17 cm from the objective. Where is the object located?