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Physics 6 Online
OpenStudy (anonymous):

An object is 10.0 mm from the objective of a certain compound microscope. The lenses are 300 mm apart, and the in-termediate image is 50.0 mm from the eyepiece. What overall magnification is produced by the instrument?

OpenStudy (anonymous):

The total magnification will be: \(M=\frac{250mm}{10mm} \times \frac{250mm}{50mm}=125times\) The first being the objective magnification, second being eyepiece magnification.

OpenStudy (anonymous):

So the answer is 125 times?

OpenStudy (anonymous):

Yup! This applies to normal eyes of near point=25cm

OpenStudy (anonymous):

Thank you so much!

OpenStudy (anonymous):

You're welcome :)

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