The distance between the eyepiece and the objective lens in a certain compound microscope is $23.0 c m$ . The focal length of the eyepiece is $2.50 c m,$ and that of the objective $0.400 c m$ . What is the overall magnification of the microscope ?

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Solution

Overall magnification of compound microscope is given by formula
$M = - \frac{L}{f_{o}} . \frac{25 c m}{f_{e}}$ where L=distance between the eyepiece and the objective lens,
$f_{o}$ and $f_{e}$ are focal lengths of objective lens and eyepiece lens respectively.
substituting values in formula
$M = - \frac{23}{0.4} . \frac{25}{2.5}$
$M = - 575$
overall magnification of given compound microscope is -575.

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The distance between the eyepiece and the objective lens in a certain compound microscope is 23.0cm . The focal length of the eyepiece is 2.50cm, and that of the objective 0.400cm. What is the overall magnification of the microscope ?

The distance between the eyepiece and the objective lens in a certain compound microscope is $23.0 c m$ . The focal length of the eyepiece is $2.50 c m,$ and that of the objective $0.400 c m$ . What is the overall magnification of the microscope ?