Question

A compound microscope has an objective of focal length 0.5 cm and a tube length of 15 cm. If it produces an angular magnification of 225 in normal adjustment, find the focal length of the eyepiece.

Answer 1

Linear magnification is $m_o=\frac{tubelength}{focallengthobjectivelength}$

$m_o=\frac{15}{0.5}=30$

Let the angular magnification be $m_e$

Total magnification is $M=m_o{m}_e$

$225=30\times m_e$

$m_e=7.5$

Let the focal length of eye piece be $f_e$.

Also $m_e=\frac{25}{f_e}$

$f_e=\frac{25}{7.5}=3.3cm$