When the image is formed at infinity, a compound microscope has a...

For a compound microscope with the final image formed at infinity, the magnifying power is given by:

$M = |m_o| \times m_e = \left(\dfrac{v_o}{|u_o|}\right) \left(\dfrac{D}{f_e}\right)<br><br>Using the lens formula for the objective lens, \dfrac{1}{v_o} - \dfrac{1}{u_o} = \dfrac{1}{f_o}:<br><br>\dfrac{v_o}{|u_o|} = \dfrac{v_o}{f_o} -$1$<br><br>The tube length$L$of the microscope when the final image is at infinity is the distance between the objective and the eyepiece, which is given by:<br><br>$L = v_o + f_e \Rightarrow v_o = L - f_e$<br><br>Substituting$v_o$into the magnification formula:<br><br>$M = \left( \dfrac{L - f_e}{f_o} - 1 \right) \left( \dfrac{D}{f_e} \right)$<br><br>Given values are:<br>$M = 100$<br>$f_o = 0.5 \text{ cm}$<br>$L = 6.5 \text{ cm}$<br>$D = 25 \text{ cm}$(standard least distance of distinct vision)<br><br>Substituting the given values:<br><br>$100 = \left( $\dfrac{6.5 - f_e}{0.5} -$1 \right) \left( \dfrac{25}{f_e} \right)<br><br>100 = \left( \dfrac{6.5 - f_e - 0.5}{0.5} \right) \left( \dfrac{25}{f_e} \right)$<br><br>$100 = \left( \dfrac{6.0 - f_e}{0.5} \right) \left( \dfrac{25}{f_e} \right)<br><br>100 = (6.0 - f_e) \times $\dfrac{50}{f_e}<br><br>$2 f_e = 6.0 - f_e$<br><br>$3 f_e = 6.0$<br><br>$f_e = 2 \text{ cm}