Discussion Forum

A compound microscope having magnifying power 35 with its eye-piece of focal length 10 cm. Assumes that the final image is at least distance of distinct vision then the magnification product by the objective is 

Answer:

Explanation:

 For  a compound microscope , magnifying power

 MP= m_e x m₀

 when the final image is at least distance of distance vision then

$M_{e}=1+\frac{D}{f_{e}}$

$\therefore$     $MP= m_{0}\left[1+\frac{D}{f_{e}}\right]$

$\Rightarrow -35= m_{0}\left[1+\frac{25}{10}\right]$

$\Rightarrow -35= m_{0}\times35$

 $\Rightarrow  m_{0}=-10$

 The negative sign shows that the image formed by the objective is inverted