Question

A compound microscope has a magnifying power 30. The focal length of its eye-piece is 5 cm. Assuming the final image to be at the least distance of distinct vision. The magnification produced by the objective will be

• A. 5
• B. -5
• C. 6
• D. -6

Answer 1

The correct option is B -5

Magnification produced by compound microscope $m=m_0\times m_e$
where $m_0=?$ and $m_e=(1+\frac{D}{f_e})=1+\frac{25}{5}=6\Rightarrow 30=-m_0\times 6\Rightarrow m_0=-5$