Question

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

• A. -4
• B. 5
• C. 10
• D. -10

Answer 1

The correct option is D -10

For a compound microscope, magnifying power
$MP=m_e\times m_0$
When the final image is at least distance of distance vision then
$M_e=1+\frac{D}{f_e}$
$\therefore MP=m_0[1+\frac{D}{f_e}]$
$\Rightarrow -35=m_0[1+\frac{25}{10}]$
$\Rightarrow -35=m_0\frac{35}{10}$
$\Rightarrow m_0=-10$
The negative sign shows that the image formed by objective is inverted.