Physics - Geometrical Optics Question with Solution | TestHub

For a compound microscope in normal adjustment (relaxed eye), the length of the microscope tube is given by:

$L=v_o+f_e$

Substituting the given values $L=15$ cm and $f_e=6$ cm:

$15=v_o+6$

$v_o=9$ cm

The magnifying power of a compound microscope for a relaxed eye is given by:

$M=\frac{v_o}{u_o}\times \frac{D}{f_e}$

where $D$ is the least distance of distinct vision ($D=25$ cm). Substituting the known values:

$25=\frac{9}{u_o}\times \frac{25}{6}$

$1=\frac{9}{6u_o}$

$u_o=\frac{9}{6}=1.5$ cm

Answer: 1.5 cm