Question

In a compound microscope, the focal length of the objective and the eye lens are 2.5 cm and 5 cm respectively. An object is placed at 3.75 cm before the objective and image is formed at the least distance of distinct vision, then the distance between two lenses will be (i.e. length of the microscopic tube)

• A. 11.67 cm
• B. 12.67 cm
• C. 13.00 cm
• D. 12.00 cm

Answer 1

The correct option is A 11.67 cm

$L_D=v_o+u_e$ and for objective lens $\frac{1}{f_o}=\frac{1}{v_o}-\frac{1}{u_o}$
Putting the values with proper sigh convention.
$\frac{1}{+2.5}=\frac{1}{v_o}-\frac{1}{(-3.57)}\Rightarrow v_o=7.5{}_{cm}$
For eye lens $\frac{1}{f_e}=\frac{1}{v_e}-\frac{1}{u_e}$
$\Rightarrow \frac{1}{+5}=\frac{1}{(-25)}-\frac{1}{u_e}\Rightarrow u_e=-4.16cm$
$\Rightarrow |u_e|=4.16cm$
Hence $L_D=7.5+4.16=11.67{}_{cm}$