Question

A compound microscope has an objective of focal length $2.0cm$ and an eyepiece of focal length $6.25cm$ separated by $15cm$. If the final image is formed at the least distance of distinct vision$\left(25cm\right)$, the distance of the object from the objective is :

• A. $1.5cm$
• B. $2.5cm$
• C. $3.0cm$
• D. $4.0cm$

Answer 1

The correct option is B $2.5cm$

Focal length of objective, $f_0=2cm$

Focal length of eye-piece, $f_e=6.25cm$

Length of the microscope, $L=15cm$

Image distance, $v_e=25cm$

Object distance, $u=?$

$L=v_0+\frac{Dfe}{D+fe}=15$

$v_0=15-\frac{25\times 6.25}{31.25}=10cm$

By lens formula, $\frac{1}{10}-\frac{1}{v_0}=\frac{1}{2}$

$-\frac{1}{v_0}=\frac{5}{10}-\frac{1}{10}=\frac{4}{10}$

$v_0=-\frac{10}{4}=-2.5cm$