Question

A compound microscope has an objective lens of $1.0cm$ focal length and an eye-piece of $2.5cm$ of focal length. When the object is placed at $1.1cm$ from the objective, the final image is at the least distance of direct vision. Thus, the magnification produced is

• A. $10$
• B. $11$
• C. $100$
• D. $110$

Answer 1

The correct option is D $110$

given,

$u_o=1.1cm,f_o=1cm,f_e=2.5cm$

by lens formula,

$\frac{1}{v_o}=\frac{1}{f_o}+\frac{1}{u_o}=1-\frac{1}{1.1}=\frac{1}{11}$

$v_o=11cm$

we know,

magnification of compound microscope$=\frac{v_o}{u_o}(1+\frac{D}{f_e})$ (when image at D)

$m=\frac{11}{1.1}(1+\frac{25}{2.5})=10\times (10+1)=110$

hence Option $\text{D}$ is correct answer.