Question

In a compound microscope, the focal lengths of two lenses are $1.5cm$ and $6.25cm$ and an object is placed at $2cm$ from objective and the final image is formed at $25cm$ from eye lens. The distance between the two lenses is

• A. $6.00cm$
• B. $7.75cm$
• C. $9.25cm$
• D. $11.00cm$

Answer 1

Answer: (D)

$f_0=1.5cm,f_e=6.25cm,u_0=-2cm,v_e=-D=-25cm$
By objective lens $\frac{1}{f_0}=\frac{1}{v_0}-\frac{1}{u_0}$
$\frac{1}{1.5}=\frac{1}{v_0}-\frac{1}{-2}$
$\Rightarrow$ $\frac{1}{v_0}=\frac{1}{1.5}-\frac{1}{2}$ or $v_0=6cm$
By eyepiece $\frac{1}{f_e}=\frac{1}{v_e}-\frac{1}{u_e}$
$\frac{1}{6.25}=\frac{1}{-25}-\frac{1}{-u_e}$
$\Rightarrow$ $\frac{1}{u_e}=\frac{1}{6.25}+\frac{1}{25}=\frac{4}{25}+\frac{1}{25}=\frac{1}{5}$
$u_e=5cm$
Length of tube $=L=v_0+u_e=6.0cm+5.0cm$
$L=11cm$