Question

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

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

Answer 1

The correct option is D $11.00$ cm

Given,
$f_0=1.5\text{cm},f_e=6.25\text{cm},u_0=-2\text{cm},$
$v_e=-D=-25\text{cm}$
For objective lens,
$\frac{1}{f_0}=\frac{1}{v_0}-\frac{1}{u_0}$
$\frac{1}{1.5}=\frac{1}{v_0}-\frac{1}{-2}\Rightarrow \frac{1}{v_0}=\frac{1}{1.5}-\frac{1}{2}{v}_0=6\text{cm}$
For eye piece,
$\frac{1}{f_e}=\frac{1}{v_e}-\frac{1}{u_e}$
$\frac{1}{6.25}=\frac{1}{-25}-\frac{1}{-u_e}\Rightarrow \frac{1}{u_e}=\frac{1}{6.25}+\frac{1}{25}=\frac{4}{25}+\frac{1}{25}=\frac{1}{5}$
$u_e=5\text{cm},$
Length of tube $=L=v_0+u_e=6.0\text{cm}+5.0\text{cm},L=11\text{cm}$