Question

A person with normal near point $25cm$ using a compound microscope with objective of focal length $8.0mm$ and an eye piece of focal length $2.5cm$ can bring an object placed at $9.0mm$ from the objective in sharp focus. The separation between two lenses and magnification respectively are

• A. $9.47cm,88$
• B. $3.36cm,44$
• C. $6.00cm,22$
• D. $7.49cm,11$

Answer 1

Answer: (A)

Here, $d=25cm$ $f_0=8.0mm,f_e=2.5cm$,
$u_0=-9.0mm=-0.9cm$
Now, $\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}$
$\therefore \frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}=\frac{1}{-25}-\frac{1}{2.5}=\frac{-11}{25}$
$\left(\because v_e=-d=-25cm\right)$
$u_e=\frac{-25}{11}=2.27cm$
Again, $\frac{1}{v_0}-\frac{1}{u_0}=\frac{1}{f_0}$
$\frac{1}{v_0}=\frac{1}{f_0}+\frac{1}{u_0}=\frac{1}{0.8}+\frac{1}{-0.9}$
$\frac{0.9-0.8}{0.72}=\frac{0.1}{0.72}$
$v_0=\frac{0.72}{0.1}=7.2cm$
Therefore, separation between two lenses
$=u_e+v_0=2.27+7.2=9.47cm$
Magnifying power, $m=\frac{v_0}{u_0}\left(1+\frac{d}{f_e}\right)$
$=\frac{7.2}{0.9}\left(1+\frac{25}{2.5}\right)=88$