The focal length of objective and eye piece lens of a microscope are 1.6 cm & 2.5 cm respectively. Distance between two lenses is 21.7 cm .If the final image is formed at infinity what is the linear magnification?

Given,

Separation between lense,$L=21.7\text{cm},$

Focal length of eye piece, $L=21.7\mathrm{c}\mathrm{m},{\mathrm{f}}_{\mathrm{e}}=2.5\mathrm{c}\mathrm{m}$

Focal length of objective lens,$f_0=1.6\mathrm{c}\mathrm{m},v_0=19.2\mathrm{c}\mathrm{m},{\mathrm{u}}_0=?$

$L=v_0+f_e$

Where $L$ is the length of the microscope and $f_e$ e the focal length of the objective

Substituting we get:

$21.7={\mathrm{v}}_0+2.5$

$\Rightarrow {\mathrm{v}}_0=21.7-2.5=19.2\mathrm{c}\mathrm{m}$

According to lens formula we know that:

$\frac{1}{f_0}=\frac{1}{v_0}-\frac{1}{u_0}$

$v_0=19.2\text{cm},{\text{u}}_0=?$

$\frac{1}{1.6}=\frac{1}{19.2}-\frac{1}{{\mathrm{u}}_0}$

$-\frac{1}{{\mathrm{u}}_0}=\frac{1}{1.6}-\frac{1}{19.2}$

$-\frac{1}{u_0}=\frac{10}{16}-\frac{10}{192}$

$-\frac{1}{u_0}=\frac{120-10}{192}=\frac{110}{192}$

$u_0=-\frac{192}{110}\mathrm{c}\mathrm{m}=-1.75\mathrm{c}\mathrm{m}$

Magnification, $m=\frac{-v}{u}=\frac{-19.2}{1.75}=10.97=11$

Hence, linear magnification is $11$