Question

In a compound microscope, the magnified virtual image is formed at a distance of $25cm$ from the eye-piece. The focal length of its objective lens is $1cm.$ If the magnification is $100$ and the tube length of the microscope is $20cm,$ then the focal length of the eye-piece lens (in $cm$ ) is __________.

Answer 1

Answer: 6.25

for first lens $=\frac{1}{v_1}-\frac{1}{-x}=\frac{1}{1}\Rightarrow v_1=\frac{x}{x-1}$
also magnification $|m_1|=|\frac{v_1}{u_1}\mid =\frac{1}{x-1}$
for $2^{\text{nd }}$ lens this is acting as object
so $u_2=-\left(20-v_1\right)=-\left(20-\frac{x}{x-1}\right)$
and $v_2=-25cm$
angular magnification $|m_A|=\left|\frac{D}{u_2}\right|=\frac{25}{|u_2|}$
Total magnification $m=m_1{m}_A=100$
$\left(\frac{1}{x-1}\right)\left(\frac{25}{20-\frac{x}{x-1}}\right)=100$
$\frac{25}{20(x-1)-x}=100$
$\Rightarrow 1=80(x-1)-4x$
$\Rightarrow 76x=81$
$\Rightarrow x=\frac{81}{76}$
$\Rightarrow u_2=-\left(20-\frac{81/76}{81/76-1}\right)=\frac{-19}{5}$
now by lens formula
$\frac{1}{-25}-\frac{1}{-19/5}=\frac{1}{f_e}$
$\Rightarrow f_e=\frac{25\times 19}{106}\approx 4.48cm$