# In a compound microscope, the magnified virtual image is formed at a distance of

Question:

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

Solution:

for first lens $=\frac{1}{v_{1}}-\frac{1}{-x}=\frac{1}{1} \Rightarrow v_{1}=\frac{x}{x-1}$

also magnification $\left|\mathrm{m}_{1}\right|=\left|\frac{\mathrm{v}_{1}}{\mathrm{u}_{1}}\right|=\frac{1}{\mathrm{x}-1}$

for $2^{\text {nd }}$ lens this is acting as object

so $\mathrm{u}_{2}=-\left(20-\mathrm{v}_{1}\right)=-\left(20-\frac{\mathrm{x}}{\mathrm{x}-1}\right)$

and $v_{2}=-25 \mathrm{~cm}$

angular magnification $\left|\mathrm{m}_{\mathrm{A}}\right|=\left|\frac{\mathrm{D}}{\mathrm{u}_{2}}\right|=\frac{25}{\left|\mathrm{u}_{2}\right|}$

Total magnification $\mathrm{m}=\mathrm{m}_{1} \mathrm{~m}_{\mathrm{A}}=100$

$\left(\frac{1}{x-1}\right)\left(\frac{25}{20-\frac{x}{x-1}}\right)=100$

$\frac{25}{20(x-1)-x}=100 \Rightarrow 1=80(x-1)-4 x$

$\Rightarrow 76 x=81 \Rightarrow x=\frac{81}{76}$

$\Rightarrow \mathrm{u}_{2}=-\left(20-\frac{81 / 76}{81 / 76-1}\right)=\frac{-19}{5}$

now by lens formula

$\frac{1}{-25}-\frac{1}{-19 / 5}=\frac{1}{f_{e}} \Rightarrow f_{e}=\frac{25 \times 19}{106} \approx 4.48 \mathrm{~cm}$