The optical instrument has $\text{f}_0=\frac{1}{25\text{D}}=0.04\text{m}=4\text{cm}$
$\text{f}_\text{e}=\frac{1}{20\text{D}}=0.05\text{m}=5\text{cm}$
tube length = 25cm (normal adjustment)
- The instrument must be a microscope as f0 < fe
- Since the final image is formed at infinity, the image produced by the objective should lie on the focal plane of the eye piece.
So, image distance for objective = v0 = 25 – 5 = 20cm
Now, using lens formula.
$\frac{1}{\text{v}_0}-\frac{1}{\text{u}_0}=\frac{1}{\text{f}_0}$
$\Rightarrow\frac{1}{\text{u}_0}=\frac{1}{\text{v}_0}-\frac{1}{\text{f}_0}=\frac{1}{20}-\frac{1}{4}$
$=\frac{-4}{20}=\frac{-1}{5}\Rightarrow\text{u}_0=-5\text{cm}$
So, angular magnification
$\text{m}=-\frac{\text{v}_0}{\text{u}_0}\times\frac{\text{D}}{\text{f}_\text{e}}$[Taking D = 25cm] $=-\frac{20}{-5}\times\frac{25}{5}=20$
