Question
Find the maximum magnifying power of a compound Microscope having a 25 diopter lens as the objective, a 5 diopter lens as the eyepiece and the separation 30cm between the two lenses. The least distance for clear vision is 25cm.
✓
Answer
For the given compound microscope$\text{f}_0=\frac{1}{25\text{ diopter}}=0.04\text{m}=4\text{cm,}$ $\text{f}_\text{e}=\frac{1}{5\text{ diopter}}=0.2\text{m}=20\text{cm}$
D = 25cm, separation between objective and eyepiece = 30cm T
he magnifying power is maximuwm hen the image is formed by the eye piece at least distance of clear vision
i.e. D = 25cm for the eye piece,
$v_e= -25cm, f_e = 20cm$ For lens formula, $\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{f}_\text{e}}$
$\Rightarrow\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{f}_\text{e}}$
$\Rightarrow\frac{1}{\text{u}_\text{e}}\frac{1}{-25}-\frac{1}{20} $
$\Rightarrow\text{u}_\text{e}=11.11\text{cm}$
So, for the objective lens,
the image distance should be $v_0 = 30 - (11.11) = 18.89cm$
Now, for the objective lens, $v_0 = +18.89cm$ (because real image is produced)
$f_0 = 4cm$ So, $\frac{1}{\text{u}_\text{o}}=\frac{1}{\text{v}_\text{o}}-\frac{1}{\text{f}_\text{o}}$
$\Rightarrow \frac{1}{18.89}-\frac{1}{4}\\=0.053-0.25=-0.197$
$\Rightarrow\text{u}_\text{o}=-5.07\text{cm}$
So, the maximum magnificent power is given by$\text{m}=-\frac{\text{v}_\text{o}}{\text{u}_\text{o}}\Big[1+\frac{\text{D}}{\text{f}_\text{e}}\Big]=\frac{18.89}{-5.07}\Big[1+\frac{25}{20}\Big]$
$=3.7225\times2.25=8.376$
D = 25cm, separation between objective and eyepiece = 30cm T
he magnifying power is maximuwm hen the image is formed by the eye piece at least distance of clear vision
i.e. D = 25cm for the eye piece,
$v_e= -25cm, f_e = 20cm$ For lens formula, $\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{f}_\text{e}}$
$\Rightarrow\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{f}_\text{e}}$
$\Rightarrow\frac{1}{\text{u}_\text{e}}\frac{1}{-25}-\frac{1}{20} $
$\Rightarrow\text{u}_\text{e}=11.11\text{cm}$
So, for the objective lens,
the image distance should be $v_0 = 30 - (11.11) = 18.89cm$
Now, for the objective lens, $v_0 = +18.89cm$ (because real image is produced)
$f_0 = 4cm$ So, $\frac{1}{\text{u}_\text{o}}=\frac{1}{\text{v}_\text{o}}-\frac{1}{\text{f}_\text{o}}$
$\Rightarrow \frac{1}{18.89}-\frac{1}{4}\\=0.053-0.25=-0.197$
$\Rightarrow\text{u}_\text{o}=-5.07\text{cm}$
So, the maximum magnificent power is given by$\text{m}=-\frac{\text{v}_\text{o}}{\text{u}_\text{o}}\Big[1+\frac{\text{D}}{\text{f}_\text{e}}\Big]=\frac{18.89}{-5.07}\Big[1+\frac{25}{20}\Big]$
$=3.7225\times2.25=8.376$
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