An eye can distinguish between two points of an object if they are separated by more than 0.22mm when the object is placed at 25cm from the eye. The object is now seen by a compound microscope having a 20D objective and 10D eyepiece separated by a distance of 20cm. The final image is formed at 25cm from the eye. What is the minimum separation between two points of the object which can now be distinguished?
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For the given compound microscope$\text{f}_0=\frac{1}{20\text{D}}=0.05\text{m}=5\text{cm},$ $\text{f}_\text{e}=\frac{1}{10\text{D}}=0.1\text{m}=10\text{cm.}$
D = 25cm, separation between objective & eyepiece= 20cm
For the minimum separation between two points which can be distinguished by eye using the microscope, the magnifying power should be maximum.
For the eyepiece$, v_0 = -25cm, f_e = 10cm$
So, $\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{f}_\text{e}}=\frac{1}{-25}-\frac{1}{10}$
$=-\Big[\frac{2+5}{50}\Big]\Rightarrow\text{u}_\text{e}=-\frac{50}{7}\text{cm}$
So, the image distance for the objective lens should be,
$\text{v}_0=20-\frac{50}{7}=\frac{90}{7}\text{cm}$
Now, for the objective lens,
$\frac{1}{\text{u}_0}=\frac{1}{\text{v}_0}-\frac{1}{\text{f}_0}=\frac{7}{90}-\frac{1}{5}=-\frac{11}{90}$
$\Rightarrow\text{u}_\text{o}=-\frac{90}{11}\text{cm}$
So, the maximum magnifying power is given by,
$\text{m}=\frac{-\text{V}_0}{\text{u}_0}\Big[1+\frac{\text{D}}{\text{f}_\text{e}}\Big]$
$=\frac{\big(\frac{90}{7}\big)}{\big(-\frac{90}{11}\big)}\Big[1+\frac{25}{10}\Big]$
$=\frac{11}{7}\times3.5=5.5$
Thus, minimum separation eye can distinguish $=\frac{0.22}{5.5}\text{mm}=0.04\text{mm}$
D = 25cm, separation between objective & eyepiece= 20cm
For the minimum separation between two points which can be distinguished by eye using the microscope, the magnifying power should be maximum.
For the eyepiece$, v_0 = -25cm, f_e = 10cm$
So, $\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{f}_\text{e}}=\frac{1}{-25}-\frac{1}{10}$
$=-\Big[\frac{2+5}{50}\Big]\Rightarrow\text{u}_\text{e}=-\frac{50}{7}\text{cm}$
So, the image distance for the objective lens should be,
$\text{v}_0=20-\frac{50}{7}=\frac{90}{7}\text{cm}$
Now, for the objective lens,
$\frac{1}{\text{u}_0}=\frac{1}{\text{v}_0}-\frac{1}{\text{f}_0}=\frac{7}{90}-\frac{1}{5}=-\frac{11}{90}$
$\Rightarrow\text{u}_\text{o}=-\frac{90}{11}\text{cm}$
So, the maximum magnifying power is given by,
$\text{m}=\frac{-\text{V}_0}{\text{u}_0}\Big[1+\frac{\text{D}}{\text{f}_\text{e}}\Big]$
$=\frac{\big(\frac{90}{7}\big)}{\big(-\frac{90}{11}\big)}\Big[1+\frac{25}{10}\Big]$
$=\frac{11}{7}\times3.5=5.5$
Thus, minimum separation eye can distinguish $=\frac{0.22}{5.5}\text{mm}=0.04\text{mm}$
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