Question

An eye can distinguish between two points of an object, if they are separated by more than $0.22\text{mm}$ when the object is placed at $25\text{cm}$ from the eye. The object is now seen by a compound microscope having a $20\text{D}$ objective and $10\text{D}$ eyepiece separated by a distance of $20\text{cm}$. The final image is formed at $25\text{cm}$ from the eye. The minimum separation between two points of the object which can now be distinguished is $4.67\times {10}^{-n}\text{mm}$. The value of $n$ rounded off to the nearest integer is

Answer 1

For the given compound microscope,

$f_o=\frac{1}{20}\text{D}=0.05\text{m}=5\text{cm}$

$f_e=\frac{1}{10}\text{D}=0.1\text{m}=10\text{cm}$

Separation between objective and eyepiece $D=20\text{cm}$

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

$v_e=-25\text{cm},f_e=10\text{cm}$

$\therefore \frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}$

$=\frac{1}{-25}-\frac{1}{10}$

$=-\left(\frac{2+5}{50}\right)$

$\Rightarrow u_e=\frac{-50}{7}\text{cm}$

So the image distance for the objective lens should be,
$v_{\circ }=20-\frac{50}{7}=\frac{90}{7}\text{cm}$

Now, for the objective lens,

$\frac{1}{u_{\circ }}=\frac{1}{v_{\circ }}-\frac{1}{f_{\circ }}$

$=\frac{7}{90}-\frac{1}{5}$

$=\frac{7-18}{90}$

$\frac{1}{u_{\circ }}=-\frac{11}{90}$

$\Rightarrow u_{\circ }=\frac{-90}{11}\text{cm}$

Also, $m=\frac{v_{\circ }}{-u_{\circ }}(1+\frac{D}{f_e})$

$=\frac{\frac{90}{7}}{\frac{90}{11}}(1+\frac{20}{10})$

$m=\frac{11}{7}\times \left(3\right)=\frac{33}{7}$

Thus, minimum separation that the eye can distinguish $=\frac{0.22}{33/7}\text{mm}$

$=\frac{14}{3}\times {10}^{-2}\text{mm}$

$=4.67\times {10}^{-2}\text{mm}$

Hence, $n=2$