Question

An eye can distinguish between two points of an object if they are separated by more than 0.22 mm when the object is placed at 25 cm from the eye. The object is now seen by a compound microscope having a 20 D objective and 10 D eyepiece separated by a distance of 20 cm. The final image is formed at 25 cm from the eye. What is the minimum separation between two points of the object which can now be distinguished?

Answer 1

For the given compound microscope

$f_0=\frac{1}{20}$ D = 0.05 m - 5 cm

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

D = 25 cm, separation between objective and eyepiece = 20 cm.

For the minimum separation between two points which can be distinguished by eye using the miscroscope, the magnifying power should be maximum.

For the eyepiece

$v_e=-25cm{f}_e=10cm$

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

$=\frac{1}{-25}-\frac{1}{10}=-\frac{2+5}{50}$

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

So, the image distance for the objective lens should be,

$v_0=20-\frac{50}{7}=\frac{90}{7}cm$

Now, for the objective lens,

$\frac{1}{u_0}=\frac{1}{v_0}-\frac{1}{f_0}$

$=\frac{7}{90}-\frac{1}{5}=\frac{7-18}{90}=-\frac{11}{90}$

$\Rightarrow u_0=\frac{-90}{11}cm$

$m=\frac{v_0}{u_0}(1+\frac{D}{f_e})$

$=-\frac{\frac{90}{7}}{\frac{-90}{11}}(1+\frac{25}{10})$

$=\frac{11}{7}\times \left(3.5\right)=5.5$

Thus, minimum separation that the eye can distinguish $=\frac{0.22}{5.5}$ mm = 0.04 mm.