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 compound microscope, we have:
Power of the objective lens = 20 D
The focal length of the objective lens is given by
$f_0=\frac{1}{20\mathrm{D}}=0.05\mathrm{m}=5\mathrm{cm}$
Power of the eyepiece = 10 D
The focal length of the eyepiece is given by
$f_e=\frac{1}{10\mathrm{D}}=0.1\mathrm{m}=10\mathrm{cm}$
Least distance for clear vision, D = 25 cm,
To distinguish the two points having minimum separation, the magnifying power should be maximum.
For the eyepiece, we have:
Image distance, v_e = − 25 cm
Focal length, f_e = 10 cm
The lens formula is given by
$\frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}$
$$\begin{gathered} =\frac{1}{-25}-\frac{1}{10} \\ =\frac{-2-5}{50}=\frac{-7}{50} \end{gathered}$$
Separation between the objective and the eyepiece = 20 cm
So, the image distance for the objective lens $\left(v_0\right)$ can be obtained as:
$v_0=20-\frac{50}{7}=\frac{90}{7}\mathrm{cm}$
The lens formula for the objective lens is given by
$$\begin{gathered} \frac{1}{u_0}=\frac{1}{v_0}-\frac{1}{f_0} \\ =\frac{7}{90}-\frac{1}{5}=\frac{7-18}{90}=-\frac{11}{90} \end{gathered}$$
$\Rightarrow u_0=\frac{-90}{11}\mathrm{cm}$
Magnification of the compound microscope:
$m=\frac{v_0}{u_0}\left(1+\frac{D}{f_e}\right)$
$$\begin{gathered} =-\frac{{\displaystyle \frac{90}{7}}}{{\displaystyle \frac{-90}{11}}}\left(1+\frac{25}{10}\right) \\ =\frac{11}{7}\times \left(3.5\right)=5.5 \end{gathered}$$
∴ Minimum separation that the eye can distinguish = $\frac{0.22}{5.5}$ mm = 0.04 mm