Resolution of a Microscope.The image formed by a microscope objective with a focal length of 5.00 mm is 160 mm from its second focal point. The eyepiece has a focal length of 26.0 mm. (a) What is the angular magnification of the microscope? (b) The unaided eye can distinguish two points as its near point as separate if they are about 0.10 mm apart. What is the minimum separation between two points that can be observed (or resolved) through this microscope?

The angular magnification of the compound microscope equals the product of two magnificationsand is also given as:

$\mathrm{M}=\mathrm{m}1{\mathrm{M}}_2=\frac{\left(250\mathrm{mm}\right){{\mathrm{s}}_1}^{\text{'}}}{{\mathrm{f}}_1{\mathrm{f}}_2}$

The minimum separation equals that distance divided by the magnification.

Minimum separation $=\frac{d}{M}$

The image of the distance${s_1}^{\text{'}}$ equals the focal length of the lens plus the image distance before magnification.

$$\begin{gathered} {s_1}^{\text{'}}=s^{\text{'}}+f_1 \\ =160mm+5.0mm \\ =165mm \end{gathered}$$

Now, plug these values into the equation of the angular magnification,

$$\begin{gathered} M=\frac{\left(250mm\right){s_1}^{\text{'}}}{f_1{f}_2} \\ =\frac{\left(250mm\right)\left(165mm\right)}{\left(5.0mm\right)\left(26.0mm\right)} \\ =317 \end{gathered}$$

The distance between two points that the eye can distinguish is d = 0.10mm.

Thus,

Minimum separation$$\begin{gathered} =\frac{d}{M}=\frac{0.10mm}{317} \\ =3.15*{10}^{-4}mm \end{gathered}$$

Therefore, the angular magnification of the microscope is 317 and the minimum separation between two points that can be observed through microscope is $3.15*{10}^{-4}$mm.