Question

A microscope has an eyepiece lens of 20X and an objective lens of 80X which are 25.0 cm apart. Calculate
(a) the total magnification
(b) the focal length of each lens
(c) where the object must be for a normal relaxed eye to see it in focus

Answer 1

The total magnification:

$$M=M_oM_e=1600\times$$

The focal length of the objective (equation 1):

$$\frac{1}{f_o}=\frac{1}{d_i}-\frac{1}{d_{ob}}$$

The magnification of the eyepiece, assuming the final image is formed at least distance of distinct vision (D=25 cm), is

$$M_e=D/f_e,\\ f_e=\frac{D}{M_e}=1.25\text{ cm}.$$

The magnification of the objective for L=25 cm:

$$M_o=L/f_o,\\ f_o=\frac{L}{M_o}=0.313\text{ cm}.$$

For the relaxed eye of a normal man, the near point is 25 cm.

$$L=d_i+f_e,\\ d_i=L-f_e.$$

Substitute this to equation 1 and calculate the object distance:

$$d_{ob}=\frac{d_if_o}{f_o-d_i}=\frac{(L-f_e)f_o}{L-f_e-f_o}=0.317\text{ cm}.$$