Question

The separation between the objective and the eyepiece of a compound microscope can be adjusted between 9.8 cm to 11.8 cm. If the focal lengths of the objective and the eyepiece are 1.0 cm and 6 cm respectively, find the range of the magnifying power if the image is always needed at 24 cm from the eye.

Answer 1

For the compound microscope, we have:
Focal length of the objective, f_o = 1.0 cm
Focal length of the eyepiece, f_e = 6 cm
Image distance from the eyepiece, v_e = $-$24 cm
Separation between the objective and the eyepiece = 9.8 cm to 11.8 cm
The lens formula is given by
$$\begin{gathered} \frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e} \\ \frac{1}{-24}-\frac{1}{u_e}=\frac{1}{6} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{1}{u_e}=-\frac{1}{6}-\frac{1}{24} \\ \Rightarrow \frac{1}{u_e}=-\frac{4+1}{24}=-\frac{5}{24} \\ \therefore u_e=-\frac{24}{5}\mathrm{cm}=-4.8\mathrm{cm} \end{gathered}$$
(a) Separation between the objective and the eyepiece = 9.8 cm
So, for the objective lens, the image distance will be
v_o = 9.8 − 4.8 = 5 cm
The lens formula for the objective lens is given by
$$\begin{gathered} \frac{1}{v_0}-\frac{1}{u_0}=\frac{1}{f_0} \\ \frac{1}{5}-\frac{1}{u_0}=\frac{1}{1} \end{gathered}$$
$$\begin{gathered} \Rightarrow -\frac{1}{u_0}=1-\frac{1}{5}=\frac{5-1}{5} \\ \Rightarrow u_0=-\frac{5}{4}\mathrm{cm}=-1.25\mathrm{cm} \end{gathered}$$
Magnifying power of the compound microscope:
$m=\frac{v_0}{u_0}\left(1+\frac{D}{f_e}\right)$
$=\frac{5}{1.25}\left(1+\frac{24}{6}\right)=20$

(b) Separation between the objective and the eyepiece = 11.8 cm
We have:
m = 30
So, the required range of the magnifying power is 20–30.