Question

An optical instrument used for angular magnification has a 25 D objective and 20 D eyepiece. The tube length is 25 cm when the eye is least strained. (a) Whether it is a microscope or a telescope? (b) What is the angular magnification produced?

Answer 1

Given:
Power of the objective lens = 25 D
The focal length of the objective lens is given by
$f_0=\frac{1}{25\mathrm{D}}=0.04\mathrm{m}=4\mathrm{cm}$
Power of the eyepiece = 5 D
The focal length of the eyepiece is given by
$f_e=\frac{1}{20\mathrm{D}}=0.05\mathrm{m}=5\mathrm{cm}$

(a) As f_o < fe, the instrument must be a microscope.

(b) Tube length, l = 25 cm
Image distance for the eye lens, v_e = 25 cm
We know:
l = $v_o+f_e$
$\Rightarrow v_0=l-f_e$=25 $-$ 5 = 20 cm
$$\begin{gathered} \mathrm{On}\mathrm{applying}\mathrm{the}\mathrm{lens}\mathrm{formula}\mathrm{for}\mathrm{the}\mathrm{objective}\mathrm{lens},\mathrm{we}\mathrm{get}: \\ \frac{1}{v_0}-\frac{1}{u_0}=\frac{1}{f_0} \end{gathered}$$
$$\begin{gathered} \Rightarrow -\frac{1}{u_o}=\frac{1}{4}-\frac{1}{20} \\ \Rightarrow -\frac{1}{u_0}=\frac{5-1}{20}=\frac{4}{20} \\ \Rightarrow u_0=\frac{20}{4}=5\mathrm{cm} \end{gathered}$$
The angular magnification of microscope $\left(m\right)$ is given by
$$\begin{gathered} m=\frac{v_0}{u_0}\left(\frac{D}{f_e}\right) \\ =\frac{20}{5}\left(\frac{25}{5}\right)=20\mathrm{cm} \end{gathered}$$
So, the required angular magnification of the microscope is 20.