Question

(a) Draw a labeled ray diagram showing the formation of a final image by a compound microscope at least distance of distinct vision.
(b) The total magnification produced by a compound microscope is 20. The magnification produced by the eye piece is 5. The microscope is focused on a certain object. The distance between the objective and eyepiece is observed to be 14 cm. If least distance of distinct vision is 20 cm, calculate the focal length of the objective and the eye piece.

Answer 1

where, AB = object, A'B' = image formed by objective and A''B'' = image formed by eyepiece
$f_o$ = focal length of objective,
$u_o$ = object distance from objective
$v_o$ = image distance from objective
$D$ = distance of least distinct vision
$l$ = length of the microscope

(b). For the least distance of clear vision, the total magnification is given by:

$m=-L{f}_o(1+D{f}_e)=m_o.m_e$ ...(1)n2
where, $L$ is the separation between the eyepiece and the objective
$f_o$ is the focal length of the objective
$f_e$ is the focal length of the eyepiece
$D$ is the least distance for clear vision
Also, the given magnification for the eyepiece:

$m_e=5=(1+D{f}_e)$
$5=1+20/f_e\Rightarrow f_e=5cm$

$\frac{1}{u_e}=-\frac{1}{20}-\frac{1}{5}=-\frac{1}{4}$

$u_e=-4$

$L=v_o{+|u}_e|$

$14=v_o+u\Rightarrow v_o=10\Rightarrow u_o=\frac{v_o}{-4}=-2.5Hence{f}_0=-\frac{u{v}_o}{5}=\frac{4\times 2.5}{5}=2cm$