Question

A small telescope has an objective lens of focal length 144cm and an eyepiece of focal length 6.0
cm. What is the magnifying power of the telescope? What is the separation between the objective
and the eyepiece?

Answer 1

Hint: For a telescope, the magnifying power can be calculated from focal length of the objective
and the focal length of the eyepiece. The formula for magnifying power; $m =
\dfrac{{{f_0}}}{{{f_e}}}$has to be applied, where ${f_0}$is the focal length of the objective
and ${f_e}$is the focal length of eyepiece. The separation between the objective and eyepiece
can be calculated by adding the focal length of objective and the focal length of eyepiece.
Complete step by step solution:
Given,
The focal length of the objective, ${f_0} = 144{\rm{ cm}}$
The focal length of the eyepiece, ${f_e} = 6.0{\rm{cm}}$
The magnifying power is the ratio of the focal length of the objective and the focal length of
eyepiece.
The formula for magnifying power is given as,
$m = \dfrac{{{f_0}}}{{{f_e}}}$ .…..(1)
The formula for separation between objective and eyepiece is the total focal length of objective
and eyepiece. It is given as,
$D = {f_0} + {f_e}$ .…..(2)
Substituting the values of $144{\rm{ cm}}$in ${f_0}$and $6{\rm{ cm}}$in ${f_e}$in equation
(1) we get,
$m = \dfrac{{144}}{6} = 24$
Hence the magnifying power is $24{\rm{ cm}}$
Substituting the values of $144{\rm{ cm}}$in ${f_0}$and $6{\rm{ cm}}$in ${f_e}$in equation
(2) we get,
$D = 144 + 6 = 150{\rm{cm}}$
Note: the students have to know the relation of magnifying power with focal length of both
objective and eyepiece. If the focal length of objective and eyepiece is directly given, then the
ratio of focal length of objective and eyepiece gives the magnifying power. Magnification of a
telescope is actually a relationship between two independent optical systems: the telescope itself
and the eyepiece you are using. The separation is the distance of the total focal length of the
objective and the eyepiece.