Question

The magnifying power of an
astronomical telescope for
relaxes vision is 16 and the
distance between the
objective and eyelens is 34
cm. The focal length of
objective and eyelens will be
respectively are​

Answer 1

In this case ∣m∣=

f

e

f

0

=16 ....(i)

and length of telescope =f

0

+f

e

=34 ....(ii)

Solving (i) and (ii), we get f

e

=20 cm,f

0

=14 cm