Question

An astronomical telescope is adjusted to form the final image at infinity. The separation between the lens is 80cm. The angular magnification is 15. Calculate the focal length of the objective and the eyepiece lens.​

Answer 1

Answer:

Hi mate here is ur answer

Explanation:

ANSWER

For astronomical telescope,

We known that, linear magnification.

m=

f

e

f

o

But Here,

m=15

15=

f

e

f

o

15f

e

=f

o

......(1)

The lengths of astronomical telescope, L=80cm

We know that, L=f

o

+f

e

.......(2)

From equation (1) and (2)

80=15f

o

+f

e

80=16f

e

f

e

=5cm

f

o

=15×5

f

e

=75cm

f

o

and f

e

are 75 and 5 respectively.....

plz m as brnlst

Answer 2

Given,

Formation of final image is at infinity, thus the separation between the objective and eyepiece will be $f_{o}+f_{e}$.

where,

$f_{o}$  is the focal length of objective

$f_{e}$ is the focal length of eyepiece.

$f_{o}+f_{e}=80cm$

Angular magnification =$-\frac{f_{o}}{f_{e}} = -15 cm$

$f_{o}=15f_{e}$

Thus by solving the above two eqns we get ,

$f_{o}=75cm\\f_{e}=5cm$

Thus the focal length of objective is 75 cm and the focal length of eyepiece is 5 cm.