Question

If the focal length of a reflection telescope is 200 cm and the focal length of the eyepiece lens is 0.25 cm, what is the magnifying power of the telescope?

Answer 1

-48cm

Step-by-step explanation:

the following data are given

focal length of telescope=200cm,

focal length of the eyepiece=0.25cm

From the genera formula used to find the magnifying power which is expressed as

`M=-(fx_(o))/(f_(e))[1+(f_(e))/(d)]`

where

`f_(e) = focal length of thr eye piece\\ f_(o) =focal length of the telescope\\`

and d=least distance of distinct vision=25cm

if we substitute values into the formula, we arrive at

`M=-(fx_(o))/(f_(e))[1+(f_(e))/(d)]\\M=-(200cm)/(0.25cm)[1+(0.25cm)/(25cm)]\\M=-800cm[1+0.01]\\M=-800cm(1.01)\\M=-808cm \\M=-808cm`

hence from the answer, we can conclude that the magnifying power of the telescope is -808cm

Answer 2

800

Step-by-step explanation:

Focal length of telescope, F = 200cm

Focal length of eyepiece, f = 0.25

The magnifying power of a telescope is given as the ratio of the focal length of the objective of the telescope to the focal length of the lens. Mathematically:

M = F/f

Therefore, when F = 200cm and f = 0.25cm:

M = 200/0.25

M = 800