Question

The magnifying power of a telescope with tube length $ 60cm $ is $ 5 $ . What is the focal length of its eye piece?
(A) $ 30cm $
(B) $ 20cm $
(C) $ 10cm $
(D) $ 40cm $

Answer 1

Hint:
We need to put the value of the magnifying power in the magnification formula for a telescope. Then using the value of tube length, we will get the focal length of the objective as well as the eyepiece.
Formula used:
The magnifying power of a telescope is given by $ m = \dfrac{{{f_0}}}{{{f_e}}} $
The total tube length of a telescope is $ L = {f_0} + {f_e} $
 $ {f_0} $ is the focal length of the objective, and $ {f_e} $ is the focal length of the eyepiece.

Complete step by step answer:
As we know that the magnification power of a telescope is given by
 $ m = \dfrac{{{f_0}}}{{{f_e}}} $ (1)
According to the question, $ m = 5 $
Putting this in (1)
 $ 5 = \dfrac{{{f_0}}}{{{f_e}}} $
Or $ {f_0} = 5{f_e} $ (2)
Also, the total tube length of the telescope $ (L) $ is the sum of the focal lengths of the objective and the eyepiece, i.e.
 $ L = {f_0} + {f_e} $ (3)
According to the question, $ L = 60cm $
Putting this in (3)
 $ 60 = {f_0} + {f_e} $ (4)
Substituting (2) in (4), we get
 $ 5{f_e} + {f_e} = 60 $
 $ 6{f_e} = 60 $
Finally, we have
 $ {f_e} = 10 $
So the focal length of the eyepiece, $ {f_e} = 10cm $
Hence, the correct answer is option (C), $ 10cm $ .

Additional Information:
Like telescopes, we have other optical instruments too in Ray Optics. They are, namely, the human eye and the microscope. The microscope is further of two types, a simple microscope and a compound microscope. They are also used to see the objects clearly. Their magnification formulas are:
Simple microscope:
- When the image is at the near point: $ m = \dfrac{D}{f} $
- When the image is at infinity: $ m = 1 + \dfrac{D}{f} $
 $ D $ is the near point of the human eye which is $ = 25cm $ and $ f $ is the focal length of the lens
Compound microscope:
 $ m = \left( {\dfrac{L}{{{f_0}}}} \right)\left( {\dfrac{D}{{{f_e}}}} \right) $
 $ {f_0} $ is the focal length of the objective lens, $ {f_e} $ is the focal length of the eyepiece, and $ L $ is the total tube length.

Note:
Don’t get confused in the expression for magnifying power, which focal length should be there in the numerator and which should be there in the denominator. Always remember, magnification is the ratio of the angle subtended by the image to that subtended by the object at the eye. As the power of a lens is inversely related to the focal length, $ {f_o} $ lies in the numerator and $ {f_e} $ lies in the denominator.