When a telescope is in normal adjusment, the distance of the objective from the eyepiece is found to `100 cm`. If the magnifying power of the telescope, at normal adjusment, is `24` focal lengths of the lenses are

Options

A. `96` cm, `4` cm
B. `48` cm, `2` cm
C. `50` cm, `50` cm
D. `80` cm, `20` cm

Answer

A

Explanation

In normal adjustment of telescope, magnifying power is
`M=(f_(0))/(f_(e))`
Where `f_(0)` and `f_(e)` be the focal lengths of the objectibe lens and eye piece respectively.
In normal adjustment of telescope, distance between objetive lens and eye piece is
`L=f_(0)+f_(e)`
According to the problem, `24=(f_(0))/(f_(e))` ...(i)
`100=f_(0)+f_(e)` ...(ii)
Solving (i) and (ii), we get
`f_(0)=96` cm, `f_(e)=4` cm