In an astronomical telescope in normal adjustment a straight black line of length `L` is drawn on inside part of objective lens. The eye piece forms a real image of this line. The length of this image is `I`. The magnification of the telescope is

Options

A. `(L+I)/(L-I)`
B. `(L)/(I)`
C. `(L)/(I)+1`
D. `(L)/(I)-1`

Answer

B

Explanation

The situation is shown in the figure.

Let `f_(0)` and `f_(e)` be the focal lengths of the objective and eyepiece respectively. For normal adjustment distance of the objective from the eyepiece (tube length)`=f_(0)+f_(e)`.
Treating the line on the objective as the object and eyepiece as the lens.
`:.u=-(f_(0)+f_(e))` and `f=f_(e)`
As `(1)/(v)-(1)/(u)=(1)/(f):.(1)/(v)(1)/(-(f_(0)+f_(e)))=(1)/(f_(e))`
`(1)/(v)=(1)/(f_(e))-(1)/(f_(0)+f_(e))=(f_(0)+f_(e)-f_(e))/(f_(e)(f_(0)+f_(e)))=(f_(0))/(f_(e)(f_(0)+f_(e)))`
or `v=(f_(e)(f_(0)+f_(e)))/(f_(0))`
Thus, `(I)/(L)=(v)/(|u|)=((f_(e)(f_(0)+f_(e)))/(f_(0)))/((f_(0)+f_(e)))=(f_(e))/(f_(0))"or"(f_(0))/(f_(e))=(L)/(I)` ...(i)
`:.` The magnification of the telescope in normal adjustment is `m=(f_(0))/(f_(e))=(L)/(I)` (using (i))