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

Answer

A

Explanation

Magnification of telescope, m = `(f_o)/(f_e)`
But, magnification m = `-("height of image")/("height of object")`
`v/u=-1/L`
`therefore (f_e)/(f_e-(f_o+f_e))=-I/L`
`therefore (f_e)/(-f_o)=-I/L`
`(f_o)/(f_e)=L/T`
`therefore` magnification `m=L/I`