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

Answer

A

Explanation

In normal adjustment
`L = f_(0)+f_(e)`
Treating line on obhective as object and eye-piece the lens
`(1)/(v)-(1)/(u)=(1)/(f) rArr (1)/(v) - (1)/(-(f_(O)+f_(e)) = (1)/(f_(e))`
`v = ((f_(O)+f_(e))f_(e))/(f_(O))`
Magnification `= |(v)/(u)| = (f_(e))/(f_(O)) = ("image size")/("object size") = (l)/(L)`
`(f_(O))/(f_(e)) = (L)/(l) =` magnification of telescope in normal adjustment