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
- L+IL-I
- LI
- LI+1
- LI-1
Answer
(2) LI
Solution
In normal adjustment the distance between the objective and the eye-piece = fo + fe
The line on the objective acts as an object for the eye-piece: u = -(fo + fe)
Lens formula: 1v - 1u = 1fe
⇒ 1v = 1fe - 1fo + fe = fofe(fo + fe)
⇒ v = fe(fo + fe)fo
Magnification of the line: IL = | vu | = fefo
Magnification of the telescope M = fofe = LI
Answer: (2)