Answer to Question #93805 in Optics for Sridhar

Question

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
(a) L+I/L-I
(b) L/I
(c) L/I+1
(d) L/I-1

Answer 1

The magnification of a telescope is

$$\lambda=\frac{F}{f}$$

where F - focus of an objective; f - focus of an ocular

From similarity of triangles we have equation

$$\frac{F}{f}+1=\frac{L}{l}$$

then

$$\lambda=\frac{L}{l}-1$$

Answer is d)