Draw a ray diagram for the formation of image of an object by an astronomical telescope,

Question

(i) Draw a ray diagram for the formation of image of an object by an astronomical telescope, in normal adjustment. Obtain the expression for its magnifying power.

(ii) The magnifying power of an astronomical telescope in normal adjustment is 2.9 and the objective and the eyepiece are separated by a distance of 150 cm. Find the focal lengths of the two lenses.

Answer 1

\(\text{Magnifying power} = \frac{\beta}{\alpha} = \frac{\left[\frac{BA}{-u_e}\right]}{\frac{BA}{f_0}} = -\frac{f_0}{u_e}\)

Since the final image is formed at the infinity hence

\(|u_e| = |f_e|\)

\(\Rightarrow \ m = \frac{-f_0}{f_e}\), for the formation of image in the normal adjustment.

(ii) \(m = -\frac{f_0}{f_e} \text{[for normal adjustment]}\)

\(\Rightarrow \text{also}, \ L = f_0 + f_e \Rightarrow 150 = f_0 + f_e \ \ ...(1)\)

and \(\frac{f_0}{f_e} = 2.9 \Rightarrow f_0 = 2.9 f_e \ \ .. (2)\)

putting the value obtained from equation (2) in equation (1)

\(\Rightarrow 2.9f_e + f_e = 150 \Rightarrow f_e = \frac{150}{3.9} = 38.46 \ cm\)

and \(f_0 = 2.9\ f_e \Rightarrow f_0 = 111.538 \ cm\) 

\(\Rightarrow \ f_0 = 111.538 \ cm \ \text{and} \ f_e = 38.46 \ cm\)