Question

(a) For the telescope described in Exercise 9.27 (a), what is the separation between the objective lens and the eyepiece? 

(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens? 

(c) What is the height of the final image of the tower if it is formed at 25cm?

Answer 1

Focal length of the objective lens, ƒ_º = 140cm
Focal length of the eyepiece, ƒ_e = 5cm
(a) In normal adjustment, the separation between the objective lens and the eyepiece = ƒ_º + ƒ_e = 140 + 5 = 145cm
(b) Height of the tower, h₁ = 100m
Distance of the tower (object) from the telescope, u = 3km = 3000m the angle subtended by the tower at the telescope is given as :\( θ = \frac{h_1 }{ u} \)

= \(\frac{100 }{ 3000} \)
=\( \frac{1 }{ 30} \)rad
The angle subtended by the image produced by the objective lens is given as : \(θ =\frac{h_2 }{ ƒ_º} = \frac{h_2 }{ 140} rad\)
Where, 
h₂ = Height of the image of the tower formed by the objective lens
\(\frac{1 }{ 30} = \frac{h_2 }{ 140} \)
\(∴ h_2 = \frac{140 }{ 30} = 4.7cm \)
Therefore, the objective lens forms a 4.7cm tall image of the tower.
(c) Image is formed at a distance, d = 25cm
The magnification of the eyepiece is given by the relation : \(m = 1 +\frac{ d }{ ƒ_e} \)
= \(1 + \frac{25 }{ 5} \)
= 1 + 5 
= 6
Height of the final image = mh₂ = 6 × 4.7 = 28.2cm
Hence, the height of the final image of the tower is 28.2cm.