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?
Given:
Focal length of the objective lens, f0 = 140 cm
Focal length of the eyepiece, fe = 5 cm
Height of the tower, h1 = 100 m
Distance of the tower (object) from the telescope, u = 3Km = 3000 m
The angle subtended by the tower at the telescope is given as:
θ = h1/ u
⇒ θ = 100/3000
⇒ θ = 1/30 rad
The angle subtended by the image produced by the objective lens is given as:
θ = h2/ fo
⇒ θ = h2 /140 rad
Where, h2 = height of the image of the tower formed by the objective lens
1/30 = h2 /140
Therefore,
h2 = 140/30
⇒ h2 = 4.7 cm
Therefore, the objective lens forms a 4.7 cm tall image of the tower.