(a) For the telescope described what is the separation between the objective lens and eye piece ?
(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 25 cm` ?

Explanation

Focal length of the objective lens,fo = 140 cm
Focal length of the eyepiece, fe=5cm
a) In normal adjustment, the separation between the objective lens and the eyepiece = `f_(0) + f_(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:
`theta=h_(1)/mu`
`=100/3000 = 1/30 rad`
The angle subtended by the image produced by the objective lens is given as:
`theta=h_(2)/f_(0)=h_(2)/(140) rad`
The angle subtended by the image produced by the objective lens is given as:
`theta= h_(2)/f_(0)=h_(2)/140 rad`
where, h2= Height of the image of the tower formed by the objective lens
`1/30=h_(2)/140`
`therefore = 140/30= 4.7cm`
Therefore, the objective lens forms a 4.7 cm tall image of the tower.
c) Image is formed at a distance