Question

(a) For the telescope described in Exercise 9.34 (a), what is theseparation 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 objectivelens? (c) What is the height of the final image of the tower if it is formed at 25 cm?

Answer 1

(a)

Given,

Focal length of the objective lens, f 0 =140 cm

Focal length of the eyepiece, f e =5 cm

In normal adjustment, separation between objective and the eyepiece is

f e + f 0 =5+140 =145 cm

(b)

Given

Height of the tower, h 1 =100 m

Distance of tower from the telescope is,

u=3 km =3000 m

Let height of the image of the tower formed by objective lens be h 2 .

Angle subtended by tower at telescope is same as the angle subtended by image produced by the objective lens.

Thus,

h 1 u = h 2 f 0

Or 100 3000 = h 2 140

Or h 2 =4.67 cm

Thus, height of the image of tower formed by objective lens is 4.67 cm.

(c)

Image is formed at a distance of d=25 cm.

Thus magnification factor is,

m=1+ d f e

Substitute the values in above expression.

m=1+ 25 5

m=6

So the height of the final image is,

h f = mh 2 =6×4.67 =28.02 cm

Thus, the height of the final image of tower will be 28 cm.