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Question
- For the telescope is in normal adjustment (i.e., when the final image is at infinity)? what is the separation between the objective lens and the eyepiece?
- 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?
- What is the height of the final image of the tower if it is formed at 25 cm?
Solution
Focal length of the objective lens, fo = 140 cm
Focal length of the eyepiece, fe = 5 cm
(a) In normal adjustment, the separation between the objective lens and the eyepiece = fo + fe = 140 + 5 = 145 cm
(b) Height of the tower, h1 = 100 m
Distance of the tower (object) from the telescope, u = 3 km = 3000 m
The angle subtended by the tower at the telescope is given as:
θ = `"h"_1/"u"`
= `100/3000`
= `1/30` rad
The angle subtended by the image produced by the objective lens is given as:
θ = `"h"_2/"f"_"o" = "h"_2/140 "rad"`
Where,
h2 = Height of the image of the tower formed by the objective lens
`1/30 = "h"_2/140`
`"h"_2 = 140/30`
∴ h2 = 4.7 cm
Therefore, the objective lens forms a 4.7 cm tall image of the tower.
(c) Image is formed at a distance, d = 25 cm
The magnification of the eyepiece is given by the relation:
`"m" = 1 + "d"/"f"_"e"`
= `1 + 25/5`
= 1 + 5
= 6
Height of the final image = mh2 = 6 × 4.7 = 28.2 cm
Hence, the height of the final image of the tower is 28.2 cm.
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