wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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?

Open in App
Solution

(a)

Given,

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

Focal length of the eyepiece, f e =5cm

In normal adjustment, separation between objective and the eyepiece is

f e + f 0 =5+140 =145cm

(b)

Given

Height of the tower, h 1 =100m

Distance of tower from the telescope is,

u=3km =3000m

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 =4.67cm

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

(c)

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

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 =28.02cm

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


flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Telescope
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon