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Karnataka Board PUCPUC Science 2nd PUC Class 12

(a) 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?

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Question

  1. 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?
  2. 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?
  3. What is the height of the final image of the tower if it is formed at 25 cm?
Numerical
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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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Chapter 9: Ray Optics and Optical Instruments - EXERCISES [Page 252]

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NCERT Physics Part I and II [English] Class 12
Chapter 9 Ray Optics and Optical Instruments
EXERCISES | Q 9.28 | Page 252
NCERT Physics Part I and II [English] Class 12
Chapter 9 Ray Optics and Optical Instruments
Exercise | Q 9.29 | Page 347

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