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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 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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प्रश्न

  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?
संख्यात्मक
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उत्तर

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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पाठ 9: Ray Optics and Optical Instruments - EXERCISES [पृष्ठ २५२]

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एनसीईआरटी Physics Part I and II [English] Class 12
पाठ 9 Ray Optics and Optical Instruments
EXERCISES | Q 9.28 | पृष्ठ २५२
एनसीईआरटी Physics Part I and II [English] Class 12
पाठ 9 Ray Optics and Optical Instruments
Exercise | Q 9.29 | पृष्ठ ३४७

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