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+1 vote
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in Optics by (64.8k points)

(a) For the telescope described, what is the separation 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 objective lens? 

(c) What is the height of the final image of the tower if it is formed at 25 cm?

1 Answer

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Best answer

(a) The separation between the objective lens and the eyepiece 

= f0 + fe = 140 + 5 = 145 cm. 

(b) height of the tower = 100 m; distance = 3 km = 3000 m.

Angle subtended by the tower = 100/3000 = 1/30 rad

Angle subtended by the image produced by the objective 

= h/f0 = h/140

∴ h/140 = 1/30 

∴ h = 140/30 = 4.7cm

(c) Magnification produced by eyepiece 

= (1 + d/fe) = (1 + 25/5) = 6

∴ Height of final image = 14/3 x 6 = 28 cm.

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