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

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#### Solution

Focal length of the objective lens, f_{o} = 140 cm

Focal length of the eyepiece, f_{e} = 5 cm

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 = mh_{2} = 6 × 4.7 = 28.2 cm

Hence, the height of the final image of the tower is 28.2 cm.

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