A small telescope has an objective lens of focal length `150 cm` and and eye piece of focal length `5 cm`. If his telescope is used to view a `100 m` high tower `3 km` away, find the height of the final image when it is formed `25 cm` away from the eye piece.

Explanation

Here, `f_(0) = 150 cm, f_(e) = 5 cm`
Angle subtended by `100 m` tall tower at `3 km`
away is
`alpha ~= tan alpha = (100)/(3 xx 10^(3)) = (1)/(30)radian` …(i)
if `h` is the heigth of image of tower, then angle suntended by the image must also be `alpha`.
`:. alpha ~= tan alpha = (h)/(f_(0)) = (h)/(150)` ...(ii)
From (i) and (ii), `(h)/(150) = (1)/(30), h = 5 cm`.
Magnification produced bt eye piece
`m_(e) = (1 + (d)/(f_(e))) = 1 + (25)/(5) = 6`
`:.` Height of final image, `h' = h xx m_(e) = 5 xx 6`
`= 30 cm`