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
`prop ~= tan prop = (100)/(3 xx 10^3) = (1)/(30)` radian …(i)
If `h` is the height of image of tower , then angle subtended by the image must also be `prop`.
`:. prop ~= tan prop = (h)/(f_0) = (h)/(150)` ...(ii)
From (i) and (ii), `(h)/(150) = (1)/(30). h = 5 cm`
Magnification produced by 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`.