A small telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the final image of the tower if it is formed at 25 cm?

30 cm

The correct answer is Option (2) → 30 cm

(M.P.) = $\frac{-f_0}{f_e}\left[1+\frac{f_e}{D}\right]$

M.P. = $\frac{-150}{5}\left[1+\frac{5}{25}\right]$

$=-30\left[\frac{6}{5}\right]=-36$

$m=\frac{\tan \beta}{\tan \alpha}$

$\tan \alpha=\frac{H}{u}=\frac{100}{3000}=\frac{1}{30}$        [∵ u = 3 km]

$\tan \beta=\frac{-36}{30}=\frac{H'}{D} \Rightarrow H'=\frac{36 \times 25}{305}$ = +30 cm