$(a)$ $A$ small telescope has an objective lens of focal length $140\;cm$ and an eyepiece of focal length $5.0\;cm$. 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 $25\;cm$?

(C) Given: Focal length of the objective lens,$f_{o} = 140\;cm$. Focal length of the eyepiece,$f_{e} = 5.0\;cm$.
$(a)$ In normal adjustment,the separation between the objective lens and the eyepiece is $L = f_{o} + f_{e} = 140 + 5 = 145\;cm$.
$(b)$ Height of the tower,$h_{1} = 100\;m$. Distance of the tower from the telescope,$u = 3\;km = 3000\;m$. The angle subtended by the tower at the telescope is $\theta = \frac{h_{1}}{u} = \frac{100}{3000} = \frac{1}{30}\;rad$. The angle subtended by the image produced by the objective lens is $\theta = \frac{h_{2}}{f_{o}}$,where $h_{2}$ is the height of the image. Thus,$\frac{h_{2}}{140} = \frac{1}{30}$,which gives $h_{2} = \frac{140}{30} \approx 4.67\;cm$.
$(c)$ The image is formed at a distance $d = 25\;cm$. The magnification of the eyepiece is $m = 1 + \frac{d}{f_{e}} = 1 + \frac{25}{5} = 6$. The height of the final image is $h_{f} = m \times h_{2} = 6 \times 4.67 = 28.02\;cm$ (or $28.2\;cm$ if using $140/30$ exactly).