Answer
See answers.
Work Step by Step
The “relaxed eye” tells us that the image is at infinity. The distance between the two lenses is the sum of the 2 focal lengths. Combine this with equation 25–3 and solve for the focal lengths.
$$M=-\frac{f_o}{f_e}$$
$$f_o=-Mf_e$$
$$\ell=f_o+f_e=(1-M)f_e$$
The magnification for an astronomical telescope is negative.
$$f_e=\frac{\ell}{1-M}=\frac{1.10m}{1-(-120)}=9.09\times10^{-3}m$$
$$f_o=-Mf_e=-(-120)( 9.09\times10^{-3}m)=1.09m$$
