What is the height of the final image of the tower if it is formed at 25cm?


Given:


Focal length of the objective lens, f0 = 140 cm


Focal length of the eyepiece, fe = 5 cm


Image is formed at a distance, d = 25cm


The magnification of the eyepiece is given by the relation:


m = 1 + d/fe


m = 1 + 25/5


m = 6


Height of the final image H, is given by,


H = m × h2 = 6 × 4.7 = 28.2 cm


Hence, the height of the final image of the tower is 28.2 cm.


35
1