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.