Practice question
Question
A telescope has an objective of focal length \( 160 \, \text{cm} \) and an eyepiece of focal length \(
8 \, \text{cm} \). What is its magnifying power?
Explanation
**Convex lens image formation**: object beyond 2F (u>2f) real inverted diminished between F and 2F, at 2F same size at 2F, between F and 2F magnified beyond 2F, at F image at infinity, within F virtual erect magnified same side. For f=15 cm, u=30 cm=2f, m = v/u =30/30=1? Actually v=30 cm, m=-1, same size inverted. Magnifying power: m = (f_o/f_e) . f_o = 160 cm , f_e = 8 cm . m = (160/8) = 20 . Substituting values gives 20, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v + 1/u (mirror),
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.