Practice question

**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. Power: P = (1/f) (in meters). P = -3 D ⇒ -3 = (1/f) ⇒ f = -(1/3) ≈ -0.333 m ≈ -33.3 cm . Substituting values gives -33 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f =