Along the principal axis of a convex lens having a focal length of 12 cm , a particle performs simple harmonic motion with an amplitude of 1.0 cm . The oscillation has its mean position located 20 cm away from the lens. Determine the amplitude of oscillation for the particle's image.

Step-by-step solution

Given the focal length of the convex lens, f = +12 cm . The mean position of the particle is at u = -20 cm and its amplitude of oscillation is A_o = 1.0 cm . The extreme positions of the oscillating particle on the principal axis are u₁ = -20 - 1 = -21 cm and u₂ = -20 + 1 = -19 cm . Using the lens formula, v = uf u+f , we find the corresponding extreme positions of the image. For u₁ = -21 cm : v₁ = -21 12 -21 + 12 = -252 -9 = 28 cm For u₂ = -19 cm : v₂ = -19 12 -19 + 12 = -228 -7 = 228 7 cm The path length of the i