Question

5. A converging lens ( 12.0 cm) is located 30.0 cm to the left of a diverging lens f -6.00 cm). A 9.00 cm tall postage stamp is located 36.0 cm to the left of the converging lens. (A) Locate the final image of the stamp relative to the diverging lens. ditem (B) Find the final image heighth?2-Aiscr (C) Is the final image real or virtual?Virtudi (D) Is the final image upright or inverted (with respect to the original object)? nverto (E) Is the final image larger or smaller than the original object? Smauur

The answers are marked but I am going wrong somewhere still. can you explain using the thin lens equation and magnification equation. thanks!!

Answer 1

Object Distance for lens (converging lens) P_1 = 36 cm, q_1 = 12 cm image Distance q_1 = alpha 1/q_1 = 1/p_1 + 1/q_1 = 1/q_1 = 1/q_1 - 1/p_1 1/q_1 = 1/12 - 1/36 = 3 - 1/36 = q_1 = 18 cm + ve = real image to the right side object Distance for lens 2, 30-18 = 12 cm P_2 = 12 cm q_2 = -6 cm q_2 = alpha 1/q_2 = 1/-6 - 1/12 = -(2 + 1)/12 q_2 = -4 cm - ve = virtual image = 4 cm to the left side of Diverging lens. = -4 cm overall magnification M = M_1 M_2 = -q_1/p_1 times -q_2/p_2 M = - 18/36 times -(-4)/12 = 1/2 times -1/3 = -1/6 M = h_i/h_o = image height h_i = mh_o = -1/6 times 9 = -1.5 cm - 4cm = virtual inverted (h_i = -1.5 cm) because (-ve) smaller (1.5 cm)