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Question

5. An object of height 5cm is placed at a distance 5cm in front of a bi-concave lens of focal length 10 cm.

a. Draw the ray diagram to locate the image formed by the lens. b. Use the lens equation to find the distance of the image formed. c. What is the height and nature (virtual/real/inverted/upright) of the image formed by the lens?

Answer 1

a) I won't attempt to draw on this screen, but it looks like an upright and virtual image between the object and the lens.

b) From the lens equation, 1/d_i = 1/f - 1/d_o = 1/(-10) - 1/5 = -0.30 because focal length is negative for a diverging lens. Image distance = -3.33 cm, i.e. on the same side of the lens as the object.

c) Magnification = -d_i /d_o = +3.33/5 = 0.667, so the height is 0.667 * 5 cm = 3.33 cm. The nature is indeed virtual and upright.