Physics

Physics

Problem

A thin concave lens has a focal length of 20 cm20\,\text{cm}. A point object is moved along the principal axis of the lens. a. (i) When the object is 40 cm40\,\text{cm} in front of the lens, calculate the position and linear magnification of the image. b. (ii) The object is now brought to a distance of 25 cm25\,\text{cm} from the lens. Calculate the new image distance and magnification. c. (iii) Using your results in parts (i) and (ii), sketch qualitatively the variation of magnification with object distance for the range 25 cm25\,\text{cm} to infinity and discuss the trend shown by the graph.
Intuitive explanation

A concave lens always produces virtual, upright, diminished images—the key is understanding how the image distance and magnification change as the object moves. Since the focal length is negative for a concave lens, you'll use the lens equation 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} to find where the image forms for different object positions, then track how magnification m=−didom = -\frac{d_i}{d_o} evolves as the object approaches or recedes from the lens.

Intuitive explanation

A concave lens always produces virtual, upright, diminished images—the key is understanding how the image distance and magnification change as the object moves. Since the focal length is negative for a concave lens, you'll use the lens equation 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} to find where the image forms for different object positions, then track how magnification m=−didom = -\frac{d_i}{d_o} evolves as the object approaches or recedes from the lens.

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