14. An object is 2 cm from a concave (diverging) lens. The resulting height of the virtual image is half as large as the
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14. An object is 2 cm from a concave (diverging) lens. The resulting height of the virtual image is half as large as the
14. An object is 2 cm from a concave (diverging) lens. The resulting height of the virtual image is half as large as the the height of the object. What is the focal length of the lens? (Hint: Use the information given to determine the lateral magnification, and from that determine the image distance. The use the thin lens equation.) (A) - cm (B) - cm (C) -1 cm (D) –2 cm
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