A 4.0-cm tall object is placed 50.0 cm from a diverging lens having a focal length of magnitude 25.0 cm.What are the nature and location of the image? The image is
A) real, 4.0 cm tall, and 20 cm from the lens on the side opposite the object.
B) virtual, 4.0 cm tall, and 20 cm from the lens on the side opposite the object.
C) virtual, 2.0 cm tall, and 10 cm from the lens on the side opposite the object.
D) virtual, 1.3 cm tall, and 16.7 cm from the lens on the same side of the object.
E) real, 1.3 cm tall, and 16.7 cm from the lens on the same side of the object.

Question

Quizplus · Accepted Answer

Diverging lenses always produce virtual images that are smaller than the object and located on the same side of the lens as the object. The image distance (v) can be found using the lens equation 1/f = 1/v + 1/u, where f is the focal length and u is the object distance. With f = -25.0 cm (negative for a diverging lens) and u = -50.0 cm (negative since the object is on the opposite side of the lens from where the image forms), solving for v gives v = 16.7 cm, indicating the image is virtual, smaller, and on the same side as the object. The magnification (m = v/u) can be used to find the image height, confirming it is smaller than the object.

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