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A 4.0-cm-tall object is placed 60.0 cm from a diverging lens having a focal length of magnitude 30.0 cm. What is the nature and location of the image?

A. A virtual image, 2.0 cm tall, 10 cm other side of the object
B. A virtual image, 1.3 cm tall, 20.0 cm same side as the object
C. A virtual image, 4.0 cm tall, 20 cm other side of the object
D. A real image, 1.3 cm tall, 20.0 cm same side as the object
E. A real image, 4.0 cm tall, 4.0 cm other side of the object

1 Answer

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Final answer:

The image formed by a diverging lens with a focal length of -30.0 cm and an object distance of 60.0 cm is virtual, 1.3 cm tall, and located 20.0 cm on the same side as the object.

Step-by-step explanation:

To find the nature and location of the image formed by a diverging lens, we use the lens formula 1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance. Since diverging lenses always have a negative focal length, we plug in f = -30.0 cm and do = 60.0 cm into the formula to find di.

After solving, we get 1/(-30) = 1/60 + 1/di, which gives us di = -20.0 cm. The negative sign indicates that the image is on the same side as the object and is virtual. To find the magnification, we use the magnification formula m = -di/do, which results in m = -20/60 = -1/3. The final image height is then the magnification multiplied by the object height, giving us an image height of 4.0 cm * -1/3 = -1.3 cm. The negative magnification indicates the image is upright.

Therefore, the correct answer is B. A virtual image, 1.3 cm tall, 20.0 cm same side as the object.