A 1.00 cm-high object is placed 3.40 cm to the left of a converging lens of...

A 1.00 cm-high object is placed 3.40 cm to the left of a converging lens of focal length 8.95 cm. A diverging lens of focal length −16.00 cm is 6.00 cm to the right of the converging lens. Find the position and height of the final image.

position
Take the image formed by the first lens to be the object for the second lens and apply the lens equation to each lens to locate the final image. cm ---Select--- behind the second lens in front of the second lens height
Calculate the magnification produced by each lens. Then consider how the magnification relates image size and object size for each lens to find the height of the final image. cm

Is the image inverted or upright?

uprightinverted

Is the image real or virtual?

realvirtual

(b)

What If? If an image of opposite characteristic, i.e., virtual if the image in part (a) is real and real if the image in part (a) is virtual, is to be obtained, what is the minimum distance (in cm), and in which direction, that the object must be moved from its original position?

distance cmdirection ---Select--- to the left to the right

Expert Solution

genius_generous answered 2 years ago

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