(a) A 1.00 cm-high object is placed 4.05 cm to the left of a converging lens of focal length 7.35 cm. A diverging lens of focal length -16.00 cm is 6.00 cm to the right of the converginglens. Find the position and height of the final image.positionheightinvertedTake 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. cmin front of the second lens ✓Is the image inverted or upright?O uprightCalculate 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 finalimage. cmrealvirtualIs the image real or virtual?(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?distancecmdirectionto the left

With sign conversation i.e, the distance measured in the direction of incident light is taken as…

Answered: (a) A 1.00 cm-high object is placed…

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