An object is placed 33.0 cm in front of a diverging lens with a focal length of 19.9 cm. (a) Find the image distance. (b) Is the image real or virtual? c) Suppose the object is brought to a distance of 11.2 cm in front of the lens. Where is the image now with respect to its previous location? (Note: Ensure the sign convention you use is consistent by treating all image distances on the object side of the lens as negative.) d) How has the height of the image changed if the object is 81.9 cm tall?
Added by Justin B.
Step 1
9\) cm (negative because it's a diverging lens), and \(d_o = -33.0\) cm (negative because it's on the same side as the object), we can solve for the image distance \(d_i\). Show more…
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