An object is placed at a distance of f2 from a convex lens. The image will be (1) At one of the foci, virtual and double its size. (2) At 3f2, real and inverted.
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Solution
1f=1f+1v 1f=2f+1v −2f=1f+1v v=−f The image is virtual and on the negative focal point. (The image is upright and magnified by 2.)