Find two positions of an object, placed in front of a concave mirror of focal length 15cm, so that the image formed is 3 times the size of the object?
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Answer:
Using mirror formula:
1/u+1/3u = -1/15 gives u = -20 cm.
OR
1/u - 1/3u = -1/15 gives u = -10 cm.
Explanation:
Given image is 3 times that of object. so there is 2 case: image is real , inverted and magnified OR the image is virtual , erect and magnified. so m= 3 or -3 . hence v = 3u or v = -3u . f = -15 cm (given).
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