A thin rod of length f3 lies along the axis of a concave mirror of focal length f. One end of its magnified image touches an end of the rod. The length of the image is
A
f
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B
f2
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C
2f
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D
f4
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Solution
The correct option is Bf2 If end A of rod acts an object for mirror then it's image will be A' and if u=2f−f3=5f3
So by using 1f=1v+1u⇒1−f=1v+1−5f3⇒v=−52f ∴ Length of image =52f−2f=f2