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An infinitely long rod lies along the ax...

An infinitely long rod lies along the axis of a concave mirror of focal length f. The near end of the rod is distance `u gt f` from the mirror. Its image will have length

A

`(uf)/(u-f)`

B

`(uf)/(u+f)`

C

`(f^(2))/(u+f)`

D

`(f^(2))/(u-f)`

Text Solution

Verified by Experts

The correct Answer is:
D


u = -u, f=-f `v_(A) = (uf)/(f-u)`
Image of last end of rod `v_(B) = -f`
Length = `v_(B) - v_(A) = -f - (uf)/(f-u) = (f^(2)+uf-uf)/(f-u)`
Length =` (f^(2))/(u-f)`, Hence [D]
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