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An infinity long rod lies along the axis...

An infinity long rod lies along the axis of a concave mirrror of focal length f.The near end of a the rod is at a distance `ugtf`from the mirror ,Its image will have a length:

A

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

B

`(uf)/(u-f)`

C

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

D

`(uf)/(u+f)`

Text Solution

Verified by Experts

The correct Answer is:
A
`(1)/(v)+(1)/(u)=(1)/(f) rArr "let"v_(1)="distance of near end of image"`
`(1)/(V_(1))+(1)/(-u)=(1)/(-f) rArr (1)/(v_(1))=(1)/(u)+(1)/(-f) rArr (1)/(v_(1))=(f-u)/(uf)`
`v_(1)=(uf)/(f-u)`
`v_(2) = "distance of last end is Image" rArr (1)/(v_(2))+(1)/(oo)=(1)/(-f) rArr v_(2)=-f`
`"Length of image"|v_(1)-v_(2)|= (uf)/(f-u)-(-f) rArr |v_(1)-v_(2\)|=(uf-uf+f^(2))/(u-f)`
`|v_(1)-v_(2)| = (f^(2))/(u-f)`
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