Correct Answer - Image will be at `-(5 R)/(6)` from near end towards souce ; image will be at - (R)/(2)` from remote end
Case- I
Applying mirror formula for remote pair
`(u=-(3R)/(2),f=-(R)/(2))`
`rArr v=(uf)/(u-f)=(-3)/(4)R`
Now consider reflection from the nearer part for nearer part
`u=(3R)/(4)-2R=-(5R)/(4) rArr v=(-5R)/(6)`
Case II
Consider reflection form nearer part first
`u=-(R)/(2)`
`v= (uf)/(u-f)=(-3)/(4)R`
Now consider reflection from the nearer part for nearer part
`u=(3R)/(4)-2R =-(5R)/(4) rArr v=(-5R)/(6)`
Case II
Consider reflection from nearer part first
`u=-(R)/(2)`
`v=(uf)/(u-f)=infty`
Reflection from remote part gives `v= - (R)/(2)`
Image will be at `-(5R)/(6)` from near end towards source , image will be at `-(R)/(2)` from remote end.