A real inverted image in a concave mirror is represented by (`u`,`v`, `f` are corrdinates)

Options

A.
B.
C.
D.

Answer

A

Explanation

For real inverted image formed by concave mirror.
`v= -ve, u= -ve, f -ve`
`implies(u)/(f)&(v)/(f)` are positive
So graph show be in 1st qudarant and `(1)/(v)+(1)/(u)=(1)/(f)`
`implies(f)/(v)+(f)/(u)=1`
This graph can also be realized by Newtons formula `x_(1)x_(2)=f^(2)` where `x_(1)`, and `x_(2)` are distance from focus so graph will be rectangular hyperbola
`implies(1)` is right answer.