Find the position of the final image after three successive reflections taking the first reflection on `m_(1)` .

Explanation

`1^(st)` reflection at `m_(1)`
`u = -15cm`
`f = -10cm`
`(1)/(v) = (1)/(f) - (1)/(u)`
`:. V = (uf)/(u-f)`
`(-15)xx(-10))/((-15)+10) = (150)/(-5)cm = -30cm`.
Thus, image is formed at a point `5cm` right of `m_(2)` which will act as an objdct for the reflection at `m_(2)`
`u = 5cm, f = 10cm`
`v = (uf)/(u-f) = (5xx10)/(5-10) = (50)/(-5) = -10cm`
`3^(rd)` reflection at `m_(1)` again.
`mu = -15cm, f = -10cm`
`v = (uf)/(u-f) = (-15xxx(-10))/((-15)+10) = -30cm`
Image is formed at 30cm` right of `m_(1)`