Size of an object `P` is four times that of `Q`. It is required that the size of the image of `P` and `Q`, placed one after the other at certain distances away from a concave mirror of radius of curvature `20 cm` should be equal. To achieve this, if the distance of `P` from the mirror is `50 cm`, what must be the distance of `Q` ?

Answer

`- 20 cm`

Explanation

For object P : `u = -50 cm, f = -10 cm`
Use mirror formula to get `v = -50//4 cm`
`:. m = (-v)/(u) = (50//4)/(-50) = -14`
As size of `P` is four times that of `q`, therefore, obtain images of equal size, magnification of `Q` must be `4` times i.e., `4 xx (-1//4) = -1`. Hence `q` should be placed at centre of curvature i.e, at `- 20 cm`.