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) = -1//4`
As size of `P` is four times that of `Q`, therefore, to 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`.