Q.A concave mirror of curvature 40 cm, used for shaving purpose produces image of double size as that of the object. Object distance must be (A) 10 cm only (B) 20 cm only (C) 30 cm only (D) 10 cm or 30 cm

[!TLDR] A concave mirror (f=-20 cm) doubles an object's size either as a real, inverted image at u=-30 cm or as a virtual, …

R = 40 cm, so f = R/2 = -20 cm (concave). Real, inverted, magnified case: m = -v/u = -2, i.e. v = 2u. Substituting into 1/f = 1/v + 1/u = 1/(2u) + 1/u = 3/(2u) gives 1/(-20) = 3/(2u), so u = -30 cm (this lies between f and 2f, consistent with a real, magnified image). Virtual, erect, magnified case: m = -v/u = +2, i.e. v = -2u. Then 1/f = 1/(-2u) + 1/u = 1/(2u) gives 1/(-20) = 1/(2u), so u = -10 cm (object inside the focus, consistent with a virtual magnified ima …

Set m = -v/u equal to -2 (real, inverted) and separately to +2 (virtual, erect), substitute v in terms of u int …

Solving only the real-image case and missing that a concave mirror can also produce a virtual, erect, doubled i …