(a) all the reflected rays meet at a point when produced backward
Explanation:
See the diagram below,
Take any ray PR which is reflected at Q and QR is the reflected ray. Since the angle of incidence is equal to the angle of reflection,
∠PQN =∠RQN=∠OPQ =α RQ, when produced back, meets PO produced at P'. In right-angled triangles PQO and P'QO, ∠QPO =∠QP'O =α
Thus both are equal in all respect. So, PO =P'O.
For any other ray, it can be proved. Hence the option (a). P' becomes a virtual image of P.