According to the given figure, we have two plane mirror, placed at 90º to each other. Since, image formed by plane mirror are virtual, erect, same size and at same distance behind mirror. Let I1 is the image of P due to mirror M1 and I2 is the image of P due to mirror M2
Therefore, the shortest distance between images i.e., between I1 and I2 I1I2=√(4a)2+(2a)2 =√16a2+4a2 =a√20=a2√5 =a×2×2.3=4.6a
