Two plane mirrors $$M_1$$ and $$M_2$$ are at right angle to each other as shown. A point source $$P$$ is placed at $$a$$ and $$2a$$ meter away from $$M_1$$ and $$M_2$$ respectively. The shortest distance between any two images thus formed is: (Take $$\sqrt{5} = 2.3$$)

We need to find the shortest distance between the images formed by two plane mirrors placed at a right angle ($$90^\circ$$) to each other, given the position of a point source $$P$$.

1. Coordinate Assignment

Let the intersection of the two perpendicular mirrors, $$M_1$$ and $$M_2$$, be the origin $$(0,0)$$.

• Let mirror $$M_1$$ lie along the x-axis.
• Let mirror $$M_2$$ lie along the y-axis.

The point source $$P$$ is placed at a distance of $$a$$ from $$M_1$$ (y-coordinate) and $$2a$$ from $$M_2$$ (x-coordinate). Therefore, the coordinates of the object source $$P$$ are:

$$P = (2a, a)$$

2. Locate the Positions of the Images

When an object is placed between two mutually perpendicular mirrors, three distinct images are formed:

1. Image $$I_1$$ (formed by reflection from $$M_1$$):

   Mirror $$M_1$$ (x-axis) flips the y-coordinate sign: $$I_1 = (2a, -a)$$

2. Image $$I_2$$ (formed by reflection from $$M_2$$):

   Mirror $$M_2$$ (y-axis) flips the x-coordinate sign: $$I_2 = (-2a, a)$$

3. Image $$I_3$$ (the third/overlapping image formed by secondary reflection):

   Both coordinates are inverted relative to the origin: $$I_3 = (-2a, -a)$$

3. Calculate the Distances Between the Images

We need to identify the shortest distance among the pairs of images ($$I_1$$, $$I_2$$, and $$I_3$$):

• Distance between $$I_1$$ and $$I_3$$:

  $$d(I_1, I_3) = \sqrt{(-2a - 2a)^2 + (-a - (-a))^2} = \sqrt{(-4a)^2 + 0} = 4a$$

• Distance between $$I_2$$ and $$I_3$$:

  $$d(I_2, I_3) = \sqrt{(-2a - (-2a))^2 + (-a - a)^2} = \sqrt{0 + (-2a)^2} = 2a$$

• Distance between $$I_1$$ and $$I_2$$:

  $$d(I_1, I_2) = \sqrt{(-2a - 2a)^2 + (a - (-a))^2} = \sqrt{(-4a)^2 + (2a)^2} = \sqrt{16a^2 + 4a^2} = \sqrt{20a^2} = 2\sqrt{5}a$$

Correct Option: C