A large square container with thin transparent vertical walls and filled with water (refractive index $$\frac{4}{3}$$) is kept on a horizontal table. A student holds a thin straight wire vertically inside the water 12 cm from one of its corners, as shown schematically in the figure. Looking at the wire from this corner, another student sees two images of the wire, located symmetrically on each side of the line of sight as shown. The separation (in cm) between these images is ____________.

Let O be the corner of the square tank from which the observer looks in. Choose the horizontal x-axis along one glass wall and the y-axis along the other, with their common origin at O. The water occupies the region $$x \gt 0,\; y \gt 0$$.

The vertical wire is held at the point $$P\,(s,\,s)$$ so that it is equidistant from the two walls. Because the straight-line distance from P to the corner is given to be 12 cm,

$$OP=\sqrt{s^{2}+s^{2}}=\sqrt{2}\,s = 12\ \text{cm}$$ $$\Rightarrow\; s=\frac{12}{\sqrt{2}}=6\sqrt{2}\ \text{cm}$$

The observer sees two virtual images, one formed by refraction through the wall at $$x=0$$ and the other through the wall at $$y=0$$.

Image through the wall $$x=0$$
For a plane surface separating water ($$n_2=\tfrac{4}{3}$$) and air ($$n_1=1$$), an object at a real perpendicular distance d inside the denser medium appears at the reduced perpendicular distance $$d'=\frac{n_1}{n_2}\,d=\frac{1}{\tfrac{4}{3}}d=\frac{3}{4}d.$$ Here the perpendicular (normal) distance of P from the plane $$x=0$$ is $$s$$, so the apparent distance is

$$x' = \frac{3}{4}s.$$

The coordinate parallel to the wall (the y-coordinate) does not change in plane refraction. Hence the first image is at $$I_1\!\left(\frac{3s}{4},\,s\right).$$

Image through the wall $$y=0$$
By the same reasoning, the second image is at $$I_2\!\left(s,\,\frac{3s}{4}\right).$$

Separation between the two images

$$\begin{aligned} I_1I_2 &= \sqrt{\left(s-\frac{3s}{4}\right)^{2}+\left(\frac{3s}{4}-s\right)^{2}} \\[4pt] &= \sqrt{\left(\frac{s}{4}\right)^{2}+\left(-\frac{s}{4}\right)^{2}} \\[4pt] &= \frac{s}{4}\,\sqrt{2}. \end{aligned}$$

Substituting $$s = \dfrac{12}{\sqrt{2}}$$:

$$I_1I_2 = \frac{1}{4}\left(\frac{12}{\sqrt{2}}\right)\sqrt{2}= \frac{12}{4}=3\ \text{cm}.$$

Therefore, the two images are separated by 3 cm.

Final Answer : 3 cm