Question

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 $12\text{cm}$ 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 $\text{cm})$ between these images is

Answer 1

We will assume that observer sees the image of object through edge, as per the given configuration of square
$\Rightarrow \alpha ={45}^{\circ }$
$AB=\frac{12d\alpha }{\cos \alpha }=\frac{xd\theta }{\cos \theta }$
By applying Snell's Law
$\frac{4}{3}\sin \alpha =1\sin \theta \dots \left(1\right)$
$\because 1\sin \theta =\frac{4}{3}\sin \alpha$
$\Rightarrow \sin \theta =\frac{2\sqrt{2}}{3}$
$\Rightarrow \cos \theta =\frac{1}{3}$
Differentiating both sides of eq $\left(1\right)$,
$\frac{4}{3}\cos \alpha d\alpha =\cos \theta d\theta$
$\Rightarrow \frac{9}{{\cos }^2\alpha }=\frac{x}{{\cos }^2\theta }$
$\Rightarrow x=18\times \frac{1}{9}=2$
$d=2x\sin (\theta -\alpha )$
Using trigonometric relation,
$d=4\times \frac{1}{\sqrt{2}}(\frac{2\sqrt{2}}{3}-\frac{1}{3})=\frac{8-2\sqrt{2}}{3}\approx 1.73\approx 2$