Question

A large square container with thin transparent vertical walls and filled with water $\left(\right.$ refractive index $\left.\frac{4}{3}\right)$ 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 _____

Answer 1

$y$ co-ordinate of first image $S^{\prime}$ is ' $y$ '
$\frac{4}{3} \sin i=\sin 45^{\circ} $
$\Rightarrow \sin i=\frac{1}{\sqrt{2}} \frac{3}{4}$

$i \approx 32^{\circ} $
$\therefore \theta=45-i$
So slope of the line passing through $(-12,0)$ is
$m=\tan (45-i)=\frac{1-\tan i}{1+\tan i}$
$\Rightarrow m=\frac{\sqrt{23}-3}{\sqrt{23}+3}$
$\therefore \text { Eq } \Rightarrow y=\frac{(\sqrt{23}-3) x}{\sqrt{23}+3}+\frac{12(\sqrt{23}-3)}{(\sqrt{23}+3)}$
Other equation $y=-x$
Now $ y=\frac{-(\sqrt{23}-3) y}{\sqrt{23}+3}+\frac{12(\sqrt{23}-3)}{(\sqrt{23}+3)} $
$\Rightarrow y=\left(\frac{2 \times 23}{\sqrt{23}+3}\right)=\frac{12(\sqrt{23}-3)}{(\sqrt{23}+3)}$
$2 y=\frac{12(\sqrt{23}-3)}{\sqrt{23}} \approx 4.5 \approx 4$
So, separation between two images is $4\, cm$.