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