Question

Sol. [Key Concept: Apply Snell's law in vector form at the interface ($\mu_{1}(incident \times \ n) = \mu_{2}(refracted \times \ n)$) to solve for the component $C = 1.6$.]

$${1(AO \times \widehat{j}) \Rightarrow (OB \times \widehat{j}) \times 1.5 }{\frac{2\widehat{k}}{\sqrt{13}} = \frac{3}{2}\left\lbrack \frac{C\widehat{k}}{\sqrt{C^{2} + 16}} \right\rbrack }{\therefore\frac{4}{13} = \frac{9C^{2}}{4\left( C^{2} + 16 \right)} \Rightarrow 16C^{2} + 256 = 117C^{2} }{256 = 101C^{2} }{C = 1.59 \approx 1.6 }$$