Light ray incident along a vector \(\vec{AO}\) \((\vec{AO} = 2\hat{i}-3\hat{j})\) emerges out along vector \(\vec{OB}\) \((\vec{OB} = C\hat{i}-4\hat{j})\) as shown in the figure below. The value of \(C\) is ________.

From the given figure, the interface is horizontal and the normal is along the y-axis.

The incident ray is along the vector \(\vec{AO} = 2\hat{i} - 3\hat{j}\).
The angle of incidence \(\alpha\) is the angle between the incident ray and the normal.
\(\sin \alpha = \dfrac{2}{\sqrt{2^2 + (-3)^2}} = \dfrac{2}{\sqrt{13}}\)

The refracted ray is along the vector \(\vec{OB} = C\hat{i} - 4\hat{j}\).
The angle of refraction \(\beta\) is the angle between the refracted ray and the normal.
\(\sin \beta = \dfrac{C}{\sqrt{C^2 + (-4)^2}} = \dfrac{C}{\sqrt{C^2 + 16}}\)

Applying Snell's law at the interface:
\(\mu_1 \sin \alpha = \mu_2 \sin \beta\)
\(1 \times \dfrac{2}{\sqrt{13}} = 1.5 \times \dfrac{C}{\sqrt{C^2 + 16}}\)
\(\dfrac{2}{\sqrt{13}} = \dfrac{3}{2} \dfrac{C}{\sqrt{C^2 + 16}}\)

Squaring both sides:
\(\dfrac{4}{13} = \dfrac{9}{4} \left( \dfrac{C^2}{C^2 + 16} \right)\)
\(16(C^2 + 16) = 117 C^2\)
\(16 C^2 + 256 = 117 C^2\)
\(101 C^2 = 256\)
\(C^2 = \dfrac{256}{101}\)
\(C = \dfrac{16}{\sqrt{101}} \approx 1.592\)

Approximating \(\sqrt{101} \approx 10\), we get \(C \approx 1.6\).

Answer: \(1.6\)