Refraction Vector Calculation for Light Ray at Interface

To find the value of $C$, we analyze the given light ray refraction at the interface using Snell's Law.

1. Representation of Vectors and Normal

From the figure, the horizontal line represents the interface separating medium 1 ($\mu_1 = 1$) and medium 2 ($\mu_2 = 1.5$), while the vertical axis represents the normal to the interface at point $O$.

• Incident Ray Vector: $\vec{AO} = 2\hat{i} - 3\hat{j}$ The magnitude of the incident vector is: $|\vec{AO}| = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}$

• Refracted Ray Vector: $\vec{OB} = C\hat{i} - 4\hat{j}$ The magnitude of the refracted vector is: $|\vec{OB}| = \sqrt{C^2 + (-4)^2} = \sqrt{C^2 + 16}$

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2. Calculating Angles with the Normal

The angle of incidence $\alpha$ is the angle between the vector $\vec{AO}$ and the vertical normal line: $\sin\alpha = \frac{|\text{x-component of } \vec{AO}|}{|\vec{AO}|} = \frac{2}{\sqrt{13}}$

Similarly, the angle of refraction $\beta$ is the angle between the vector $\vec{OB}$ and the vertical normal line: $\sin\beta = \frac{|\text{x-component of } \vec{OB}|}{|\vec{OB}|} = \frac{C}{\sqrt{C^2 + 16}}$

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3. Applying Snell's Law

According to Snell's Law at the interface: $\mu_1 \sin\alpha = \mu_2 \sin\beta$

Substituting the known values ($\mu_1 = 1$ and $\mu_2 = 1.5 = \frac{3}{2}$): $1 \cdot \left(\frac{2}{\sqrt{13}}\right) = \frac{3}{2} \cdot \left(\frac{C}{\sqrt{C^2 + 16}}\right)$

Rearranging to isolate the fraction containing $C$: $\frac{C}{\sqrt{C^2 + 16}} = \frac{4}{3\sqrt{13}}$

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4. Solving for $C$

Squaring both sides of the equation: $\frac{C^2}{C^2 + 16} = \frac{16}{9 \times 13} = \frac{16}{117}$

Cross-multiplying to solve for $C^2$: $117 C^2 = 16 (C^2 + 16)$ $117 C^2 = 16 C^2 + 256$ $101 C^2 = 256$ $C^2 = \frac{256}{101}$ $C = \frac{16}{\sqrt{101}} \approx \frac{16}{10.05} \approx 1.592 \approx 1.6$

Thus, the value of $C$ is 1.6.

Correct Option: A