91Ó°ÊÓ

To find the incident angle, we first calculate the direction cosine of vector \( \mathbf{A} = 6 \sqrt{3} \hat{\mathbf{i}} + 8 \sqrt{3} \hat{\mathbf{j}} - 10 \hat{\mathbf{k}} \). The incident angle \( \theta_i \) is the angle between \( \mathbf{A} \) and the positive \( z \)-axis \( \hat{\mathbf{k}} \). The cosine of \( \theta_i \) is computed by finding the dot product of \( \mathbf{A} \) and \( \hat{\mathbf{k}} \), and then dividing by their magnitudes:\[ \cos \theta_i = \frac{\mathbf{A} \cdot \hat{\mathbf{k}}}{||\mathbf{A}||} = \frac{-10}{||\mathbf{A}||} \]Calculate \( ||\mathbf{A}|| = \sqrt{(6\sqrt{3})^2 + (8\sqrt{3})^2 + (-10)^2} = \sqrt{108 + 192 + 100} = \sqrt{400} = 20 \).Then, \[ \cos \theta_i = \frac{-10}{20} = -0.5 \] which gives \( \theta_i = \cos^{-1}(-0.5) = 120^\circ \). However, since the angle needs to be between the ray and normal (z axis), adjust this to \( \theta_i' = 180^\circ - 120^\circ = 60^\circ \).

Snell's law relates the angles and refractive indices of the two media. The law is given by:\[ n_1 \sin \theta_i = n_2 \sin \theta_t \]Here, \( n_1 = \sqrt{2} \), \( \theta_i = 60^\circ \), \( n_2 = \sqrt{3} \). Plug in these values to find \( \theta_t \):\[ \sqrt{2} \sin(60^\circ) = \sqrt{3} \sin \theta_t \]Since \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \), the equation becomes:\[ \sqrt{2} \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \sin \theta_t \]Solving this, we get:\[ \frac{3}{2} = \sqrt{3} \sin \theta_t \]\[ \sin \theta_t = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2} \]Thus, \( \theta_t = 60^\circ \).

From the previous calculations, \( \sin \theta_t = \frac{\sqrt{3}}{2} \), which corresponds to an angle of \( \theta_t = 60^\circ \) in medium 2. Thus, the correct answer is option (b) \( 60^\circ \).