Let the x-z plane be the boundary between two transparent media. Media 1 in `zge0` has refractive index of `sqrt2` and medium 2 with `zlt0` has a refractive index of `sqrt3`. A ray of light in medium 1 given by the vector `vecA=6sqrt3hati+8sqrt3hatj-10hatk` in incident on the plane of separation. The angle of refraction in medium 2 is

Options

A. `30^@`
B. `45^@`
C. `60^@`
D. `75^@`

Explanation

To solve the problem, we need to find the angle of refraction of a ray of light as it passes from medium 1 to medium 2 at the boundary defined by the x-z plane. We will use Snell's Law for this purpose.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Refractive index of medium 1, \( \mu_1 = \sqrt{2} \)
- Refractive index of medium 2, \( \mu_2 = \sqrt{3} \)
- Incident ray vector in medium 1:
\[ \vec{A} = 6\sqrt{3} \hat{i} + 8\sqrt{3} \hat{j} - 10 \hat{k} \]

2. **Calculate the Direction Cosines:**
- The direction cosines \( l, m, n \) are given by:
\[ l = \frac{A_x}{|\vec{A}|}, \quad m = \frac{A_y}{|\vec{A}|}, \quad n = \frac{A_z}{|\vec{A}|} \]
- First, calculate the magnitude of the vector \( |\vec{A}| \):
\[ |\vec{A}| = \sqrt{(6\sqrt{3})^2 + (8\sqrt{3})^2 + (-10)^2} \]
\[ = \sqrt{108 + 192 + 100} = \sqrt{400} = 20 \]
- Now calculate the direction cosines:
\[ l = \frac{6\sqrt{3}}{20} = \frac{3\sqrt{3}}{10}, \quad m = \frac{8\sqrt{3}}{20} = \frac{2\sqrt{3}}{5}, \quad n = \frac{-10}{20} = -\frac{1}{2} \]

3. **Apply Snell's Law:**
- Snell's Law states:
\[ \mu_1 \sin \theta_1 = \mu_2 \sin \theta_2 \]
- Here, \( \theta_1 \) is the angle of incidence and \( \theta_2 \) is the angle of refraction. The angle with respect to the normal (z-axis) is given by:
\[ \sin \theta_1 = \sqrt{1 - n^2} = \sqrt{1 - \left(-\frac{1}{2}\right)^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

4. **Substitute into Snell's Law:**
- Substitute the values into Snell's Law:
\[ \sqrt{2} \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \sin \theta_2 \]
- Simplifying gives:
\[ \frac{\sqrt{6}}{2} = \sqrt{3} \sin \theta_2 \]
- Rearranging for \( \sin \theta_2 \):
\[ \sin \theta_2 = \frac{\sqrt{6}}{2\sqrt{3}} = \frac{\sqrt{2}}{2} \]

5. **Find the Angle of Refraction:**
- The angle \( \theta_2 \) can be found using:
\[ \theta_2 = \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) \]
- This gives:
\[ \theta_2 = 45^\circ \]

### Final Answer:
The angle of refraction in medium 2 is \( 45^\circ \).