The `x-y` plane is the boundary between two transparent media. Medium-1 with `zle0` has a refractive index `sqrt(2)` and medium -2 with `zle0` has a refractive index `sqrt(2)`. A ray of light in medium -1 given by vector `A=sqrt(3)hati-hatk` is incident on the plane of separation. The unit vector in the direction of the refracted ray in medium-2 is

Options

A. `1/(sqrt(2))(hatk-hati)`
B. `1/(sqrt(2))(hati+hatj)`
C. `1/(sqrt(2))(hati-hatk)`
D. `1/(sqrt(2))(hati+hatk)`

Answer

To solve the problem, we need to find the unit vector in the direction of the refracted ray in medium-2 when a ray of light is incident from medium-1.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Refractive index of medium-1, \( n_1 = \sqrt{2} \)
- Refractive index of medium-2, \( n_2 = \sqrt{2} \)
- Incident ray vector in medium-1, \( \mathbf{A} = \sqrt{3} \hat{i} - \hat{k} \)

2. **Determine the Angle of Incidence:**
- The direction of the incident ray can be expressed as a vector. The incident ray vector is given by \( \mathbf{A} = \sqrt{3} \hat{i} - \hat{k} \).
- The components of the incident ray are:
- \( A_x = \sqrt{3} \)
- \( A_y = 0 \)
- \( A_z = -1 \)
- The angle of incidence \( i \) can be found using the tangent function:
\[ \tan(i) = \frac{A_z}{A_x} = \frac{-1}{\sqrt{3}} \]
- Therefore, \( i = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) \), which corresponds to \( -45^\circ \) or \( 135^\circ \).

3. **Apply Snell's Law:**
- According to Snell's law:
\[ n_1 \sin(i) = n_2 \sin(r) \]
- Since \( n_1 = n_2 \), we have:
\[ \sin(i) = \sin(r) \]
- This implies that \( r = i \). Therefore, the angle of refraction \( r \) is also \( -45^\circ \) or \( 135^\circ \).

4. **Determine the Direction of the Refracted Ray:**
- Let the unit vector of the refracted ray be \( \mathbf{B} = x \hat{i} + y \hat{j} + z \hat{k} \).
- Since the angle of refraction \( r = -45^\circ \), we have:
\[ \tan(r) = \frac{y}{x} = -1 \implies y = -x \]
- The z-component of the refracted ray will be determined by the symmetry of the media. Since the media are symmetrical, we can assume \( z \) remains the same as the incident ray, which is \( -1 \).

5. **Normalize the Refracted Ray Vector:**
- The refracted ray vector can be expressed as:
\[ \mathbf{B} = x \hat{i} - x \hat{j} - \hat{k} \]
- To find the magnitude:
\[ |\mathbf{B}| = \sqrt{x^2 + (-x)^2 + 1^2} = \sqrt{2x^2 + 1} \]
- To find the unit vector, we need to normalize \( \mathbf{B} \):
\[ \text{Unit vector} = \frac{\mathbf{B}}{|\mathbf{B}|} \]

6. **Choose a Value for \( x \):**
- Let \( x = \frac{1}{\sqrt{2}} \) (to simplify calculations):
\[ \mathbf{B} = \frac{1}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{j} - \hat{k} \]
- The magnitude becomes:
\[ |\mathbf{B}| = \sqrt{2 \left(\frac{1}{\sqrt{2}}\right)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \]
- Therefore, the unit vector in the direction of the refracted ray is:
\[ \mathbf{B}_{\text{unit}} = \frac{1}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{j} - \frac{1}{\sqrt{2}} \hat{k} \]

### Final Answer:
The unit vector in the direction of the refracted ray in medium-2 is:
\[ \mathbf{B}_{\text{unit}} = \frac{1}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{j} - \hat{k} \]