A ray of light moving along a vector `(3sqrt(2) hat(i)-3 hat(j)-3 hat(k))` undergoes refraction at an interface of two media which is y-z plane. The refractive index for `x le 0` is 1 while for `x ge 0` it is `sqrt(2)`. Then,

Options

A. Refracted ray bend towards y-axis
B. Refracted ray bend towards x-axis
C. The unit vector along the refracted ray is `(sqrt(3)hat(i)-hat(j)-hat(k))/(2)`
D. The unit vector along the refracted ray is `(sqrt(6)hat(i)-hat(j)-hat(k))/(sqrt(8))`

Answer

To solve the problem of a ray of light undergoing refraction at the interface of two media in the y-z plane, we will follow these steps:

### Step 1: Identify the incident ray vector
The incident ray is given as:
\[ \vec{I} = 3\sqrt{2} \hat{i} - 3 \hat{j} - 3 \hat{k} \]

### Step 2: Determine the angle of incidence
The interface is the y-z plane, which means the normal to the surface is along the x-axis. The angle of incidence \( i \) can be determined using the direction of the incident ray. The projection of the incident ray on the y-z plane is:
\[ \vec{I}_{yz} = -3 \hat{j} - 3 \hat{k} \]
The angle \( i \) can be calculated using the tangent of the angle:
\[ \tan(i) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sqrt{(-3)^2 + (-3)^2}}{3\sqrt{2}} = \frac{3\sqrt{2}}{3\sqrt{2}} = 1 \]
Thus, \( i = 45^\circ \).

### Step 3: Apply Snell's Law
According to Snell's Law:
\[ n_1 \sin(i) = n_2 \sin(r) \]
Where:
- \( n_1 = 1 \) (for \( x < 0 \))
- \( n_2 = \sqrt{2} \) (for \( x \geq 0 \))
- \( i = 45^\circ \)

Substituting the values:
\[ 1 \cdot \sin(45^\circ) = \sqrt{2} \cdot \sin(r) \]
\[ \frac{1}{\sqrt{2}} = \sqrt{2} \cdot \sin(r) \]
\[ \sin(r) = \frac{1}{2} \]
Thus, \( r = 30^\circ \).

### Step 4: Determine the direction of the refracted ray
The refracted ray will bend towards the x-axis. The angle with respect to the normal (x-axis) is \( 30^\circ \). The angle with respect to the y-z plane will be \( 90^\circ - 30^\circ = 60^\circ \).

### Step 5: Calculate the components of the refracted ray
The magnitude of the incident ray can be calculated as:
\[ |\vec{I}| = \sqrt{(3\sqrt{2})^2 + (-3)^2 + (-3)^2} = \sqrt{18 + 9 + 9} = \sqrt{36} = 6 \]
The components of the refracted ray can be calculated using the angles:
- The x-component:
\[ x = 6 \cos(30^\circ) = 6 \cdot \frac{\sqrt{3}}{2} = 3\sqrt{3} \]
- The y-component:
\[ y = 6 \sin(30^\circ) = 6 \cdot \frac{1}{2} = 3 \]
- The z-component will be the same as the incident ray in the y-z plane, thus:
\[ z = -3 \]

### Step 6: Write the refracted ray vector
The refracted ray vector can be written as:
\[ \vec{R} = 3\sqrt{3} \hat{i} + 3 \hat{j} - 3 \hat{k} \]

### Step 7: Normalize the refracted ray vector to find the unit vector
The magnitude of the refracted ray vector is:
\[ |\vec{R}| = \sqrt{(3\sqrt{3})^2 + 3^2 + (-3)^2} = \sqrt{27 + 9 + 9} = \sqrt{45} = 3\sqrt{5} \]
The unit vector \( \hat{R} \) is:
\[ \hat{R} = \frac{1}{3\sqrt{5}}(3\sqrt{3} \hat{i} + 3 \hat{j} - 3 \hat{k}) = \frac{\sqrt{3}}{\sqrt{5}} \hat{i} + \frac{1}{\sqrt{5}} \hat{j} - \frac{1}{\sqrt{5}} \hat{k} \]

### Final Answer
The unit vector along the refracted ray is:
\[ \hat{R} = \frac{\sqrt{3}}{\sqrt{5}} \hat{i} + \frac{1}{\sqrt{5}} \hat{j} - \frac{1}{\sqrt{5}} \hat{k} \]