A ray of light moving along the vector (`-i-2j`)undergoes refraction at an interface two media,which is the x-zplane. The refracive index for `ygt0` is `2` and below it is`sqrt(5)//2`.the unit vector along which the refracted ray moves is:

Options

A. `(-3hat(i)-5hat(j))/(sqrt(34))`
B. `(-4hat(i)-3hat(j))/(5)`
C. `(-3hat(i)-4hat(j))/(5)`
D. `(3hat(i)+5hat(j))/(sqrt(34))`

Answer

To find the unit vector along which the refracted ray moves, we can use Snell's law and the given information about the refractive indices of the two media. Let's break down the solution step by step.

### Step 1: Identify the given information
- The incident ray is represented by the vector: \(-\hat{i} - 2\hat{j}\).
- The refractive index for the medium above the x-z plane (where \(y > 0\)) is \(n_1 = 2\).
- The refractive index for the medium below the x-z plane (where \(y < 0\)) is \(n_2 = \frac{\sqrt{5}}{2}\).

### Step 2: Apply Snell's Law
According to Snell's law:
\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]
Where:
- \(\theta_1\) is the angle of incidence,
- \(\theta_2\) is the angle of refraction.

### Step 3: Determine the direction of the incident ray
The direction of the incident ray can be expressed as:
\[ \vec{I} = -\hat{i} - 2\hat{j} \]
To find the unit vector of the incident ray, we calculate its magnitude:
\[ |\vec{I}| = \sqrt{(-1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \]
Thus, the unit vector for the incident ray is:
\[ \hat{I} = \frac{\vec{I}}{|\vec{I}|} = \frac{-\hat{i} - 2\hat{j}}{\sqrt{5}} = -\frac{1}{\sqrt{5}}\hat{i} - \frac{2}{\sqrt{5}}\hat{j} \]

### Step 4: Set up the equation using Snell's law
Using Snell's law in vector form:
\[ n_1 \hat{I} \cdot \hat{n} = n_2 \hat{R} \cdot \hat{n} \]
Where \(\hat{n}\) is the normal to the interface (which is along the y-axis, \(\hat{j}\)).

### Step 5: Calculate the refracted ray vector
Let the refracted ray be represented as:
\[ \vec{R} = a\hat{i} + b\hat{j} \]
To find the unit vector of the refracted ray, we need to ensure that:
\[ n_1 \cdot (-\frac{1}{\sqrt{5}}) = n_2 \cdot \frac{b}{\sqrt{a^2 + b^2}} \]
Substituting the values:
\[ 2 \cdot (-\frac{1}{\sqrt{5}}) = \frac{\sqrt{5}}{2} \cdot \frac{b}{\sqrt{a^2 + b^2}} \]

### Step 6: Solve for \(b\)
This leads to:
\[ -\frac{2}{\sqrt{5}} = \frac{\sqrt{5}}{2} \cdot \frac{b}{\sqrt{a^2 + b^2}} \]
Cross-multiplying gives:
\[ -4 = 5 \cdot \frac{b}{\sqrt{a^2 + b^2}} \]
Thus, we can express \(b\) in terms of \(a\).

### Step 7: Normalize the refracted ray vector
After determining \(a\) and \(b\), we can find the unit vector of the refracted ray:
\[ \hat{R} = \frac{a\hat{i} + b\hat{j}}{\sqrt{a^2 + b^2}} \]

### Step 8: Finalize the unit vector
After performing the calculations, we find:
\[ \hat{R} = \frac{4\hat{i} - 3\hat{j}}{5} \]

### Conclusion
The unit vector along which the refracted ray moves is:
\[ \hat{R} = \frac{4\hat{i} - 3\hat{j}}{5} \]