The `xz` plane separates two media `A` and `B` with refractive indices `mu_(1)` & `mu_(2)` respectively. A ray of light travels from `A` to `B`. Its directions in the two media are given by the unit vectors, `vec(r)_(A)=a hat i+ b hat j` & `vec(r)_(B)= alpha hat i + beta hat j` respectively where `hat i` & `hat j` are unit vectors in the `x` & `y` directions. Then :

Options

A. `mu_(1) a = mu_(2) alpha`
B. `mu_(1) alpha = mu_(2) a`
C. `mu_(1) b mu_(2) beta`
D. `mu_(1) beta = mu_(2) b`

Explanation

To solve the problem, we need to apply Snell's Law, which relates the angles of incidence and refraction to the refractive indices of the two media. The steps are as follows:

### Step 1: Understand the given information
We have two media A and B separated by the `xz` plane. The refractive indices are given as \( \mu_1 \) for medium A and \( \mu_2 \) for medium B. The direction of the light ray in medium A is represented by the unit vector \( \vec{r}_A = a \hat{i} + b \hat{j} \) and in medium B by \( \vec{r}_B = \alpha \hat{i} + \beta \hat{j} \).

### Step 2: Determine the magnitudes of the unit vectors
Since \( \vec{r}_A \) and \( \vec{r}_B \) are unit vectors, their magnitudes must equal 1. Therefore, we can write:
\[ |\vec{r}_A| = \sqrt{a^2 + b^2} = 1 \]
\[ |\vec{r}_B| = \sqrt{\alpha^2 + \beta^2} = 1 \]

### Step 3: Apply Snell's Law
According to Snell's Law:
\[ \mu_1 \sin I = \mu_2 \sin R \]
Where \( I \) is the angle of incidence and \( R \) is the angle of refraction.

### Step 4: Calculate \( \sin I \) and \( \sin R \)
From the geometry of the unit vectors, we can express \( \sin I \) and \( \sin R \):
- For medium A:
\[ \sin I = \frac{b}{\sqrt{a^2 + b^2}} = b \quad \text{(since } \sqrt{a^2 + b^2} = 1\text{)} \]

- For medium B:
\[ \sin R = \frac{\beta}{\sqrt{\alpha^2 + \beta^2}} = \beta \quad \text{(since } \sqrt{\alpha^2 + \beta^2} = 1\text{)} \]

### Step 5: Substitute into Snell's Law
Substituting the values of \( \sin I \) and \( \sin R \) into Snell's Law gives:
\[ \mu_1 b = \mu_2 \beta \]

### Step 6: Rearranging the equation
Rearranging the equation, we find the relation between the refractive indices and the components of the direction vectors:
\[ \mu_1 a = \mu_2 \alpha \]

This is the required relation.

### Final Answer
The relation between the refractive indices and the direction components is:
\[ \mu_1 a = \mu_2 \alpha \]