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 unut 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. None of the above

Explanation

To solve the problem, we will use 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: Define the vectors and angles
The direction of the ray in medium A is given by the unit vector:
\[ \vec{r}_A = a \hat{i} + b \hat{j} \]
The direction of the ray in medium B is given by the unit vector:
\[ \vec{r}_B = \alpha \hat{i} + \beta \hat{j} \]
The angles of incidence \( i \) and refraction \( r \) are defined with respect to the normal to the interface (the \( xz \) plane).

### Step 2: Calculate the cosine of the angles
Using the definition of cosine in terms of dot products, we can find:
- For medium A:
\[ \cos i = \frac{\vec{r}_A \cdot \hat{j}}{|\vec{r}_A|} = \frac{b}{\sqrt{a^2 + b^2}} \]
Since \(\vec{r}_A\) is a unit vector, \(|\vec{r}_A| = 1\), hence:
\[ \cos i = b \]

- For medium B:
\[ \cos r = \frac{\vec{r}_B \cdot \hat{j}}{|\vec{r}_B|} = \frac{\beta}{\sqrt{\alpha^2 + \beta^2}} \]
Again, since \(\vec{r}_B\) is a unit vector, \(|\vec{r}_B| = 1\), hence:
\[ \cos r = \beta \]

### Step 3: Apply Snell's Law
Snell's Law states:
\[ \mu_1 \sin i = \mu_2 \sin r \]
We can express \(\sin i\) and \(\sin r\) in terms of \(\cos i\) and \(\cos r\):
\[ \sin^2 i = 1 - \cos^2 i \quad \text{and} \quad \sin^2 r = 1 - \cos^2 r \]
Thus:
\[ \sin i = \sqrt{1 - b^2} \quad \text{and} \quad \sin r = \sqrt{1 - \beta^2} \]
Substituting these into Snell's Law gives:
\[ \mu_1 \sqrt{1 - b^2} = \mu_2 \sqrt{1 - \beta^2} \]

### Step 4: Substitute the values
Using the relationships \(1 - b^2 = a^2\) and \(1 - \beta^2 = \alpha^2\) (since both \(\vec{r}_A\) and \(\vec{r}_B\) are unit vectors), we can rewrite the equation:
\[ \mu_1 \sqrt{a^2} = \mu_2 \sqrt{\alpha^2} \]
This simplifies to:
\[ \mu_1 a = \mu_2 \alpha \]

### Conclusion
Thus, the relationship we derived is:
\[ \mu_1 a = \mu_2 \alpha \]
This matches with option A from the given choices.