A light ray incident along vector `2hati+4hatj+sqrt(5)hatk` strikes in the `x-z` from medium of refractive index `sqrt(3)` and enters into medium II of refractive index is `mu_(2)`. The value of `mu_(2)` for which the value of angle of refraction becomes `90^(@)` is

Options

A. `(4sqrt(3))/5`
B. `(3sqrt(3))/5`
C. `(2sqrt(3))/5`
D. `(sqrt(3))/5`

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 law is given by:

\[ \mu_1 \sin i = \mu_2 \sin r \]

where:
- \(\mu_1\) is the refractive index of the first medium,
- \(\mu_2\) is the refractive index of the second medium,
- \(i\) is the angle of incidence,
- \(r\) is the angle of refraction.

### Step 1: Identify the incident vector and its components
The incident light ray is given by the vector:

\[ \vec{A} = 2\hat{i} + 4\hat{j} + \sqrt{5}\hat{k} \]

### Step 2: Calculate the angle of incidence
The angle of incidence \(i\) can be determined using the normal to the interface. Since the light ray strikes the \(x-z\) plane, the normal vector is along the \(y\) direction, which is \(\hat{j}\).

To find the angle of incidence, we first calculate the magnitude of the incident vector:

\[ |\vec{A}| = \sqrt{(2)^2 + (4)^2 + (\sqrt{5})^2} = \sqrt{4 + 16 + 5} = \sqrt{25} = 5 \]

Next, we find the component of the incident vector along the normal direction (which is along \(\hat{j}\)):

\[ A_y = 4 \]

Using the definition of sine in terms of the angle of incidence:

\[ \sin i = \frac{A_y}{|\vec{A}|} = \frac{4}{5} \]

### Step 3: Set up Snell's Law
Given that the angle of refraction \(r\) is \(90^\circ\), we know that:

\[ \sin r = \sin 90^\circ = 1 \]

Substituting into Snell's Law:

\[ \mu_1 \sin i = \mu_2 \sin r \]

Substituting the known values:

\[ \sqrt{3} \cdot \frac{4}{5} = \mu_2 \cdot 1 \]

### Step 4: Solve for \(\mu_2\)
Rearranging the equation gives us:

\[ \mu_2 = \sqrt{3} \cdot \frac{4}{5} \]

### Step 5: Calculate \(\mu_2\)
Now we can compute the value:

\[ \mu_2 = \frac{4\sqrt{3}}{5} \]

### Final Answer
Thus, the value of \(\mu_2\) for which the angle of refraction becomes \(90^\circ\) is:

\[ \mu_2 = \frac{4\sqrt{3}}{5} \]