A ray of light strikes a transparent rectangular slab of refractive index `sqrt(2)` at an angle of incidence of `45^(@)`. The angle betweent the reflected and refracted rays is

Options

A. `75^(@)`
B. `90^(@)`
C. `105^(@)`
D. `120^(@)`

Explanation

To solve the problem of finding the angle between the reflected and refracted rays when a ray of light strikes a transparent rectangular slab of refractive index \(\sqrt{2}\) at an angle of incidence of \(45^\circ\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Angle of incidence, \(I = 45^\circ\)
- Refractive index, \(\mu = \sqrt{2}\)

2. **Apply the Law of Reflection**:
- According to the law of reflection, the angle of incidence is equal to the angle of reflection.
- Therefore, \(R_1 = I = 45^\circ\).

3. **Use Snell's Law for Refraction**:
- Snell's law states that \(\mu_1 \sin I = \mu_2 \sin R_2\).
- Here, \(\mu_1 = 1\) (for air), \(\mu_2 = \sqrt{2}\), and \(R_2\) is the angle of refraction.
- Thus, we can write:
\[ 1 \cdot \sin(45^\circ) = \sqrt{2} \cdot \sin(R_2) \]
- Since \(\sin(45^\circ) = \frac{\sqrt{2}}{2}\), we have:
\[ \frac{\sqrt{2}}{2} = \sqrt{2} \cdot \sin(R_2) \]
- Dividing both sides by \(\sqrt{2}\):
\[ \frac{1}{2} = \sin(R_2) \]

4. **Calculate the Angle of Refraction**:
- The angle \(R_2\) can be found by taking the inverse sine:
\[ R_2 = \sin^{-1}\left(\frac{1}{2}\right) = 30^\circ \]

5. **Determine the Angle Between Reflected and Refracted Rays**:
- To find the angle \(\theta\) between the reflected ray and the refracted ray, we can visualize the situation:
- The angle between the reflected ray and the normal is \(R_1 = 45^\circ\).
- The angle between the refracted ray and the normal is \(R_2 = 30^\circ\).
- Therefore, the angle \(\theta\) between the reflected ray and the refracted ray is given by:
\[ \theta = R_1 + R_2 = 45^\circ + 30^\circ = 75^\circ \]

### Final Answer:
The angle between the reflected and refracted rays is \(75^\circ\).