For a ray of light refracted through a prism of angle `60^(@)`, the angle of incidence is equal to the angle of emergence, each equal to `45^(@)`. Find the refractive index of the material of the prism.

Answer

To find the refractive index of the material of a prism with an angle of 60 degrees, given that the angle of incidence and the angle of emergence are both 45 degrees, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Geometry of the Prism**:
- We have a prism with an apex angle \( A = 60^\circ \).
- The angle of incidence \( i = 45^\circ \) and the angle of emergence \( e = 45^\circ \).

2. **Applying Snell's Law at Point A (Incident Ray)**:
- According to Snell's Law:
\[ n_1 \sin(i) = n_2 \sin(R_1) \]
- Here, \( n_1 = 1 \) (refractive index of air), \( i = 45^\circ \), and \( n_2 = n \) (refractive index of the prism).
- Thus, we can write:
\[ 1 \cdot \sin(45^\circ) = n \cdot \sin(R_1) \]
- Since \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ \frac{1}{\sqrt{2}} = n \cdot \sin(R_1) \]
- Rearranging gives:
\[ \sin(R_1) = \frac{1}{\sqrt{2} n} \quad \text{(Equation 1)} \]

3. **Applying Snell's Law at Point B (Emergent Ray)**:
- Again using Snell's Law:
\[ n \sin(R_2) = n_1 \sin(e) \]
- Here, \( e = 45^\circ \):
\[ n \cdot \sin(R_2) = 1 \cdot \sin(45^\circ) \]
- This simplifies to:
\[ n \cdot \sin(R_2) = \frac{1}{\sqrt{2}} \]
- Rearranging gives:
\[ \sin(R_2) = \frac{1}{\sqrt{2} n} \quad \text{(Equation 2)} \]

4. **Relating Angles R1 and R2**:
- The sum of the angles \( R_1 \) and \( R_2 \) is equal to the angle of the prism:
\[ R_1 + R_2 = A = 60^\circ \]
- Therefore, we can express \( R_2 \) in terms of \( R_1 \):
\[ R_2 = 60^\circ - R_1 \]

5. **Using the Relationship of Sines**:
- From Equations 1 and 2, we have:
\[ \sin(R_1) = \sin(60^\circ - R_1) \]
- Since \( R_1 \) and \( R_2 \) are both less than \( 90^\circ \), we can equate:
\[ R_1 = R_2 \]
- Thus, \( 2R_1 = 60^\circ \) leading to:
\[ R_1 = 30^\circ \]
- Consequently, \( R_2 = 30^\circ \).

6. **Finding the Refractive Index**:
- Substitute \( R_1 = 30^\circ \) back into Equation 1:
\[ \sin(30^\circ) = \frac{1}{\sqrt{2} n} \]
- Since \( \sin(30^\circ) = \frac{1}{2} \):
\[ \frac{1}{2} = \frac{1}{\sqrt{2} n} \]
- Rearranging gives:
\[ n = \frac{1}{\sqrt{2} \cdot \frac{1}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]
- Therefore, the refractive index \( n \) is:
\[ n \approx 1.414 \]

### Final Answer:
The refractive index of the material of the prism is approximately \( 1.414 \).