The reflected and refracted rays are observed to be perpendicular to each other, when ray of light is incident at an angle of 60∘ on a transparent block. The refractive index of the block is

Options

A. `3/2`
B. `1/2`
C. `2/sqrt(3)`
D. `sqrt(3)`

Explanation

To find the refractive index of the transparent block when the reflected and refracted rays are perpendicular to each other, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- When light strikes the surface of a transparent block, it gets reflected and refracted. According to the problem, the angle of incidence (i) is given as 60°.
- The reflected ray and the refracted ray are perpendicular to each other. Therefore, if the angle of refraction is r, then we have:
\[ i + r = 90^\circ \]
- This implies:
\[ r = 90^\circ - i = 90^\circ - 60^\circ = 30^\circ \]

2. **Applying Snell's Law**:
- Snell's Law states:
\[ n_1 \sin(i) = n_2 \sin(r) \]
- Here, \( n_1 \) is the refractive index of air (approximately 1), \( n_2 \) is the refractive index of the block, \( i = 60^\circ \), and \( r = 30^\circ \).
- Substituting the known values into Snell's Law:
\[ 1 \cdot \sin(60^\circ) = n_2 \cdot \sin(30^\circ) \]

3. **Calculating the Sine Values**:
- We know:
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \quad \text{and} \quad \sin(30^\circ) = \frac{1}{2} \]
- Substituting these values into the equation:
\[ \frac{\sqrt{3}}{2} = n_2 \cdot \frac{1}{2} \]

4. **Solving for the Refractive Index \( n_2 \)**:
- Rearranging the equation to solve for \( n_2 \):
\[ n_2 = \frac{\sqrt{3}/2}{1/2} = \sqrt{3} \]

5. **Conclusion**:
- Therefore, the refractive index of the block is:
\[ n_2 = \sqrt{3} \]

### Final Answer:
The refractive index of the block is \( \sqrt{3} \).

---