A light ray is incident at an angle 45∘ on parallel sided glass slab and emerges out grazing the vertical surface. The refractive index of the slab is

Options

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

Explanation

To solve the problem of finding the refractive index of a parallel-sided glass slab through which a light ray grazes the vertical surface, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- A light ray is incident at an angle of \( 45^\circ \) on a parallel-sided glass slab.
- The ray emerges grazing the vertical surface, which means it exits the slab at an angle of \( 90^\circ \) to the normal at the point of emergence.

2. **Identify the Angles**:
- Let \( \theta_1 = 45^\circ \) (the angle of incidence).
- The angle of refraction \( \theta_2 \) when the light enters the slab can be found using Snell's Law:
\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]
- Here, \( n_1 = 1 \) (the refractive index of air) and \( n_2 = \mu \) (the refractive index of the glass slab).

3. **Using Snell's Law**:
- Substituting the known values into Snell's Law:
\[ 1 \cdot \sin(45^\circ) = \mu \cdot \sin(\theta_2) \]
- Since \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \), we have:
\[ \frac{1}{\sqrt{2}} = \mu \cdot \sin(\theta_2) \]

4. **Emergence Condition**:
- When the ray emerges grazing the vertical surface, it means that the angle of incidence at the second surface is \( 90^\circ \).
- Therefore, the angle of refraction at the second surface must be \( 0^\circ \) (grazing condition).

5. **Applying Snell's Law at Emergence**:
- At the second surface, we apply Snell's Law again:
\[ \mu \cdot \sin(\theta_2) = 1 \cdot \sin(0^\circ) \]
- Since \( \sin(0^\circ) = 0 \), this condition indicates that the angle of incidence \( \theta_2 \) must be such that it leads to a critical angle condition.

6. **Finding the Critical Angle**:
- The critical angle \( \theta_c \) can be defined as:
\[ \sin(\theta_c) = \frac{1}{\mu} \]
- Since the ray grazes the vertical surface, we can set \( \theta_2 = 90^\circ - \theta_c \).

7. **Relating Angles**:
- From the geometry, we have:
\[ \theta_2 = 45^\circ \]
- Therefore, we can set:
\[ \sin(45^\circ) = \frac{1}{\mu} \]
- This gives us:
\[ \frac{1}{\sqrt{2}} = \frac{1}{\mu} \]

8. **Solving for the Refractive Index**:
- Rearranging gives:
\[ \mu = \sqrt{2} \]

### Final Answer:
The refractive index of the slab is \( \mu = \sqrt{2} \).

---