A ray of light strikes a glass slab (R.l. = `mu`) of finite thickness t. The emerging light is

Options

A. parallel to. the incident ray
B. perpendicular to the incident ray
C. rotates by an angle`[sin^(-1)(1/mu)]`
D. totally internally reflected

Answer

To solve the problem of a ray of light striking a glass slab of finite thickness \( t \) and refractive index \( \mu \), we will analyze the behavior of the light ray as it passes through the slab.

### Step-by-Step Solution:

1. **Identify the Incident Ray and the Mediums**:
- The incident ray strikes the glass slab at an angle of incidence \( I \). The medium before the glass slab is air (a rarer medium) and the glass slab itself is a denser medium.

2. **Apply Snell's Law at the First Interface**:
- According to Snell's Law, when light travels from a rarer medium (air) to a denser medium (glass), it bends towards the normal.
- The relationship is given by:
\[ n_1 \sin I = n_2 \sin r \]
where \( n_1 = 1 \) (refractive index of air), \( n_2 = \mu \) (refractive index of glass), and \( r \) is the angle of refraction.

3. **Determine the Angle of Refraction**:
- Rearranging Snell's Law gives:
\[ \sin r = \frac{\sin I}{\mu} \]
- This indicates that the angle of refraction \( r \) is less than the angle of incidence \( I \) since \( \mu > 1 \).

4. **Ray Propagation Through the Glass Slab**:
- The ray travels through the thickness \( t \) of the glass slab and reaches the second interface (glass-air boundary).

5. **Apply Snell's Law at the Second Interface**:
- Now, the light ray is moving from the denser medium (glass) to the rarer medium (air). According to Snell's Law:
\[ \mu \sin r = n_1 \sin R \]
where \( R \) is the angle of emergence.

6. **Determine the Angle of Emergence**:
- Rearranging gives:
\[ \sin R = \mu \sin r \]
- Substituting \( \sin r = \frac{\sin I}{\mu} \):
\[ \sin R = \mu \left(\frac{\sin I}{\mu}\right) = \sin I \]
- This implies that \( R = I \).

7. **Conclusion**:
- Since the angle of incidence \( I \) is equal to the angle of emergence \( R \), the emergent ray is parallel to the incident ray.

### Final Answer:
The emerging light is **parallel to the incident ray**.