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 determining the nature of the emerging light, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a glass slab with a refractive index \( \mu \) and a thickness \( t \).
- A ray of light strikes the glass slab at an angle of incidence \( i \).

2. **Refraction at the First Surface**:
- When the ray enters the glass slab, it refracts according to Snell's Law:
\[ n_1 \sin(i) = n_2 \sin(r) \]
- Here, \( n_1 = 1 \) (refractive index of air) and \( n_2 = \mu \) (refractive index of glass).
- Therefore, we have:
\[ \sin(i) = \mu \sin(r) \]

3. **Refraction at the Second Surface**:
- As the ray exits the glass slab, it refracts again at the second surface:
\[ n_2 \sin(r) = n_1 \sin(e) \]
- This gives us:
\[ \mu \sin(r) = 1 \sin(e) \]
- Thus, we can write:
\[ \sin(e) = \mu \sin(r) \]

4. **Equating Angles**:
- From the two equations derived from Snell's Law, we can equate the terms:
\[ \sin(i) = \mu \sin(r) \]
\[ \sin(e) = \mu \sin(r) \]
- This implies:
\[ \sin(i) = \sin(e) \]
- Therefore, the angle of incidence \( i \) is equal to the angle of emergence \( e \):
\[ i = e \]

5. **Conclusion**:
- Since the angle of incidence is equal to the angle of emergence, the incident ray is parallel to the emergent ray.
- Thus, the emerging light is parallel to the incident ray.

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