The critical angle between and equilateral prism and air is `45^(@)`. If the incident ray is perpendicular to the refracting surface, then

Options

A. After deviation it will emerge from the second refracting surface
B. It is totally reflected on the second surface and emerges out perpendicularly from third surface in air
C. It is totally reflected from the second and third refracting surfaces and finally emerges out from the first surface.
D. It is totally reflected from all the three sides of prism and never emerges out

Explanation

To solve the problem regarding the behavior of light in an equilateral prism when the incident ray is perpendicular to the refracting surface, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Prism Geometry**:
- An equilateral prism has angles of 60 degrees each.
- The critical angle for the prism-air interface is given as 45 degrees.

2. **Identify the Incident Ray**:
- The incident ray is coming perpendicular to the refracting surface of the prism. This means that the angle of incidence (i) at the first surface is 0 degrees.

3. **Apply Snell's Law at the First Surface**:
- Since the incident ray is perpendicular to the surface, the angle of refraction (r) can be calculated using Snell's Law:
\[ n_1 \sin(i) = n_2 \sin(r) \]
- Here, \( n_1 \) is the refractive index of air (approximately 1), \( n_2 \) is the refractive index of the prism, and \( i = 0 \). Therefore,
\[ 1 \cdot \sin(0) = n_2 \cdot \sin(r) \implies \sin(r) = 0 \implies r = 0 \]
- The ray will continue straight into the prism without deviation.

4. **Determine the Path Inside the Prism**:
- Inside the prism, the ray travels straight until it reaches the second surface. The angle of incidence at the second surface is now 60 degrees (since the internal angle of the prism is 60 degrees).

5. **Apply Snell's Law at the Second Surface**:
- Now, we apply Snell's Law again at the second surface:
\[ n_2 \sin(60^\circ) = n_{air} \sin(r') \]
- Given that the critical angle is 45 degrees, if the angle of incidence exceeds this, total internal reflection (TIR) will occur. Since \( \sin(60^\circ) > \sin(45^\circ) \), TIR will occur.

6. **Result of Total Internal Reflection**:
- The ray will reflect back inside the prism at the second surface at an angle of 60 degrees to the normal.

7. **Final Emergence from the Prism**:
- The ray will then travel to the third surface, where it will again be at 60 degrees incidence.
- Since it is still within the prism, it will undergo TIR again and reflect back.

8. **Conclusion**:
- The ray will not emerge from the prism but will continue to reflect internally.

### Final Answer:
The incident ray will undergo total internal reflection at both the second and third surfaces of the prism and will not emerge into the air.