If the critical angle for total internal reflection, from a medium to vacuum is `30^(@)`, then velocity of light in the medium is

Options

A. `1.5xx10^(8)m//s`
B. `2xx10^(8)m//s`
C. `3xx10^(8)m//s`
D. `6xx10^(8)m//s`

Answer

To find the velocity of light in the medium given the critical angle for total internal reflection, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Critical Angle**:
The critical angle (C) is the angle of incidence in a denser medium (with refractive index μ) at which the angle of refraction in a rarer medium (with refractive index 1, for vacuum) is 90 degrees.

2. **Use the Relationship Between Refractive Index and Critical Angle**:
The relationship between the refractive index (μ) and the critical angle (C) is given by:
\[ μ = \frac{1}{\sin C} \]

3. **Substitute the Given Critical Angle**:
We are given that the critical angle \( C = 30^\circ \). Therefore, we can calculate \( \sin 30^\circ \):
\[ \sin 30^\circ = \frac{1}{2} \]
Now substituting this value into the formula for refractive index:
\[ μ = \frac{1}{\sin 30^\circ} = \frac{1}{\frac{1}{2}} = 2 \]

4. **Relate Refractive Index to Velocity of Light**:
The refractive index is also related to the speed of light in vacuum (c) and the speed of light in the medium (v) by the formula:
\[ μ = \frac{c}{v} \]
Rearranging this gives us:
\[ v = \frac{c}{μ} \]

5. **Substitute the Values**:
We know that the speed of light in vacuum \( c = 3 \times 10^8 \) m/s and \( μ = 2 \):
\[ v = \frac{3 \times 10^8}{2} = 1.5 \times 10^8 \text{ m/s} \]

6. **Conclusion**:
The velocity of light in the medium is \( 1.5 \times 10^8 \) m/s.

### Final Answer:
The velocity of light in the medium is \( 1.5 \times 10^8 \) m/s.

---