The velocities of light in two different media are 2*`10^8`m/s and 2.5 * `10^8`m/s respectively.The critical angle for these media is

Options

A. `sin^(-1)(1/5)`
B. `sin^(-1)(4/5)`
C. `sin^(-1)(1/2)`
D. `sin^(-1)(2/3)`

Answer

To find the critical angle for the two media with given velocities of light, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Velocities of Light**:
- Let the velocity of light in the first medium (denser) be \( V_1 = 2 \times 10^8 \, \text{m/s} \).
- Let the velocity of light in the second medium (rarer) be \( V_2 = 2.5 \times 10^8 \, \text{m/s} \).

2. **Understand the Concept of Critical Angle**:
- The critical angle (\( C \)) is defined as the angle of incidence in the denser medium at which the angle of refraction in the rarer medium is \( 90^\circ \).

3. **Apply Snell's Law**:
- According to Snell's Law:
\[ n_1 \sin(i) = n_2 \sin(r) \]
- Here, \( n_1 \) and \( n_2 \) are the refractive indices of the two media, \( i \) is the angle of incidence (critical angle \( C \)), and \( r \) is the angle of refraction (which is \( 90^\circ \) at the critical angle).

4. **Calculate the Refractive Indices**:
- The refractive index \( n \) is given by:
\[ n = \frac{c}{V} \]
- For the first medium (denser):
\[ n_1 = \frac{c}{V_1} \]
- For the second medium (rarer):
\[ n_2 = \frac{c}{V_2} \]

5. **Substituting into Snell's Law**:
- Since \( \sin(90^\circ) = 1 \), we can rewrite Snell's Law as:
\[ n_1 \sin(C) = n_2 \]
- Therefore:
\[ \frac{c}{V_1} \sin(C) = \frac{c}{V_2} \]

6. **Cancel Out \( c \)**:
- The speed of light in vacuum \( c \) cancels out:
\[ \frac{\sin(C)}{V_1} = \frac{1}{V_2} \]

7. **Rearranging the Equation**:
- Rearranging gives:
\[ \sin(C) = \frac{V_1}{V_2} \]

8. **Substituting the Values**:
- Substitute \( V_1 = 2 \times 10^8 \, \text{m/s} \) and \( V_2 = 2.5 \times 10^8 \, \text{m/s} \):
\[ \sin(C) = \frac{2 \times 10^8}{2.5 \times 10^8} = \frac{2}{2.5} = \frac{4}{5} \]

9. **Finding the Critical Angle**:
- Now, to find the critical angle \( C \):
\[ C = \sin^{-1}\left(\frac{4}{5}\right) \]

### Final Answer:
The critical angle \( C \) for the two media is \( \sin^{-1}\left(\frac{4}{5}\right) \).

---