Two transparent media A and B are separated by a plane boundary. The speed of light in medium A is `2.0xx10^(8)ms^(-1)` and in medium B is `2.5xx10^(8)ms^(-1)`. The critical angle for which a ray of light going from A to B suffers total internal reflection is

Options

A. `sin^(-1)(1/(sqrt(2)))`
B. `sin^(-1)(1/2)`
C. `sin^(-1)(4/5)`
D. 90

Answer

To find the critical angle for a ray of light going from medium A to medium B, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given data:**
- Speed of light in medium A, \( v_A = 2.0 \times 10^8 \, \text{m/s} \)
- Speed of light in medium B, \( v_B = 2.5 \times 10^8 \, \text{m/s} \)
- Speed of light in vacuum, \( c = 3.0 \times 10^8 \, \text{m/s} \)

2. **Calculate the refractive index of medium A (\( n_A \)):**
\[ n_A = \frac{c}{v_A} = \frac{3.0 \times 10^8}{2.0 \times 10^8} = 1.5 \]

3. **Calculate the refractive index of medium B (\( n_B \)):**
\[ n_B = \frac{c}{v_B} = \frac{3.0 \times 10^8}{2.5 \times 10^8} = 1.2 \]

4. **Use Snell's Law to find the critical angle (\( \theta_c \)):**
- For total internal reflection, the angle of refraction is \( 90^\circ \).
- According to Snell's Law:
\[ n_A \sin(\theta_c) = n_B \sin(90^\circ) \]
- Since \( \sin(90^\circ) = 1 \), we can simplify this to:
\[ n_A \sin(\theta_c) = n_B \]

5. **Substituting the values:**
\[ 1.5 \sin(\theta_c) = 1.2 \]

6. **Solve for \( \sin(\theta_c) \):**
\[ \sin(\theta_c) = \frac{1.2}{1.5} = \frac{12}{15} = \frac{4}{5} \]

7. **Calculate the critical angle (\( \theta_c \)):**
\[ \theta_c = \sin^{-1}\left(\frac{4}{5}\right) \]

### Final Answer:
The critical angle for which a ray of light going from medium A to B suffers total internal reflection is \( \theta_c = \sin^{-1}\left(\frac{4}{5}\right) \).