Light travels in two media A and B with speeds `1.8 × 10^(8) m s^(–1) and2.4 × 10^(8) m s^(–1)` respectively. Then the critical angle between them is

Options

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

Explanation

To find the critical angle between two media A and B, we can use the relationship between the speeds of light in the two media and their refractive indices.

### Step-by-Step Solution:

1. **Identify the speeds of light in the two media:**
- Speed of light in medium A, \( V_A = 1.8 \times 10^8 \, \text{m/s} \)
- Speed of light in medium B, \( V_B = 2.4 \times 10^8 \, \text{m/s} \)

2. **Calculate the refractive indices of the two media:**
- The refractive index \( n \) of a medium is given by the formula:
\[ n = \frac{c}{V} \]
where \( c \) is the speed of light in vacuum (approximately \( 3 \times 10^8 \, \text{m/s} \)).
- For medium A:
\[ n_A = \frac{c}{V_A} = \frac{3 \times 10^8}{1.8 \times 10^8} = \frac{3}{1.8} = \frac{5}{3} \]
- For medium B:
\[ n_B = \frac{c}{V_B} = \frac{3 \times 10^8}{2.4 \times 10^8} = \frac{3}{2.4} = \frac{5}{4} \]

3. **Use the formula for critical angle:**
- The critical angle \( \theta_c \) can be found using Snell's law:
\[ \sin \theta_c = \frac{n_B}{n_A} \]
- Substituting the values of the refractive indices:
\[ \sin \theta_c = \frac{\frac{5}{4}}{\frac{5}{3}} = \frac{5}{4} \times \frac{3}{5} = \frac{3}{4} \]

4. **Calculate the critical angle:**
- To find the critical angle \( \theta_c \), take the inverse sine:
\[ \theta_c = \sin^{-1}\left(\frac{3}{4}\right) \]

### Final Answer:
The critical angle between the two media is:
\[ \theta_c = \sin^{-1}\left(\frac{3}{4}\right) \]