If light travels a distance `x` in `t_(1)` sec in air and `10x` distance in `t_(2)` sec in a medium, the critical angle of the medium will be

Options

A. `tan^(-1)((t_(1))/(t_(2)))`
B. `sin^(-1)((t_(1))/(t_(2)))`
C. `sin^(-1)((10t_(1))/(t_(2)))`
D. `tan^(-1)((10t_(1))/(t_(2)))`

Explanation

To find the critical angle of the medium given the distances and times, we can follow these steps:

### Step 1: Determine the speed of light in air and the medium
- **Speed of light in air (v1)**:
\[ v_1 = \frac{x}{t_1} = c \]
where \( c \) is the speed of light in a vacuum (approximately \( 3 \times 10^8 \) m/s).

- **Speed of light in the medium (v2)**:
\[ v_2 = \frac{10x}{t_2} \]

### Step 2: Calculate the refractive index (μ) of the medium
The refractive index \( \mu \) is given by the ratio of the speed of light in air to the speed of light in the medium:
\[ \mu = \frac{c}{v_2} \]
Substituting the expression for \( v_2 \):
\[ \mu = \frac{c}{\frac{10x}{t_2}} = \frac{c \cdot t_2}{10x} \]

### Step 3: Relate the refractive index to the critical angle
The critical angle \( \theta_c \) is related to the refractive index by the formula:
\[ \sin(\theta_c) = \frac{1}{\mu} \]
Substituting the expression for \( \mu \):
\[ \sin(\theta_c) = \frac{10x}{c \cdot t_2} \]

### Step 4: Solve for the critical angle
To find the critical angle, we take the inverse sine:
\[ \theta_c = \sin^{-1}\left(\frac{10x}{c \cdot t_2}\right) \]

### Step 5: Final expression for the critical angle
Since \( c \) is a constant, we can express the critical angle in terms of \( t_1 \) and \( t_2 \):
\[ \theta_c = \sin^{-1}\left(\frac{10 t_1}{t_2}\right) \]

### Conclusion
Thus, the critical angle of the medium is given by:
\[ \theta_c = \sin^{-1}\left(\frac{10 t_1}{t_2}\right) \]