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

Explanation

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

### Step 1: Calculate the velocities of light in both media
The velocity of light in air (v1) can be calculated using the formula:
\[ v_1 = \frac{x}{t_1} \]
The velocity of light in the medium (v2) can be calculated as:
\[ v_2 = \frac{10x}{t_2} \]

### Step 2: Use Snell's Law to relate the refractive indices
According to Snell's Law:
\[ \frac{\mu_2}{\mu_1} = \frac{v_1}{v_2} \]
Where:
- \(\mu_1\) is the refractive index of air, which is approximately 1.
- \(\mu_2\) is the refractive index of the medium.

### Step 3: Substitute the velocities into Snell's Law
Substituting the expressions for \(v_1\) and \(v_2\) into Snell's Law gives:
\[ \frac{\mu_2}{1} = \frac{\frac{x}{t_1}}{\frac{10x}{t_2}} \]
This simplifies to:
\[ \mu_2 = \frac{t_2}{10t_1} \]

### Step 4: Relate the critical angle to the refractive indices
The critical angle (C) can be found using the formula:
\[ \sin C = \frac{\mu_1}{\mu_2} \]
Since \(\mu_1 = 1\) (for air), we have:
\[ \sin C = \frac{1}{\mu_2} \]
Substituting the expression for \(\mu_2\):
\[ \sin C = \frac{1}{\frac{t_2}{10t_1}} = \frac{10t_1}{t_2} \]

### Step 5: Find the critical angle
To find the critical angle, we take the inverse sine:
\[ C = \sin^{-1}\left(\frac{10t_1}{t_2}\right) \]

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

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