Light takes `t_(1)` second to travel a distance x cm in vacuum and the same light takes `t_(2)` second to travel 10x cm in medium. The critical angle for the corresponding medium is

Options

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

Answer

To solve the problem, we need to find the critical angle for the medium given the time taken for light to travel certain distances in vacuum and in the medium. Let's break down the solution step by step:

### Step 1: Understanding the problem
We know that light takes `t1` seconds to travel `x` cm in vacuum and `t2` seconds to travel `10x` cm in a medium. We need to find the critical angle for the medium.

### Step 2: Calculate the speed of light in vacuum
The speed of light in vacuum (denoted as `c`) can be calculated using the formula:
\[ c = \frac{\text{distance}}{\text{time}} \]
For the distance `x` cm and time `t1` seconds, we have:
\[ c = \frac{x}{t1} \]

### Step 3: Calculate the speed of light in the medium
The speed of light in a medium (denoted as `v`) is related to the speed of light in vacuum and the refractive index (denoted as `μ`) of the medium by the formula:
\[ v = \frac{c}{μ} \]

### Step 4: Calculate the time taken in the medium
For the distance `10x` cm and time `t2` seconds in the medium, we can express this as:
\[ t2 = \frac{10x}{v} \]
Substituting the expression for `v` from Step 3:
\[ t2 = \frac{10x}{\frac{c}{μ}} = \frac{10xμ}{c} \]

### Step 5: Relate `t1` and `t2`
Now we have two expressions:
1. \( t1 = \frac{x}{c} \)
2. \( t2 = \frac{10xμ}{c} \)

Taking the ratio of \( t1 \) and \( t2 \):
\[ \frac{t1}{t2} = \frac{\frac{x}{c}}{\frac{10xμ}{c}} \]
This simplifies to:
\[ \frac{t1}{t2} = \frac{1}{10μ} \]

### Step 6: Expressing the refractive index in terms of critical angle
From the equation above, we can rearrange it to find \( μ \):
\[ μ = \frac{1}{10} \cdot \frac{t2}{t1} \]

### Step 7: Finding the critical angle
The critical angle \( θ_c \) is related to the refractive index by the formula:
\[ \sin(θ_c) = \frac{1}{μ} \]
Substituting our expression for \( μ \):
\[ \sin(θ_c) = 10 \cdot \frac{t1}{t2} \]

### Step 8: Final expression for the critical angle
Thus, the critical angle \( θ_c \) can be expressed as:
\[ θ_c = \sin^{-1}\left(10 \cdot \frac{t1}{t2}\right) \]

### Conclusion
The critical angle for the corresponding medium is:
\[ θ_c = \sin^{-1}\left(10 \cdot \frac{t1}{t2}\right) \]