The critical angle for light going from medium X into medium Y is `theta` . The speed of light in medium X is v. The speed of light in medium Y is

Options

A. `v/(costheta)`
B. `vsintheta`
C. `v/(sintheta)`
D. `vcostheta`

Answer

To solve the problem, we need to find the speed of light in medium Y (denoted as \( v' \)) given the critical angle \( \theta \) for light transitioning from medium X to medium Y, and the speed of light in medium X (denoted as \( v \)).

### Step-by-Step Solution:

1. **Understanding Critical Angle**:
The critical angle \( \theta \) is defined as the angle of incidence in medium X at which the angle of refraction in medium Y is 90 degrees. This means that at this angle, light travels along the boundary between the two media.

2. **Applying Snell's Law**:
According to Snell's Law, we have:
\[ \mu_1 \sin(\theta) = \mu_2 \sin(90^\circ) \]
Here, \( \mu_1 \) is the refractive index of medium X, \( \mu_2 \) is the refractive index of medium Y, and \( \sin(90^\circ) = 1 \).

3. **Refractive Index Definition**:
The refractive index \( \mu \) is defined as:
\[ \mu = \frac{c}{v} \]
where \( c \) is the speed of light in vacuum and \( v \) is the speed of light in the medium.

4. **Expressing Refractive Indices**:
For medium X:
\[ \mu_1 = \frac{c}{v} \]
For medium Y:
\[ \mu_2 = \frac{c}{v'} \]

5. **Substituting Refractive Indices into Snell's Law**:
Substituting these expressions into Snell's Law gives:
\[ \frac{c}{v} \sin(\theta) = \frac{c}{v'} \cdot 1 \]

6. **Canceling \( c \)**:
Since \( c \) appears on both sides of the equation, we can cancel it out:
\[ \frac{\sin(\theta)}{v} = \frac{1}{v'} \]

7. **Rearranging to Find \( v' \)**:
Rearranging the equation to solve for \( v' \):
\[ v' = \frac{v}{\sin(\theta)} \]

8. **Conclusion**:
Thus, the speed of light in medium Y is:
\[ v' = \frac{v}{\sin(\theta)} \]

### Final Answer:
The speed of light in medium Y is \( \frac{v}{\sin(\theta)} \).