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 cos`theta`
B. `v//costheta`
C. vsin`theta`
D. `v//sintheta`

Explanation

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

### Step-by-Step Solution:

1. **Understanding the Critical Angle**:
- The critical angle \( \theta \) is defined as the angle of incidence in the denser medium (medium X) at which light is refracted at an angle of \( 90^\circ \) in the rarer medium (medium Y). This means that at this angle, the light does not enter medium Y but instead undergoes total internal reflection.

2. **Applying Snell's Law**:
- According to Snell's Law, we have:
\[ n_1 \sin \theta = n_2 \sin r \]
where \( n_1 \) is the refractive index of medium X, \( n_2 \) is the refractive index of medium Y, \( \theta \) is the critical angle, and \( r = 90^\circ \) (the angle of refraction).
- Since \( \sin 90^\circ = 1 \), we can simplify this to:
\[ n_1 \sin \theta = n_2 \]

3. **Relating Refractive Index to Speed of Light**:
- 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 and \( v \) is the speed of light in the medium.
- Therefore, for medium X and medium Y, we can write:
\[ n_1 = \frac{c}{v} \quad \text{and} \quad n_2 = \frac{c}{v_y} \]

4. **Substituting Refractive Indices**:
- Substituting these into the equation from Snell's Law gives:
\[ \frac{c}{v} \sin \theta = \frac{c}{v_y} \]
- The \( c \) cancels out from both sides:
\[ \frac{\sin \theta}{v} = \frac{1}{v_y} \]

5. **Solving for Speed in Medium Y**:
- Rearranging the equation to solve for \( v_y \):
\[ v_y = \frac{v}{\sin \theta} \]

### Final Result:
The speed of light in medium Y is given by:
\[ v_y = \frac{v}{\sin \theta} \]