The critical angle of light from medium A to medium B is `theta`. The speed of light in medium A is v. the speed of light in medium B is

Options

A. `v/(sin theta)`
B. `v/(cos theta)`
C. `v sin theta`
D. `v cos theta`

Answer

To find the speed of light in medium B given the critical angle \( \theta \) and the speed of light in medium A (\( v \)), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Critical Angle**: The critical angle \( \theta \) is defined as the angle of incidence in medium A at which light is refracted along the boundary between the two media. Beyond this angle, total internal reflection occurs.

2. **Using Snell's Law**: According to Snell's Law, the relationship between the angles and the refractive indices of the two media is given by:
\[ n_A \sin(\theta) = n_B \sin(90^\circ) \]
Since \( \sin(90^\circ) = 1 \), we can simplify this to:
\[ n_A \sin(\theta) = n_B \]

3. **Refractive Index Relation**: The refractive index \( n \) of a medium is defined as:
\[ n = \frac{c}{v} \]
where \( c \) is the speed of light in vacuum and \( v \) is the speed of light in the medium. Therefore, we can express the refractive indices of the two media as:
\[ n_A = \frac{c}{v_A} \quad \text{and} \quad n_B = \frac{c}{v_B} \]

4. **Substituting Refractive Indices**: Substituting the expressions for \( n_A \) and \( n_B \) into the equation from Snell's Law gives us:
\[ \frac{c}{v_A} \sin(\theta) = \frac{c}{v_B} \]

5. **Cancelling \( c \)**: Since \( c \) appears on both sides of the equation, we can cancel it out:
\[ \frac{\sin(\theta)}{v_A} = \frac{1}{v_B} \]

6. **Rearranging for \( v_B \)**: Rearranging the equation to solve for \( v_B \) gives:
\[ v_B = \frac{v_A}{\sin(\theta)} \]

7. **Final Expression**: Since we are given that the speed of light in medium A is \( v \), we can replace \( v_A \) with \( v \):
\[ v_B = \frac{v}{\sin(\theta)} \]

### Conclusion:
The speed of light in medium B is given by:
\[ v_B = \frac{v}{\sin(\theta)} \]