Critical angle for light going from medium (i) to (ii) is `theta` . The speed of light in medium (i) is v then speed in medium (ii) is

Options

A. `v(1-cos theta)`
B. `v//sin theta`
C. `v//cos theta`
D. `v(1-sin theta)`

Answer

To solve the problem step by step, let's analyze the situation using Snell's Law and the relationship between the speed of light in different media.

### Step-by-Step Solution:

1. **Understanding the Critical Angle**:
- The critical angle \( \theta \) is the angle of incidence in medium 1 (with speed \( v \)) at which light is refracted at an angle of 90 degrees in medium 2.

2. **Applying Snell's Law**:
- According to Snell's Law, we have:
\[ \mu_1 \sin(\theta) = \mu_2 \sin(90^\circ) \]
- Since \( \sin(90^\circ) = 1 \), this simplifies to:
\[ \mu_1 \sin(\theta) = \mu_2 \]

3. **Relating Refractive Index to Speed**:
- 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. Thus, we can express the refractive indices for both media:
\[ \mu_1 = \frac{c}{v_1} \quad \text{and} \quad \mu_2 = \frac{c}{v_2} \]

4. **Substituting the Refractive Indices**:
- From the relationship derived from Snell's Law:
\[ \frac{c}{v_1} \sin(\theta) = \frac{c}{v_2} \]
- We can cancel \( c \) from both sides (assuming \( c \neq 0 \)):
\[ \frac{\sin(\theta)}{v_1} = \frac{1}{v_2} \]

5. **Rearranging for \( v_2 \)**:
- Rearranging the equation gives:
\[ v_2 = \frac{v_1}{\sin(\theta)} \]

6. **Substituting \( v_1 \)**:
- Since we know that \( v_1 = v \) (the speed of light in medium 1), we can substitute this into the equation:
\[ v_2 = \frac{v}{\sin(\theta)} \]

### Final Answer:
The speed of light in medium 2 is:
\[ v_2 = \frac{v}{\sin(\theta)} \]