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-costheta)`
B. `(v)/(sintheta)`
C. `v/(costheta)`
D. `v/((1-sintheta))`

Explanation

To solve the problem, we need to find the speed of light in medium (ii) given the critical angle (θ) for light going from medium (i) to medium (ii) and the speed of light in medium (i) (v).

### Step-by-Step Solution:

1. **Understanding the Concept of Critical Angle**:
The critical angle is the angle of incidence in medium (i) at which light is refracted along the boundary and does not enter medium (ii). Beyond this angle, total internal reflection occurs.

2. **Using Snell's Law**:
According to Snell's Law:
\[ \frac{\sin(\theta_1)}{\sin(\theta_2)} = \frac{v_1}{v_2} \]
where:
- \( \theta_1 \) is the angle of incidence (which is the critical angle θ in this case),
- \( \theta_2 \) is the angle of refraction (which is 90 degrees at the critical angle),
- \( v_1 \) is the speed of light in medium (i),
- \( v_2 \) is the speed of light in medium (ii).

3. **Substituting the Values**:
Since at the critical angle, \( \theta_2 = 90^\circ \), we have:
\[ \sin(90^\circ) = 1 \]
Therefore, Snell's Law simplifies to:
\[ \frac{\sin(\theta)}{1} = \frac{v_1}{v_2} \]

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

5. **Substituting the Known Values**:
We know that the speed of light in medium (i) is \( v \). Thus, substituting \( v_1 \) with \( v \):
\[ v_2 = \frac{v}{\sin(\theta)} \]

6. **Final Answer**:
Therefore, the speed of light in medium (ii) is:
\[ v_2 = \frac{v}{\sin(\theta)} \]

### Conclusion:
The speed of light in medium (ii) is \( \frac{v}{\sin(\theta)} \).