A ray of light passes from vaccum into a medium of refractive index n. If the angle of incidence is twice the angle of refraction, then the angle of incidence is

Options

A. `cos^-1(n//2)`
B. `sin^-1(n//2)`
C. `2cos^-1(n//2)`
D. `2 sin^-1(n//2)`

Explanation

To solve the problem step by step, we will use Snell's law and the relationship between the angles of incidence and refraction.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- A ray of light is passing from vacuum (refractive index \( \mu_1 = 1 \)) into a medium with refractive index \( n \).
- The angle of incidence \( i \) is twice the angle of refraction \( r \). Therefore, we can express this relationship as:
\[ i = 2r \]

2. **Applying Snell's Law:**
- Snell's law states that:
\[ \mu_1 \sin(i) = \mu_2 \sin(r) \]
- Substituting the values we have:
\[ 1 \cdot \sin(i) = n \cdot \sin(r) \]
- This simplifies to:
\[ \sin(i) = n \cdot \sin(r) \]

3. **Substituting the Relationship Between Angles:**
- Since \( i = 2r \), we can substitute \( i \) in the equation:
\[ \sin(2r) = n \cdot \sin(r) \]

4. **Using the Double Angle Formula:**
- We can use the double angle formula for sine:
\[ \sin(2r) = 2 \sin(r) \cos(r) \]
- Substituting this into our equation gives:
\[ 2 \sin(r) \cos(r) = n \cdot \sin(r) \]

5. **Dividing by \( \sin(r) \):**
- Assuming \( \sin(r) \neq 0 \), we can divide both sides by \( \sin(r) \):
\[ 2 \cos(r) = n \]

6. **Finding the Angle of Incidence:**
- Rearranging gives:
\[ \cos(r) = \frac{n}{2} \]
- Now, we can find \( r \) using the inverse cosine function:
\[ r = \cos^{-1}\left(\frac{n}{2}\right) \]
- Since \( i = 2r \), we substitute for \( r \):
\[ i = 2 \cos^{-1}\left(\frac{n}{2}\right) \]

### Final Answer:
The angle of incidence \( i \) is:
\[ i = 2 \cos^{-1}\left(\frac{n}{2}\right) \]

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