A ray of light entering from air to glas `(mu=1.5)` is partly reflected and partly refracted. lf the reflected and refracted rays are at right angles to each other, the angle of refraction is

Options

A. `sin^(-1)(sqrt(2/13))`
B. `sin^(-1)((sqrt(2))/13)`
C. `sin^(-1)(2/(sqrt(13)))`
D. `sin^(-1)(3/(sqrt(13)))`

Answer

To solve the problem, we need to find the angle of refraction when a ray of light enters from air into glass, given that the reflected and refracted rays are at right angles to each other.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- When a ray of light travels from air (refractive index \( \mu_1 = 1 \)) into glass (refractive index \( \mu_2 = 1.5 \)), it undergoes partial reflection and refraction.
- Let \( i \) be the angle of incidence, \( r \) be the angle of refraction, and \( r' \) be the angle of reflection. According to the law of reflection, \( i = r' \).

2. **Setting Up the Angles**:
- The problem states that the reflected and refracted rays are at right angles to each other. Therefore, we can write:
\[ r' + r = 90^\circ \]
- This implies:
\[ r = 90^\circ - r' \]

3. **Using the Law of Reflection**:
- Since \( r' = i \), we can substitute this into the equation:
\[ r = 90^\circ - i \]

4. **Applying Snell's Law**:
- Snell's law states:
\[ \mu_1 \sin(i) = \mu_2 \sin(r) \]
- Substituting the values of refractive indices:
\[ 1 \cdot \sin(i) = 1.5 \cdot \sin(r) \]

5. **Substituting for \( r \)**:
- From step 3, we know \( r = 90^\circ - i \). Therefore:
\[ \sin(r) = \sin(90^\circ - i) = \cos(i) \]
- Now substituting this into Snell's law gives:
\[ \sin(i) = 1.5 \cdot \cos(i) \]

6. **Using the Identity**:
- We can use the identity \( \sin(i) = \sqrt{1 - \cos^2(i)} \):
\[ \sqrt{1 - \cos^2(i)} = 1.5 \cdot \cos(i) \]

7. **Squaring Both Sides**:
- Squaring both sides:
\[ 1 - \cos^2(i) = 2.25 \cdot \cos^2(i) \]
- Rearranging gives:
\[ 1 = 3.25 \cdot \cos^2(i) \]
- Thus:
\[ \cos^2(i) = \frac{1}{3.25} = \frac{4}{13} \]

8. **Finding \( \sin(i) \)**:
- Now, using \( \sin^2(i) + \cos^2(i) = 1 \):
\[ \sin^2(i) = 1 - \frac{4}{13} = \frac{9}{13} \]
- Therefore:
\[ \sin(i) = \frac{3}{\sqrt{13}} \]

9. **Finding \( \sin(r) \)**:
- Now substituting back into the equation \( \sin(i) = 1.5 \cdot \cos(i) \):
\[ \sin(r) = \frac{1}{1.5} \cdot \sin(i) = \frac{2}{3} \cdot \frac{3}{\sqrt{13}} = \frac{2}{\sqrt{13}} \]

10. **Calculating the Angle of Refraction**:
- Finally, we find the angle of refraction \( r \):
\[ r = \sin^{-1}\left(\frac{2}{\sqrt{13}}\right) \]

### Final Answer:
The angle of refraction \( r \) is \( \sin^{-1}\left(\frac{2}{\sqrt{13}}\right) \).

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