Unpolarized light is incident on a plane glass surface. The angle of incidence so that reflected and refracted rays and perpendicular to each other, then

Options

A. `"tan"i_(beta)=(mu)/(2)`
B. `"tan"i_(beta)=mu`
C. `"sin"i_(beta)=mu`
D. `"cos"i_(beta)=mu`

Explanation

To solve the problem, we need to find the relationship between the angle of incidence (I) at which the reflected and refracted rays are perpendicular to each other when unpolarized light strikes a plane glass surface.

### Step-by-Step Solution:

1. **Understanding the Situation**:
- We have unpolarized light incident on a plane glass surface.
- The light will be partially reflected and partially refracted at the surface.

2. **Condition for Perpendicular Rays**:
- We need to find the angle of incidence (I) such that the reflected ray and the refracted ray are perpendicular to each other.
- This means that if the angle of reflection is R and the angle of refraction is r, then R + r = 90°.

3. **Using Snell's Law**:
- According to Snell's Law, we have:
\[ n_1 \sin I = n_2 \sin r \]
- Here, \( n_1 \) is the refractive index of air (approximately 1), and \( n_2 \) is the refractive index of glass.

4. **Relating Angles**:
- Since R + r = 90°, we can express r as:
\[ r = 90° - R \]
- Therefore, we can substitute this into Snell's Law:
\[ \sin r = \sin(90° - R) = \cos R \]
- This gives us:
\[ n_1 \sin I = n_2 \cos R \]

5. **Using the Reflection Law**:
- From the law of reflection, we know that the angle of reflection R is equal to the angle of incidence I:
\[ R = I \]
- Substituting this into our earlier equation gives:
\[ n_1 \sin I = n_2 \cos I \]

6. **Rearranging the Equation**:
- Rearranging the equation leads to:
\[ \frac{\sin I}{\cos I} = \frac{n_2}{n_1} \]
- This simplifies to:
\[ \tan I = \frac{n_2}{n_1} \]

7. **Identifying the Polarizing Angle**:
- The angle of incidence I at which the reflected and refracted rays are perpendicular is known as the polarizing angle (θp). Therefore, we can write:
\[ \tan θ_p = n_2 \]
- If we denote the refractive index of air as \( n_1 = 1 \), we have:
\[ \tan θ_p = n_2 \]

### Final Relation:
Thus, the relationship we are looking for is:
\[ \tan I_b = n \]
where \( I_b \) is the angle of incidence at which the reflected and refracted rays are perpendicular.