A ray of light-is incident on glass at the polarising angle ip. Find the angle between the reflected and refracted rays.

Answer

To solve the problem of finding the angle between the reflected and refracted rays when a ray of light is incident on glass at the polarizing angle (also known as Brewster's angle), we can follow these steps:

### Step-by-Step Solution:

1. **Understand Brewster's Angle**: Brewster's angle (denoted as \( i_p \)) is the angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection. This angle can be calculated using the formula:
\[ \tan(i_p) = \frac{n_2}{n_1} \]
where \( n_1 \) is the refractive index of the first medium (air, typically taken as 1) and \( n_2 \) is the refractive index of the second medium (glass).

2. **Use Snell's Law**: According to Snell's law, the relationship between the angle of incidence \( i \) and the angle of refraction \( r \) is given by:
\[ n_1 \sin(i) = n_2 \sin(r) \]

3. **Relation at Brewster's Angle**: At Brewster's angle, the angle of incidence \( i_p \) and the angle of refraction \( r \) are related such that:
\[ i_p + r = 90^\circ \]
This means that:
\[ r = 90^\circ - i_p \]

4. **Determine the Angle Between Reflected and Refracted Rays**: The angle between the reflected ray and the refracted ray can be found by observing that the reflected ray makes an angle of \( i_p \) with the normal, and the refracted ray makes an angle of \( r \) with the normal. Therefore, the angle \( \theta \) between the reflected and refracted rays is given by:
\[ \theta = i_p + r \]

5. **Substituting the Relation**: Since we established that \( r = 90^\circ - i_p \), we can substitute this into our equation for \( \theta \):
\[ \theta = i_p + (90^\circ - i_p) = 90^\circ \]

### Final Answer:
The angle between the reflected and refracted rays when a ray of light is incident on glass at the polarizing angle is \( 90^\circ \).

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