A beam of parallel rays of diameter `'b'` propagates in glass at an angle `theta` to its plane. Find the diameter of the beam when it goes to air through this face. `(n_(glass)=n)`

Explanation

To solve the problem of finding the diameter of a beam of parallel rays when it propagates from glass to air, we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Setup**:
- We have a beam of parallel rays of diameter \( b \) propagating in glass with a refractive index \( n \) at an angle \( \theta \) to the normal of the surface.

2. **Identify the Angles**:
- The angle of incidence \( i \) is given as \( \theta \).
- According to Snell's law, the relationship between the angles and the refractive indices is given by:
\[ n \sin(i) = \sin(r) \]
- Here, \( r \) is the angle of refraction when the beam exits into air (where the refractive index is approximately 1).

3. **Using Snell's Law**:
- From Snell's law, we can express:
\[ n \sin(\theta) = \sin(r) \]
- This implies:
\[ \sin(r) = n \sin(\theta) \]

4. **Finding the Diameter After Refraction**:
- The diameter of the beam after refraction can be expressed in terms of the original diameter and the angles involved.
- For the triangle formed by the rays, we can use the cosine of the angle of refraction:
\[ CD = AB \cdot \cos(r) \]
- Since \( AB = b \), we have:
\[ CD = b \cdot \cos(r) \]

5. **Relating \( \cos(r) \) to \( \theta \)**:
- We can find \( \cos(r) \) using the identity:
\[ \cos^2(r) = 1 - \sin^2(r) \]
- Substituting \( \sin(r) = n \sin(\theta) \):
\[ \cos^2(r) = 1 - (n \sin(\theta))^2 \]
- Thus:
\[ \cos(r) = \sqrt{1 - n^2 \sin^2(\theta)} \]

6. **Substituting Back into the Diameter Equation**:
- Now substituting \( \cos(r) \) back into the diameter equation:
\[ CD = b \cdot \sqrt{1 - n^2 \sin^2(\theta)} \]

7. **Final Expression**:
- Therefore, the diameter of the beam when it goes to air through the face is:
\[ CD = b \sqrt{1 - n^2 \sin^2(\theta)} \]