A ray of light is refracted through a sphere whose material has a refractive index `mu` in such a way that it passes through the extremities of two radii which make an angle `beta` with each other. Prove that if `alpha` is the deviation of the ray caused by its passage through the sphere, then
`cos((beta-alpha)/2)=mu cos(beta/alpha)`

Explanation

To solve the problem, we need to prove that:

\[ \cos\left(\frac{\beta - \alpha}{2}\right) = \mu \cos\left(\frac{\beta}{2}\right) \]

where:
- \(\alpha\) is the angle of deviation,
- \(\beta\) is the angle between the two radii,
- \(\mu\) is the refractive index of the sphere.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- Consider a ray of light passing through a sphere. The ray is incident at point A and refracted at point B, making an angle \(\beta\) with the radius at point A and point B.
- The angles of incidence and refraction at the first surface are \(i\) and \(r\) respectively.

2. **Using the Angle of Deviation**:
- The angle of deviation \(\alpha\) can be expressed as:
\[ \alpha = i + e - 2r \]
where \(e\) is the angle of emergence.

3. **Relating Angles**:
- From the geometry of the sphere, we know that:
\[ \beta + 2r = 180^\circ \]
This implies:
\[ r = 90^\circ - \frac{\beta}{2} \]

4. **Substituting for \(r\)**:
- Substitute \(r\) into the deviation formula:
\[ \alpha = i + e - 2\left(90^\circ - \frac{\beta}{2}\right) \]
Simplifying this gives:
\[ \alpha = i + e - 180^\circ + \beta \]
Rearranging yields:
\[ \beta - \alpha = i + e - 180^\circ \]

5. **Finding the Average of \(i\) and \(e\)**:
- The average of the angles of incidence and emergence can be expressed as:
\[ \frac{i + e}{2} = 90^\circ - \frac{\beta - \alpha}{2} \]

6. **Using Snell's Law**:
- Applying Snell's Law at the first surface:
\[ \mu_1 \sin i = \mu_2 \sin r \]
Given \(\mu_1 = 1\) (air) and \(\mu_2 = \mu\), we have:
\[ \sin i = \mu \sin r \]
Since \(r = 90^\circ - \frac{\beta}{2}\), we can write:
\[ \sin r = \cos\left(\frac{\beta}{2}\right) \]
Thus:
\[ \sin i = \mu \cos\left(\frac{\beta}{2}\right) \]

7. **Expressing \(\sin i\)**:
- We can also express \(\sin i\) in terms of \(\alpha\):
\[ \sin i = \sin\left(90^\circ - \frac{\beta - \alpha}{2}\right) = \cos\left(\frac{\beta - \alpha}{2}\right) \]

8. **Equating the Two Expressions for \(\sin i\)**:
- We have:
\[ \cos\left(\frac{\beta - \alpha}{2}\right) = \mu \cos\left(\frac{\beta}{2}\right) \]

9. **Final Result**:
- Thus, we arrive at the desired proof:
\[ \cos\left(\frac{\beta - \alpha}{2}\right) = \mu \cos\left(\frac{\beta}{2}\right) \]