A ray of light falls on the surface of a spherical glass paper weight making an angle `alpha` with the normal and is refracted in the medium at an angle `beta`. The angle of deviation of the emergent ray from the direction of the incident ray is :

Options

A. `(alpha-beta)`
B. `2(alpha-beta)`
C. `(alpha-beta)//2`
D. `(beta-alpha)`

Answer

To solve the problem, we need to determine the angle of deviation of the emergent ray from the direction of the incident ray when a ray of light passes through a spherical glass paperweight. Here’s the step-by-step solution:

### Step 1: Understand the Geometry of the Problem
- A ray of light strikes the surface of a spherical glass paperweight at an angle \( \alpha \) with respect to the normal.
- The ray is refracted inside the glass and emerges at an angle \( \beta \) with respect to the normal.

### Step 2: Apply Snell's Law
- At the point of incidence, we can apply Snell's Law, which states:
\[ n_1 \sin(\alpha) = n_2 \sin(\beta) \]
where \( n_1 \) is the refractive index of air (approximately 1) and \( n_2 \) is the refractive index of glass (greater than 1).

### Step 3: Determine the Angles of Deviation
- The angle of deviation \( D \) is defined as the angle between the direction of the incident ray and the direction of the emergent ray.
- The deviation can be calculated as:
\[ D = (\alpha - \beta) + (\beta - \alpha) \]
This is because the ray is refracted twice: once at the entry and once at the exit of the glass.

### Step 4: Calculate Total Deviation
- The total deviation \( D \) can be expressed as:
\[ D = (\alpha - \beta) + (\alpha - \beta) = 2(\alpha - \beta) \]
- This means the total angle of deviation of the emergent ray from the direction of the incident ray is:
\[ D = 2(\alpha - \beta) \]

### Conclusion
- Therefore, the angle of deviation of the emergent ray from the direction of the incident ray is:
\[ D = 2(\alpha - \beta) \]