A ray incident at a point at an angle of incidence of `60^(@)` enters a glass sphere with refractive index `sqrt(3)` and it is reflected and refracted at the farther surface of the sphere. The angle between the reflected and refracted rays at this surface is:

Options

A. `50^(@)`
B. `60^(@)`
C. `90^(@)`
D. `40^(@)`

Explanation

To solve the problem step by step, we will analyze the situation involving the incident ray, the glass sphere, and the angles of incidence, reflection, and refraction.

### Step 1: Identify the given data
- Angle of incidence (i) = 60°
- Refractive index of glass (n) = √3

### Step 2: Apply Snell's Law at the first interface
Using Snell's Law, we have:
\[ n_1 \sin(i) = n_2 \sin(r) \]
Where:
- \( n_1 = 1 \) (refractive index of air)
- \( n_2 = \sqrt{3} \)
- \( r \) is the angle of refraction.

Substituting the known values:
\[ 1 \cdot \sin(60°) = \sqrt{3} \cdot \sin(r) \]
\[ \sin(60°) = \frac{\sqrt{3}}{2} \]
Thus,
\[ \frac{\sqrt{3}}{2} = \sqrt{3} \cdot \sin(r) \]
Dividing both sides by √3:
\[ \frac{1}{2} = \sin(r) \]
Therefore,
\[ r = 30° \]

### Step 3: Analyze the geometry of the situation
Since the ray enters the sphere and refracts, we denote the angle of refraction as \( r = 30° \). The ray will continue traveling inside the sphere until it reaches the farther surface.

### Step 4: Apply Snell's Law at the second interface
At the second interface (the farther surface of the sphere), we again apply Snell's Law:
\[ n_2 \sin(r) = n_1 \sin(E) \]
Where:
- \( E \) is the angle of incidence at the second interface.

Substituting the known values:
\[ \sqrt{3} \cdot \sin(30°) = 1 \cdot \sin(E) \]
Since \( \sin(30°) = \frac{1}{2} \):
\[ \sqrt{3} \cdot \frac{1}{2} = \sin(E) \]
Thus,
\[ \sin(E) = \frac{\sqrt{3}}{2} \]
Therefore,
\[ E = 60° \]

### Step 5: Determine the angle between the reflected and refracted rays
At the second interface, the angle of reflection (which equals the angle of incidence) is also \( E = 60° \). The angle between the reflected ray and the refracted ray can be calculated as follows:

1. The angle of reflection is \( 60° \) (from the normal).
2. The angle of refraction is also \( 30° \) (from the normal).

Now, the angle between the reflected ray and the refracted ray is:
\[ \text{Angle between reflected and refracted rays} = \text{Angle of reflection} + \text{Angle of refraction} \]
\[ = 60° + 30° = 90° \]

### Final Answer
The angle between the reflected and refracted rays at the surface of the sphere is **90°**.

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