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. `120^(@)`
B. `90^(@)`
C. `60^(@)`
D. `150^(@)`

Answer

To solve the problem step by step, we will apply Snell's law and the principles of reflection and refraction.

### Step 1: Identify the given values
- 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:
\[ n_1 \sin(i) = n_2 \sin(r_1) \]
Where:
- \( n_1 = 1 \) (refractive index of air)
- \( n_2 = \sqrt{3} \)
- \( r_1 \) is the angle of refraction at the first interface.

Substituting the values:
\[ 1 \cdot \sin(60°) = \sqrt{3} \cdot \sin(r_1) \]
\[ \sin(60°) = \frac{\sqrt{3}}{2} \]
Thus,
\[ \frac{\sqrt{3}}{2} = \sqrt{3} \cdot \sin(r_1) \]
Dividing both sides by √3:
\[ \frac{1}{2} = \sin(r_1) \]
This gives:
\[ r_1 = 30° \]

### Step 3: Determine the angle of incidence at the second interface
At the second interface (the farther surface of the sphere), the angle of incidence \( i_2 \) is equal to the angle of refraction \( r_1 \):
\[ i_2 = r_1 = 30° \]

### Step 4: Apply Snell's Law at the second interface
Using Snell's law again:
\[ n_2 \sin(i_2) = n_3 \sin(r_2) \]
Where:
- \( n_2 = \sqrt{3} \)
- \( n_3 = 1 \) (refractive index of air)
- \( r_2 \) is the angle of refraction at the second interface.

Substituting the values:
\[ \sqrt{3} \cdot \sin(30°) = 1 \cdot \sin(r_2) \]
Since \( \sin(30°) = \frac{1}{2} \):
\[ \sqrt{3} \cdot \frac{1}{2} = \sin(r_2) \]
Thus,
\[ \sin(r_2) = \frac{\sqrt{3}}{2} \]
This gives:
\[ r_2 = 60° \]

### Step 5: Determine the angle between the reflected and refracted rays
The angle of reflection (r') at the second interface is equal to the angle of incidence \( i_2 \):
\[ r' = i_2 = 30° \]

Now, the angle between the reflected ray and the refracted ray at the second interface can be calculated as:
\[ \text{Angle between reflected and refracted rays} = r' + r_2 \]
\[ = 30° + 60° = 90° \]

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

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