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 reflection and refraction.

### Step 1: Understand the scenario
We have a ray of light incident on a glass sphere at an angle of incidence of \(60^\circ\). The refractive index of the glass sphere is \(\sqrt{3}\).

### Step 2: Draw the diagram
Draw a diagram of the glass sphere with the incident ray, the normal at the point of incidence, and label the angles:
- Angle of incidence \(i = 60^\circ\)
- Angle of reflection \(r\)
- Angle of refraction \(e\)

### Step 3: Apply Snell's Law at the first interface
Using Snell's Law, we can relate the angles of incidence and refraction:
\[ \frac{\sin i}{\sin r} = n \]
Where \(n\) is the refractive index. Here, \(n = \sqrt{3}\).

Substituting the known values:
\[ \frac{\sin 60^\circ}{\sin r} = \sqrt{3} \]
Since \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), we have:
\[ \frac{\frac{\sqrt{3}}{2}}{\sin r} = \sqrt{3} \]

### Step 4: Solve for \(\sin r\)
Cross-multiplying gives:
\[ \frac{\sqrt{3}}{2} = \sqrt{3} \cdot \sin r \]
Dividing both sides by \(\sqrt{3}\):
\[ \sin r = \frac{1}{2} \]
Thus, \(r = 30^\circ\).

### Step 5: Apply Snell's Law at the second interface
Now we consider the ray that has been refracted into the sphere. At the second interface (the farther surface of the sphere), we apply Snell's Law again:
\[ \frac{\sin e}{\sin r} = n \]
Where \(r = 30^\circ\) and \(n = \sqrt{3}\):
\[ \frac{\sin e}{\sin 30^\circ} = \sqrt{3} \]
Since \(\sin 30^\circ = \frac{1}{2}\), we have:
\[ \frac{\sin e}{\frac{1}{2}} = \sqrt{3} \]

### Step 6: Solve for \(\sin e\)
Cross-multiplying gives:
\[ \sin e = \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2} \]
Thus, \(e = 60^\circ\).

### Step 7: Find the angle between the reflected and refracted rays
The angle between the reflected ray and the refracted ray is given by:
\[ \alpha = 180^\circ - r - e \]
Substituting the values we found:
\[ \alpha = 180^\circ - 30^\circ - 60^\circ = 90^\circ \]

### Final Answer
The angle between the reflected and refracted rays at the surface of the sphere is \(90^\circ\).

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