A beam of light travelling with speed `c` in vacuum encounters a glass medium at an angle of incidence `60^(@)`. If the speed of light in the medium is `(c)/(sqrt(3))`, then the angle between the reflected and the refracted beams is :

Options

A. `30^(@)`
B. `60^(@)`
C. `90^(@)`
D. `120^(@)`

Explanation

To solve the problem, we need to find the angle between the reflected and refracted beams when a beam of light travels from vacuum into a glass medium.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Angle of incidence, \( i = 60^\circ \)
- Speed of light in vacuum, \( c \)
- Speed of light in glass, \( v = \frac{c}{\sqrt{3}} \)

2. **Calculate the Refractive Index:**
The refractive index \( \mu \) of the glass can be calculated using the formula:
\[ \mu = \frac{c}{v} \]
Substituting the value of \( v \):
\[ \mu = \frac{c}{\frac{c}{\sqrt{3}}} = \sqrt{3} \]

3. **Apply Snell's Law:**
Snell's law states:
\[ \mu_1 \sin(i) = \mu_2 \sin(r) \]
Here, \( \mu_1 = 1 \) (for vacuum), \( \mu_2 = \sqrt{3} \), and \( r \) is the angle of refraction. Substituting the known values:
\[ 1 \cdot \sin(60^\circ) = \sqrt{3} \cdot \sin(r) \]
We know that \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \):
\[ \frac{\sqrt{3}}{2} = \sqrt{3} \cdot \sin(r) \]

4. **Solve for \( \sin(r) \):**
Dividing both sides by \( \sqrt{3} \):
\[ \frac{1}{2} = \sin(r) \]
Therefore, we find:
\[ r = \sin^{-1}\left(\frac{1}{2}\right) = 30^\circ \]

5. **Determine the Angle Between Reflected and Refracted Beams:**
The angle of reflection \( r' \) is equal to the angle of incidence \( i \):
\[ r' = 60^\circ \]
The angle between the reflected beam and the refracted beam is given by:
\[ \text{Angle between} = r' + r = 60^\circ + 30^\circ = 90^\circ \]

### Final Answer:
The angle between the reflected and the refracted beams is \( 90^\circ \).