A ray of light is incident on one face of a transparent glass slab of thickness 9cm at an angle of incidence `60^@`. Lateral displacement of the ray on emerging from the opposite parallel face is `3sqrt3`cm. Refractive index of the material of slab is:

Options

A. (a)`sqrt2`
B. (b)`sqrt3/2`
C. (c)`sqrt(3/2)`
D. (d)`sqrt3`

Explanation

To solve the problem, we need to find the refractive index of the material of the glass slab given the thickness, angle of incidence, and lateral displacement.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Thickness of the glass slab, \( t = 9 \, \text{cm} \)
- Angle of incidence, \( i = 60^\circ \)
- Lateral displacement, \( \Delta = 3\sqrt{3} \, \text{cm} \)

2. **Use the Formula for Lateral Displacement:**
The formula for lateral displacement \( \Delta \) when light passes through a slab is given by:
\[ \Delta = t \left( \sin i - \sin r \right) \frac{1}{\cos r} \]
where \( r \) is the angle of refraction.

3. **Apply Snell's Law:**
According to Snell's Law:
\[ n = \frac{\sin i}{\sin r} \]
where \( n \) is the refractive index of the material.

4. **Calculate \( \sin i \):**
For \( i = 60^\circ \):
\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \]

5. **Substituting Values into the Lateral Displacement Formula:**
We can rearrange the formula for lateral displacement:
\[ \Delta = t \left( \sin i - \sin r \right) \frac{1}{\cos r} \]
Substituting the known values:
\[ 3\sqrt{3} = 9 \left( \frac{\sqrt{3}}{2} - \sin r \right) \frac{1}{\cos r} \]

6. **Simplifying the Equation:**
Divide both sides by 9:
\[ \frac{\sqrt{3}}{3} = \left( \frac{\sqrt{3}}{2} - \sin r \right) \frac{1}{\cos r} \]

7. **Cross-Multiplying:**
Rearranging gives:
\[ \sqrt{3} \cos r = 3 \left( \frac{\sqrt{3}}{2} - \sin r \right) \]

8. **Distributing the Right Side:**
\[ \sqrt{3} \cos r = \frac{3\sqrt{3}}{2} - 3\sin r \]

9. **Rearranging the Equation:**
\[ 3\sin r + \sqrt{3} \cos r = \frac{3\sqrt{3}}{2} \]

10. **Dividing by \( \cos r \):**
\[ 3\tan r + \sqrt{3} = \frac{3\sqrt{3}}{2\cos r} \]

11. **Finding \( \tan r \):**
From the equation, we can isolate \( \tan r \):
\[ \tan r = \frac{\frac{3\sqrt{3}}{2} - \sqrt{3}}{3} = \frac{3\sqrt{3} - 2\sqrt{3}}{6} = \frac{\sqrt{3}}{6} \]

12. **Finding \( r \):**
Since \( \tan r = \frac{\sqrt{3}}{3} \), we find that:
\[ r = 30^\circ \]

13. **Using Snell's Law to Find Refractive Index:**
Now, applying Snell's Law:
\[ n = \frac{\sin 60^\circ}{\sin 30^\circ} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \]

### Final Answer:
The refractive index of the material of the slab is \( n = \sqrt{3} \).