A ray of light is incident on one face of a transparent slab of thickness 15 cm. The angle of incidence is `60^(@)`. If the lateral displacement of the ray on emerging from the parallel plane is `5sqrt(3)` cm, the refractive index of the material of the slab is

Options

A. 1.414
B. 1.532
C. 1.732
D. None of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
- Thickness of the slab (t) = 15 cm
- Angle of incidence (i) = 60°
- Lateral displacement (x) = 5√3 cm

### Step 2: Use the formula for lateral displacement
The formula for lateral displacement (x) is given by:
\[ x = t \cdot \frac{\sin(i - r)}{\cos(r)} \]
where \( r \) is the angle of refraction.

### Step 3: Rearrange the formula
We can rearrange the formula to solve for \( \sin(i - r) \):
\[ x \cdot \cos(r) = t \cdot \sin(i - r) \]
Thus,
\[ \sin(i - r) = \frac{x \cdot \cos(r)}{t} \]

### Step 4: Substitute the known values
Substituting the known values into the equation:
\[ 5\sqrt{3} \cdot \cos(r) = 15 \cdot \sin(60° - r) \]
We know that \( \sin(60°) = \frac{\sqrt{3}}{2} \).

### Step 5: Calculate \( \sin(60° - r) \)
Using the sine subtraction formula:
\[ \sin(60° - r) = \sin(60°) \cos(r) - \cos(60°) \sin(r) \]
Substituting the known values:
\[ \sin(60° - r) = \frac{\sqrt{3}}{2} \cos(r) - \frac{1}{2} \sin(r) \]

### Step 6: Set up the equation
Now we can set up the equation:
\[ 5\sqrt{3} \cdot \cos(r) = 15 \left( \frac{\sqrt{3}}{2} \cos(r) - \frac{1}{2} \sin(r) \right) \]

### Step 7: Simplify the equation
Expanding the right side:
\[ 5\sqrt{3} \cdot \cos(r) = \frac{15\sqrt{3}}{2} \cos(r) - \frac{15}{2} \sin(r) \]
Rearranging gives:
\[ 5\sqrt{3} \cdot \cos(r) - \frac{15\sqrt{3}}{2} \cos(r) = -\frac{15}{2} \sin(r) \]

### Step 8: Combine like terms
Combining the terms on the left side:
\[ \left(5\sqrt{3} - \frac{15\sqrt{3}}{2}\right) \cos(r) = -\frac{15}{2} \sin(r) \]
Calculating the left side:
\[ \left(\frac{10\sqrt{3}}{2} - \frac{15\sqrt{3}}{2}\right) \cos(r) = -\frac{15}{2} \sin(r) \]
This simplifies to:
\[ -\frac{5\sqrt{3}}{2} \cos(r) = -\frac{15}{2} \sin(r) \]

### Step 9: Solve for \( \tan(r) \)
Dividing both sides by \(-\frac{5}{2}\):
\[ \sqrt{3} \cos(r) = 3 \sin(r) \]
Thus:
\[ \tan(r) = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} \]
This implies:
\[ r = 30° \]

### Step 10: Calculate the refractive index
Using Snell's law:
\[ n = \frac{\sin(i)}{\sin(r)} = \frac{\sin(60°)}{\sin(30°)} \]
Substituting the known values:
\[ n = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \approx 1.732 \]

### Final Answer
The refractive index of the material of the slab is approximately \( 1.732 \).

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