A ray of light is incident at `65^(@)` on one face of a prism of angle of `30^(@)` and the emergent ray makes `35^(@)` with the incident ray. The refractive index of the prism is:

Options

A. 1.812
B. 1.414
C. 1.5
D. 1.33

Answer

To find the refractive index of the prism, we will follow these steps:

### Step 1: Identify the given values
- Angle of incidence (i) = 65°
- Angle of the prism (A) = 30°
- Angle of deviation (D) = 35°

### Step 2: Use the formula for angle of deviation
The angle of deviation (D) can be expressed as:
\[ D = i + e - A \]
where \( e \) is the angle of emergence.

### Step 3: Rearranging the formula
From the above equation, we can rearrange it to find the angle of emergence (e):
\[ e = D - i + A \]

### Step 4: Substitute the known values
Substituting the known values into the equation:
\[ e = 35° - 65° + 30° \]
\[ e = 35° - 65° + 30° = 0° \]

### Step 5: Interpret the result
Since the angle of emergence (e) is 0°, it means that the emergent ray is normal to the surface of the prism.

### Step 6: Apply Snell's Law at the first surface
Using Snell's Law at the first surface of the prism:
\[ n_1 \sin(i) = n_2 \sin(r_1) \]
Where:
- \( n_1 \) = refractive index of air = 1
- \( n_2 \) = refractive index of the prism
- \( r_1 \) = angle of refraction at the first surface

### Step 7: Determine the angle of refraction (r1)
Since the angle of emergence is 0°, the angle of refraction (r1) at the first surface can be calculated as:
\[ r_1 = A - e = 30° - 0° = 30° \]

### Step 8: Substitute values into Snell's Law
Now substituting the values into Snell's Law:
\[ 1 \cdot \sin(65°) = n_2 \cdot \sin(30°) \]
\[ \sin(65°) = n_2 \cdot \frac{1}{2} \]

### Step 9: Solve for the refractive index (n2)
Now we can solve for \( n_2 \):
\[ n_2 = \frac{\sin(65°)}{\sin(30°)} \]
\[ n_2 = \frac{\sin(65°)}{0.5} \]
\[ n_2 = 2 \cdot \sin(65°) \]

### Step 10: Calculate the value
Using the value of \( \sin(65°) \approx 0.906 \):
\[ n_2 = 2 \cdot 0.906 \]
\[ n_2 \approx 1.812 \]

Thus, the refractive index of the prism is approximately **1.812**.

---