The angle of minimum deviation is an equilateral prism of refractive index `1.414` is

Options

A. `60^(@)`
B. `30^(@)`
C. `90^(@)`
D. `45^(@)`

Explanation

To find the angle of minimum deviation (Δm) for an equilateral prism with a refractive index (μ) of 1.414, we can use the formula:

\[ \mu = \frac{\sin\left(\frac{A + \Delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

Where:
- \( \mu \) is the refractive index of the prism,
- \( A \) is the angle of the prism,
- \( \Delta_m \) is the angle of minimum deviation.

### Step-by-Step Solution:

1. **Identify the angle of the prism (A)**:
Since it is an equilateral prism, each angle is 60 degrees.
\[ A = 60^\circ \]

2. **Substitute the values into the formula**:
Substitute \( \mu = 1.414 \) and \( A = 60^\circ \) into the formula:
\[ 1.414 = \frac{\sin\left(\frac{60^\circ + \Delta_m}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} \]

3. **Calculate \(\sin\left(\frac{A}{2}\right)\)**:
\[ \sin\left(\frac{60^\circ}{2}\right) = \sin(30^\circ) = \frac{1}{2} \]

4. **Rearranging the equation**:
Substitute \(\sin(30^\circ)\) into the equation:
\[ 1.414 = \frac{\sin\left(30^\circ + \frac{\Delta_m}{2}\right)}{\frac{1}{2}} \]
This simplifies to:
\[ 1.414 \times \frac{1}{2} = \sin\left(30^\circ + \frac{\Delta_m}{2}\right) \]
\[ 0.707 = \sin\left(30^\circ + \frac{\Delta_m}{2}\right) \]

5. **Finding the angle whose sine is 0.707**:
We know that:
\[ \sin(45^\circ) = 0.707 \]
Therefore:
\[ 30^\circ + \frac{\Delta_m}{2} = 45^\circ \]

6. **Solving for \(\Delta_m\)**:
Rearranging gives:
\[ \frac{\Delta_m}{2} = 45^\circ - 30^\circ \]
\[ \frac{\Delta_m}{2} = 15^\circ \]
Multiplying both sides by 2:
\[ \Delta_m = 30^\circ \]

### Final Answer:
The angle of minimum deviation (Δm) for the equilateral prism is:
\[ \Delta_m = 30^\circ \]