Refractive index of an equilateral prism is `sqrt(2)`.

Options

A. minimum deviation from this prism can be `30^(@)`
B. minimum deviation from this prism can be `45^(@)`
C. at angle of incidence `= 45^(@)`, deviation is minimum
D. at angle of incidence `= 60^(@)`, deviation is minimum

Answer

To solve the problem regarding the refractive index of an equilateral prism and its minimum deviation, we will follow these steps:

### Step 1: Understand the Geometry of the Prism
An equilateral prism has all its angles equal to 60 degrees. This means that the angle of the prism (A) is 60°.

### Step 2: Identify the Given Refractive Index
The refractive index (μ) of the prism is given as √2.

### Step 3: Use the Formula for Minimum Deviation
For a prism, the relationship between the refractive index (μ), the angle of the prism (A), and the minimum deviation (Δ) is given by the formula:
\[ μ = \frac{\sin\left(\frac{A + Δ}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

### Step 4: Substitute the Known Values
Substituting the known values into the formula:
- A = 60°
- μ = √2

The equation becomes:
\[ \sqrt{2} = \frac{\sin\left(\frac{60° + Δ}{2}\right)}{\sin\left(30°\right)} \]
Since \(\sin(30°) = \frac{1}{2}\), we can rewrite the equation as:
\[ \sqrt{2} = 2 \cdot \sin\left(\frac{60° + Δ}{2}\right) \]

### Step 5: Simplify the Equation
This simplifies to:
\[ \sin\left(\frac{60° + Δ}{2}\right) = \frac{\sqrt{2}}{2} \]
Since \(\sin(45°) = \frac{\sqrt{2}}{2}\), we can set:
\[ \frac{60° + Δ}{2} = 45° \]

### Step 6: Solve for Minimum Deviation (Δ)
Multiplying both sides by 2 gives:
\[ 60° + Δ = 90° \]
Subtracting 60° from both sides results in:
\[ Δ = 30° \]

### Step 7: Determine the Angle of Incidence
At minimum deviation for an equilateral prism, the angle of incidence (i) is equal to the angle of emergence (e). Using Snell's law at the first surface:
\[ \frac{\sin(i)}{\sin(r)} = μ \]
Where r is the angle of refraction. Since the prism is equilateral, the angle of refraction (r) at the first surface is:
\[ r = \frac{A + Δ}{2} = \frac{60° + 30°}{2} = 45° \]
Thus, we have:
\[ \frac{\sin(i)}{\sin(45°)} = \sqrt{2} \]
Since \(\sin(45°) = \frac{\sqrt{2}}{2}\), we can write:
\[ \sin(i) = \sqrt{2} \cdot \frac{\sqrt{2}}{2} = 1 \]
This implies:
\[ i = 90° \]
However, we need to find the angle of incidence that gives minimum deviation. We can find the angle of incidence at which the deviation is minimum using the earlier established relationship:
\[ i = 45° \]

### Conclusion
Thus, the minimum deviation for the prism is 30° and occurs at an angle of incidence of 45°.