For a prism kept in air, it is found that for an angle of incidence `60^(@)`, the angle of refraction A, angle of deviation `delta` and anble of emergence e become equal. Then, the refractive index of the prism is

Options

A. 1.73
B. 1.15
C. 1.5
D. 1.33

Explanation

To find the refractive index of the prism given that the angle of incidence (I) is 60 degrees and the angle of refraction (A), angle of deviation (Δ), and angle of emergence (e) are all equal, we can follow these steps:

### Step 1: Understand the relationships
We know from the problem that:
- \( I = 60^\circ \)
- \( A = Δ = e \)

### Step 2: Use the prism formula
In a prism, the relationship between the angles can be expressed as:
\[ I + e = A + Δ \]
Since \( A = Δ = e \), we can substitute \( e \) in the equation:
\[ I + e = A + e \]
This simplifies to:
\[ I = A \]
Thus, we have:
\[ A = 60^\circ \]

### Step 3: Identify the case of minimum deviation
Since \( I = e \) and \( A = Δ \), this indicates that we are at the condition of minimum deviation. In this case, the angle of deviation \( Δ \) can be expressed as:
\[ Δ_{min} = I \]
Thus:
\[ Δ_{min} = 60^\circ \]

### Step 4: Use the formula for refractive index
The refractive index \( n \) of a prism can be calculated using the formula:
\[ n = \frac{\sin\left(\frac{A + Δ_{min}}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
Substituting the known values:
- \( A = 60^\circ \)
- \( Δ_{min} = 60^\circ \)

We have:
\[ n = \frac{\sin\left(\frac{60^\circ + 60^\circ}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} \]
This simplifies to:
\[ n = \frac{\sin(60^\circ)}{\sin(30^\circ)} \]

### Step 5: Calculate the sine values
Using known values:
- \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \)
- \( \sin(30^\circ) = \frac{1}{2} \)

Substituting these values into the equation:
\[ n = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \]

### Step 6: Final answer
Thus, the refractive index of the prism is:
\[ n = \sqrt{3} \approx 1.732 \]