What is the angle of incidence for an equilateral prism of refractive index `sqrt(3)` so that the ray si parallel to the base inside the prism?

Options

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

Explanation

To solve the problem of finding the angle of incidence for an equilateral prism with a refractive index of \(\sqrt{3}\) such that the ray is parallel to the base inside the prism, we can follow these steps:

### Step 1: Understand the Geometry of the Prism
The prism is equilateral, meaning each angle is \(60^\circ\). We denote the angle of the prism as \(A = 60^\circ\).

### Step 2: Use the Formula for Refractive Index
The refractive index (\(\mu\)) of the prism can be expressed in terms of the angle of deviation (\(\delta\)) and the angle of the prism (\(A\)) using the formula:
\[ \mu = \frac{\sin\left(\frac{A + \delta}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
Since the ray is parallel to the base inside the prism, the angle of deviation (\(\delta\)) can be determined.

### Step 3: Substitute Known Values
Given that \(\mu = \sqrt{3}\) and \(A = 60^\circ\):
- \(\frac{A}{2} = \frac{60^\circ}{2} = 30^\circ\)
- \(\sin(30^\circ) = \frac{1}{2}\)

Now, substituting these values into the refractive index formula:
\[ \sqrt{3} = \frac{\sin\left(\frac{60^\circ + \delta}{2}\right)}{\sin(30^\circ)} = \frac{\sin\left(\frac{60^\circ + \delta}{2}\right)}{\frac{1}{2}} \]
This simplifies to:
\[ \sqrt{3} = 2 \sin\left(\frac{60^\circ + \delta}{2}\right) \]

### Step 4: Solve for \(\delta\)
Rearranging gives:
\[ \sin\left(\frac{60^\circ + \delta}{2}\right) = \frac{\sqrt{3}}{2} \]
The angle whose sine is \(\frac{\sqrt{3}}{2}\) is \(60^\circ\). Thus:
\[ \frac{60^\circ + \delta}{2} = 60^\circ \]
Multiplying both sides by 2:
\[ 60^\circ + \delta = 120^\circ \]
Solving for \(\delta\):
\[ \delta = 120^\circ - 60^\circ = 60^\circ \]

### Step 5: Apply the Angle of Deviation Formula
Using the relationship for prisms:
\[ I + E = A + \delta \]
Where \(I\) is the angle of incidence and \(E\) is the angle of emergence. For the case where the angle of incidence is equal to the angle of emergence (\(I = E\)):
\[ 2I = A + \delta \]
Substituting the known values:
\[ 2I = 60^\circ + 60^\circ = 120^\circ \]
Thus:
\[ I = \frac{120^\circ}{2} = 60^\circ \]

### Final Answer
The angle of incidence \(I\) is \(60^\circ\).

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