Find the (a) angle of minimum and (b) the maximum deviation for an equilateral prism having refractive index `n=sqrt(3)`.

Answer

To solve the problem of finding the angle of minimum deviation and the maximum deviation for an equilateral prism with a refractive index \( n = \sqrt{3} \), we can follow these steps:

### Step 1: Identify the Angle of the Prism
For an equilateral prism, the angle \( A \) is given as:
\[ A = 60^\circ \]

### Step 2: Use the Formula for Minimum Deviation
The formula for the angle of minimum deviation \( D_{min} \) in terms of the refractive index \( n \) and the angle of the prism \( A \) is given by:
\[ n = \frac{\sin\left(\frac{A + D_{min}}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

### Step 3: Substitute Known Values
Substituting \( n = \sqrt{3} \) and \( A = 60^\circ \):
\[ \sqrt{3} = \frac{\sin\left(\frac{60^\circ + D_{min}}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} \]
Since \( \sin(30^\circ) = \frac{1}{2} \), we have:
\[ \sqrt{3} = \frac{\sin\left(\frac{60^\circ + D_{min}}{2}\right)}{\frac{1}{2}} \]
This simplifies to:
\[ \sqrt{3} \cdot \frac{1}{2} = \sin\left(\frac{60^\circ + D_{min}}{2}\right) \]
\[ \frac{\sqrt{3}}{2} = \sin\left(\frac{60^\circ + D_{min}}{2}\right) \]

### Step 4: Find the Angle Corresponding to the Sine Value
The sine value \( \frac{\sqrt{3}}{2} \) corresponds to:
\[ \frac{60^\circ + D_{min}}{2} = 60^\circ \]
Thus, we can solve for \( D_{min} \):
\[ 60^\circ + D_{min} = 120^\circ \]
\[ D_{min} = 120^\circ - 60^\circ = 60^\circ \]

### Step 5: Calculate the Maximum Deviation
The formula for the maximum deviation \( D_{max} \) is given by:
\[ D_{max} = 2I_{min} - A \]
Where \( I_{min} \) is the angle of incidence at minimum deviation. For an equilateral prism, \( I_{min} = 60^\circ \):
\[ D_{max} = 2 \times 60^\circ - 60^\circ = 120^\circ - 60^\circ = 60^\circ \]

### Step 6: Verify the Maximum Deviation
Alternatively, we can use:
\[ D_{max} = I_{min} + 90^\circ - A \]
Substituting the values:
\[ D_{max} = 60^\circ + 90^\circ - 60^\circ = 90^\circ \]

### Final Answers
(a) The angle of minimum deviation \( D_{min} \) is \( 60^\circ \).

(b) The maximum deviation \( D_{max} \) is \( 90^\circ \).

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