The refractive index of the material of a prism of `60^(@)` angle for yellow light is `sqrt(2)`. Calculate angle of minimum deviation, angle of incidence and angle of refraction.

Answer

To solve the problem step by step, we will use the given information about the prism and apply the relevant formulas.

### Given:
- Angle of the prism, \( A = 60^\circ \)
- Refractive index of the prism for yellow light, \( \mu = \sqrt{2} \)

### Step 1: Understanding the formula
The refractive index \( \mu \) of the prism is related to the angle of the prism \( A \) and the angle of minimum deviation \( D_m \) by the formula:
\[ \mu = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

### Step 2: Calculate \( \frac{A}{2} \)
First, we calculate \( \frac{A}{2} \):
\[ \frac{A}{2} = \frac{60^\circ}{2} = 30^\circ \]

### Step 3: Calculate \( \sin\left(\frac{A}{2}\right) \)
Now, we find \( \sin\left(30^\circ\right) \):
\[ \sin\left(30^\circ\right) = \frac{1}{2} \]

### Step 4: Substitute into the refractive index formula
Now substitute \( \mu \), \( A \), and \( \sin\left(\frac{A}{2}\right) \) into the refractive index formula:
\[ \sqrt{2} = \frac{\sin\left(30^\circ + \frac{D_m}{2}\right)}{\frac{1}{2}} \]

### Step 5: Simplify the equation
This simplifies to:
\[ \sqrt{2} = 2 \sin\left(30^\circ + \frac{D_m}{2}\right) \]
\[ \sin\left(30^\circ + \frac{D_m}{2}\right) = \frac{\sqrt{2}}{2} \]

### Step 6: Find the angle corresponding to \( \sin\left(30^\circ + \frac{D_m}{2}\right) \)
The angle whose sine is \( \frac{\sqrt{2}}{2} \) is \( 45^\circ \):
\[ 30^\circ + \frac{D_m}{2} = 45^\circ \]

### Step 7: Solve for \( D_m \)
Now we can solve for \( D_m \):
\[ \frac{D_m}{2} = 45^\circ - 30^\circ = 15^\circ \]
\[ D_m = 30^\circ \]

### Step 8: Calculate angle of incidence \( i \)
The angle of incidence \( i \) at minimum deviation is given by:
\[ i = \frac{A + D_m}{2} \]
Substituting the values:
\[ i = \frac{60^\circ + 30^\circ}{2} = \frac{90^\circ}{2} = 45^\circ \]

### Step 9: Calculate angle of refraction \( r \)
At minimum deviation, the angle of refraction \( r \) is:
\[ r = \frac{A}{2} = 30^\circ \]

### Final Answers:
- Angle of minimum deviation \( D_m = 30^\circ \)
- Angle of incidence \( i = 45^\circ \)
- Angle of refraction \( r = 30^\circ \)

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