A ray of light passing through a prism having refractive index `sqrt(2)` suffers minimum devitation. It is found that the angle of incidence is double the angle of refraction within the prism. Then angle of prism is

Options

A. `45^(@)`
B. `60^(@)`
C. `75^(@)`
D. `90^(@)`

Answer

To solve the problem step by step, we will use the information provided about the prism, the refractive index, and the relationship between the angle of incidence and the angle of refraction.

### Step 1: Understand the given information
We know that:
- The refractive index \( n = \sqrt{2} \)
- The angle of incidence \( i \) is double the angle of refraction \( r \), i.e., \( i = 2r \).

### Step 2: Use the formula for refractive index
The formula for refractive index is given by:
\[ n = \frac{\sin i}{\sin r} \]
Substituting \( i = 2r \) into the equation, we get:
\[ n = \frac{\sin(2r)}{\sin r} \]

### Step 3: Apply the double angle formula
Using the double angle formula for sine, we have:
\[ \sin(2r) = 2 \sin r \cos r \]
Substituting this into the refractive index equation, we get:
\[ n = \frac{2 \sin r \cos r}{\sin r} \]
This simplifies to:
\[ n = 2 \cos r \]

### Step 4: Substitute the value of the refractive index
Now, substituting \( n = \sqrt{2} \) into the equation:
\[ \sqrt{2} = 2 \cos r \]

### Step 5: Solve for \( \cos r \)
Rearranging the equation gives:
\[ \cos r = \frac{\sqrt{2}}{2} \]

### Step 6: Find the angle \( r \)
The value of \( \cos r = \frac{\sqrt{2}}{2} \) corresponds to:
\[ r = 45^\circ \]

### Step 7: Use the condition for minimum deviation
At minimum deviation, the angle of refraction \( r \) is related to the angle of the prism \( A \) by the formula:
\[ r = \frac{A}{2} \]
Substituting \( r = 45^\circ \):
\[ 45^\circ = \frac{A}{2} \]

### Step 8: Solve for the angle of the prism \( A \)
Multiplying both sides by 2 gives:
\[ A = 90^\circ \]

### Final Answer
The angle of the prism is \( 90^\circ \).

---