A ray of light suffers a minimum deviation when incident on equilateral prism of refractive index `sqrt2` The angle of deviation is

Options

A. `30^@`
B. `45^@`
C. `60^@`
D. `50^@`

Answer

To solve the problem of finding the angle of deviation when a ray of light passes through an equilateral prism with a refractive index of \(\sqrt{2}\), we can follow these steps:

### Step 1: Understand the Problem
We are given:
- The prism is equilateral, which means the angle \(A\) of the prism is \(60^\circ\).
- The refractive index \(\mu\) is \(\sqrt{2}\).
- We need to find the minimum angle of deviation \(\Delta_m\).

### Step 2: Use the Formula for Minimum Deviation
For a prism, the relationship between the refractive index \(\mu\), the angle of the prism \(A\), and the minimum angle of deviation \(\Delta_m\) is given by the formula:
\[ \mu = \frac{\sin\left(\frac{A + \Delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

### Step 3: Substitute Known Values
Since \(A = 60^\circ\), we can substitute this value into the formula:
\[ \mu = \frac{\sin\left(\frac{60^\circ + \Delta_m}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} \]
This simplifies to:
\[ \mu = \frac{\sin\left(\frac{60^\circ + \Delta_m}{2}\right)}{\sin(30^\circ)} \]

### Step 4: Calculate \(\sin(30^\circ)\)
We know that:
\[ \sin(30^\circ) = \frac{1}{2} \]
Thus, the equation becomes:
\[ \mu = 2 \cdot \sin\left(\frac{60^\circ + \Delta_m}{2}\right) \]

### Step 5: Substitute the Value of \(\mu\)
Now, substituting \(\mu = \sqrt{2}\):
\[ \sqrt{2} = 2 \cdot \sin\left(\frac{60^\circ + \Delta_m}{2}\right) \]

### Step 6: Solve for \(\sin\left(\frac{60^\circ + \Delta_m}{2}\right)\)
Dividing both sides by 2 gives:
\[ \sin\left(\frac{60^\circ + \Delta_m}{2}\right) = \frac{\sqrt{2}}{2} \]
We know that \(\frac{\sqrt{2}}{2} = \sin(45^\circ)\), so:
\[ \frac{60^\circ + \Delta_m}{2} = 45^\circ \]

### Step 7: Solve for \(\Delta_m\)
Multiplying both sides by 2:
\[ 60^\circ + \Delta_m = 90^\circ \]
Subtracting \(60^\circ\) from both sides gives:
\[ \Delta_m = 30^\circ \]

### Conclusion
The minimum angle of deviation \(\Delta_m\) is \(30^\circ\).