The angle of a prism is `60^(@)` and its refractive index is `sqrt(2)`. The angle of minumum deviation sufferred by a ray of light in passing through it is

Options

A. About `20^@`
B. `30^@`
C. `60^@`
D. `45^@`

Answer

To find the angle of minimum deviation (Δm) for a prism with an angle of 60° and a refractive index (μ) of √2, we can use the formula for minimum deviation in a prism:

\[ \mu = \frac{\sin\left(\frac{A + \Delta m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

where:
- μ is the refractive index,
- A is the angle of the prism,
- Δm is the angle of minimum deviation.

### Step 1: Identify the given values
- Angle of the prism (A) = 60°
- Refractive index (μ) = √2

### Step 2: Calculate A/2
\[ \frac{A}{2} = \frac{60°}{2} = 30° \]

### Step 3: Substitute values into the formula
Substituting the known values into the formula:
\[ \sqrt{2} = \frac{\sin\left(\frac{60° + \Delta m}{2}\right)}{\sin(30°)} \]

### Step 4: Calculate sin(30°)
We know that:
\[ \sin(30°) = \frac{1}{2} \]

### Step 5: Rewrite the equation
Now substituting sin(30°) into the equation:
\[ \sqrt{2} = \frac{\sin\left(\frac{60° + \Delta m}{2}\right)}{\frac{1}{2}} \]

This simplifies to:
\[ \sqrt{2} = 2 \cdot \sin\left(\frac{60° + \Delta m}{2}\right) \]

### Step 6: Solve for sin(60° + Δm/2)
Rearranging gives:
\[ \sin\left(\frac{60° + \Delta m}{2}\right) = \frac{\sqrt{2}}{2} \]

### Step 7: Determine the angle
We know that:
\[ \sin(45°) = \frac{\sqrt{2}}{2} \]
Thus:
\[ \frac{60° + \Delta m}{2} = 45° \]

### Step 8: Solve for Δm
Multiplying both sides by 2:
\[ 60° + \Delta m = 90° \]

Now, subtracting 60° from both sides:
\[ \Delta m = 90° - 60° = 30° \]

### Conclusion
The angle of minimum deviation (Δm) is:
\[ \Delta m = 30° \]