A light ray going through a prism with the angle of prism `60^@,` is founded to deviate at least by `30^@.` what Is the range of the refractive index of the prism?

Explanation

To solve the problem, we need to find the range of the refractive index of a prism given that the angle of the prism (A) is 60° and the light ray deviates at least by 30° (Δm).

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Angle of prism, \( A = 60^\circ \)
- Minimum deviation, \( \Delta_m \geq 30^\circ \)

2. **Use the Formula for Refractive Index**:
The relationship between the refractive index (n), the angle of the prism (A), and the minimum deviation (Δm) is given by:
\[ n = \frac{\sin\left(\frac{A + \Delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

3. **Substitute the Known Values**:
We substitute \( A = 60^\circ \) into the formula:
\[ n = \frac{\sin\left(\frac{60^\circ + \Delta_m}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} = \frac{\sin\left(\frac{60^\circ + \Delta_m}{2}\right)}{\sin(30^\circ)} \]
Since \( \sin(30^\circ) = \frac{1}{2} \), we can simplify:
\[ n = 2 \sin\left(\frac{60^\circ + \Delta_m}{2}\right) \]

4. **Consider the Minimum Deviation Condition**:
Since the problem states that the deviation is at least 30°, we can analyze the case when \( \Delta_m = 30^\circ \):
\[ n = 2 \sin\left(\frac{60^\circ + 30^\circ}{2}\right) = 2 \sin\left(\frac{90^\circ}{2}\right) = 2 \sin(45^\circ) \]
Knowing that \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \):
\[ n = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2} \]

5. **Determine the Range of Refractive Index**:
Since the deviation can be greater than 30°, the refractive index must be less than or equal to \( \sqrt{2} \). Therefore, the range of the refractive index is:
\[ n \leq \sqrt{2} \]

### Final Answer:
The range of the refractive index of the prism is:
\[ n \leq \sqrt{2} \]