A light ray, going through a prism with the angle of prism `60^@`, is found to deviate by ` 30^@`. What limit on the refractive index can be put from these data ?

Explanation

To find the limit on the refractive index of a prism given the angle of the prism and the angle of deviation, we can follow these steps:

### Step-by-Step Solution:

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

2. **Understand the Formula for Minimum Refractive Index**:
The formula for the minimum refractive index \( \mu \) of a prism is given by:
\[ \mu_{\text{min}} = \frac{\sin\left(\frac{A + \delta_{\text{min}}}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
where \( \delta_{\text{min}} \) is the minimum angle of deviation.

3. **Determine the Relationship of Deviation**:
Since the deviation given is \( 30^\circ \), we know that the minimum deviation \( \delta_{\text{min}} \) can be equal to or less than \( 30^\circ \). Therefore:
\[ \delta_{\text{min}} \leq 30^\circ \]

4. **Calculate the Sine Values**:
- Calculate \( \frac{A}{2} = \frac{60^\circ}{2} = 30^\circ \)
- Thus, \( \sin\left(\frac{A}{2}\right) = \sin(30^\circ) = \frac{1}{2} \)

5. **Substitute the Maximum Deviation**:
For the maximum limit of the refractive index, we can substitute \( \delta_{\text{min}} = 30^\circ \) into the formula:
\[ \mu_{\text{min}} = \frac{\sin\left(\frac{60^\circ + 30^\circ}{2}\right)}{\sin(30^\circ)} \]

6. **Calculate the Sine of the Sum**:
- Calculate \( \frac{60^\circ + 30^\circ}{2} = \frac{90^\circ}{2} = 45^\circ \)
- Thus, \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \)

7. **Final Calculation**:
Substitute the sine values into the equation:
\[ \mu_{\text{min}} = \frac{\sin(45^\circ)}{\sin(30^\circ)} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

8. **Conclusion**:
Therefore, the limit on the refractive index \( \mu \) is:
\[ \mu \leq \sqrt{2} \]