For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index

Options

A. lies between 2 and `sqrt2`
B. is less than 1
C. is greater than 2
D. lies between`sqrt `2 and 1

Explanation

To solve the problem, we need to find the refractive index (μ) of a prism for which the angle of minimum deviation (D) is equal to its refracting angle (A).

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to establish the relationship between the refractive index (μ), the angle of the prism (A), and the angle of minimum deviation (D). The condition given is that D = A.

2. **Using the Formula for Minimum Deviation**:
The formula relating the refractive index (μ) to the angle of the prism (A) and the angle of minimum deviation (D) is given by:
\[ \mu = \frac{\sin\left(\frac{A + D}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
Since we are given that D = A, we can substitute D with A in the formula.

3. **Substituting D = A**:
Replacing D with A in the formula gives us:
\[ \mu = \frac{\sin\left(\frac{A + A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin(A)}{\sin\left(\frac{A}{2}\right)} \]

4. **Using the Double Angle Identity**:
We can use the double angle identity for sine:
\[ \sin(A) = 2 \sin\left(\frac{A}{2}\right) \cos\left(\frac{A}{2}\right) \]
Substituting this into our equation for μ gives:
\[ \mu = \frac{2 \sin\left(\frac{A}{2}\right) \cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

5. **Simplifying the Expression**:
The \(\sin\left(\frac{A}{2}\right)\) terms cancel out (assuming \(A \neq 0\)):
\[ \mu = 2 \cos\left(\frac{A}{2}\right) \]

6. **Finding the Range of μ**:
The angle A can vary from 0 to 90 degrees. Therefore, \(\frac{A}{2}\) will vary from 0 to 45 degrees. The cosine function decreases from 1 to 0 in this range:
- When \(A = 0\), \(\mu = 2 \cos(0) = 2\)
- When \(A = 90\), \(\mu = 2 \cos(45) = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}\)

7. **Conclusion**:
Therefore, the refractive index μ lies between \(\sqrt{2}\) and 2:
\[ \sqrt{2} < \mu < 2 \]

### Final Answer:
The prism must be made of a material whose refractive index (μ) lies between \(\sqrt{2}\) and 2.