Which one of the following is/are correct relation?
I.`n_21 = 1/n_12`
II. `n_32 = n_31 xx n_12`
III. `n_21 = (n_1alpha)/(n_2alpha)`
IV. `n_21 = (n_2aalpha)/(n_1alpha)`.

Options

A. I, II and IV
B. I, III and IV
C. II, III and IV
D. All of the above

Answer

To solve the question regarding the correctness of the given relations involving refractive indices, we will analyze each statement step by step.

### Step-by-Step Solution:

1. **Understanding Refractive Indices**:
- The refractive index \( n_{21} \) represents the refractive index of medium 2 with respect to medium 1. It can be defined as:
\[ n_{21} = \frac{n_2}{n_1} \]
- Conversely, the refractive index \( n_{12} \) (medium 1 with respect to medium 2) is given by:
\[ n_{12} = \frac{n_1}{n_2} \]
- From these definitions, we can derive that:
\[ n_{21} = \frac{1}{n_{12}} \]
- **Conclusion**: Relation I: \( n_{21} = \frac{1}{n_{12}} \) is **correct**.

2. **Analyzing Relation II**:
- The relation \( n_{32} \) represents the refractive index of medium 3 with respect to medium 2, and \( n_{31} \) represents the refractive index of medium 3 with respect to medium 1.
- We can express these as:
\[ n_{32} = \frac{n_3}{n_2} \quad \text{and} \quad n_{31} = \frac{n_3}{n_1} \]
- The relation \( n_{12} \) is defined as:
\[ n_{12} = \frac{n_1}{n_2} \]
- To check if \( n_{32} = n_{31} \times n_{12} \):
\[ n_{31} \times n_{12} = \left(\frac{n_3}{n_1}\right) \times \left(\frac{n_1}{n_2}\right) = \frac{n_3}{n_2} = n_{32} \]
- **Conclusion**: Relation II: \( n_{32} = n_{31} \times n_{12} \) is **correct**.

3. **Analyzing Relation III**:
- Relation III states \( n_{21} = \frac{n_1 \alpha}{n_2 \alpha} \).
- Simplifying this gives:
\[ n_{21} = \frac{n_1}{n_2} \quad \text{(since } \alpha \text{ cancels out)} \]
- However, we previously established that \( n_{21} = \frac{n_2}{n_1} \), which contradicts this relation.
- **Conclusion**: Relation III is **incorrect**.

4. **Analyzing Relation IV**:
- Relation IV states \( n_{21} = \frac{n_2 \alpha}{n_1 \alpha} \).
- Simplifying this gives:
\[ n_{21} = \frac{n_2}{n_1} \quad \text{(since } \alpha \text{ cancels out)} \]
- This is consistent with our earlier definition of \( n_{21} \).
- **Conclusion**: Relation IV is **correct**.

### Final Answer:
The correct relations are:
- I: \( n_{21} = \frac{1}{n_{12}} \) (Correct)
- II: \( n_{32} = n_{31} \times n_{12} \) (Correct)
- III: \( n_{21} = \frac{n_1 \alpha}{n_2 \alpha} \) (Incorrect)
- IV: \( n_{21} = \frac{n_2 \alpha}{n_1 \alpha} \) (Correct)

Thus, the correct relations are I, II, and IV.