`""_(i)u_(j)` represents refractive index when a light ray goes from medium I to medium j, then the product `""_(2)µ_(1)xx``""_(3)µ_(2)xx``""_(4)µ_(3)` is equal to

Options

A. `""_(3)µ_(1)`
B. `""_(3)µ_(2)`
C. `(1)/(""_(1)µ_(4))`
D. `""_(4)µ_(2)`

Explanation

To solve the problem, we need to analyze the refractive indices given in the question. The refractive index \( \mu_{ij} \) is defined as the ratio of the refractive index of medium \( j \) to that of medium \( i \).

Let's break down the problem step by step:

### Step 1: Understand the definition of refractive index
The refractive index \( \mu_{ij} \) is defined as:
\[ \mu_{ij} = \frac{n_j}{n_i} \]
where \( n_j \) is the refractive index of medium \( j \) and \( n_i \) is the refractive index of medium \( i \).

### Step 2: Write down the refractive indices for the given products
We need to find the product \( \mu_{21} \mu_{32} \mu_{43} \):
- \( \mu_{21} = \frac{n_2}{n_1} \)
- \( \mu_{32} = \frac{n_3}{n_2} \)
- \( \mu_{43} = \frac{n_4}{n_3} \)

### Step 3: Substitute the definitions into the product
Now, substituting these definitions into the product:
\[ \mu_{21} \mu_{32} \mu_{43} = \left(\frac{n_2}{n_1}\right) \left(\frac{n_3}{n_2}\right) \left(\frac{n_4}{n_3}\right) \]

### Step 4: Simplify the expression
When we multiply these fractions, we notice that \( n_2 \) in the numerator of \( \mu_{21} \) and the denominator of \( \mu_{32} \) cancels out, and \( n_3 \) in the numerator of \( \mu_{32} \) and the denominator of \( \mu_{43} \) also cancels out:
\[ \mu_{21} \mu_{32} \mu_{43} = \frac{n_4}{n_1} \]

### Step 5: Relate the result to the refractive index
The expression \( \frac{n_4}{n_1} \) can be rewritten in terms of the refractive index \( \mu_{14} \):
\[ \frac{n_4}{n_1} = \mu_{14} \]

### Conclusion
Thus, we conclude that:
\[ \mu_{21} \mu_{32} \mu_{43} = \mu_{14} \]

### Final Answer
The product \( \mu_{21} \mu_{32} \mu_{43} \) is equal to \( \mu_{14} \).