The wave of wavelength `5900 Å` emitted by any atom or molecule must have some finite total length which known as coherence length. For sodium light, this length is `2.4cm`. The number of oscillation in this length will be,

Options

A. (a)`4.068xx18^(8)`
B. (b)`4.068xx10^(4)`
C. (c)`4.068xx10^(6)`
D. (d)`4.068xx10^(5)`

Answer

To solve the problem of finding the number of oscillations in the coherence length of sodium light, we can follow these steps:

### Step 1: Understand the relationship
The number of oscillations (N) in a given length (L) can be calculated using the formula:
\[ N = \frac{L}{\lambda} \]
where \(L\) is the coherence length and \(\lambda\) is the wavelength.

### Step 2: Convert the given values
We need to convert the given coherence length and wavelength into compatible units (meters).

- Coherence length \(L = 2.4 \, \text{cm} = 2.4 \times 10^{-2} \, \text{m}\)
- Wavelength \(\lambda = 5900 \, \text{Å} = 5900 \times 10^{-10} \, \text{m} = 5.9 \times 10^{-7} \, \text{m}\)

### Step 3: Substitute the values into the formula
Now substitute the values of \(L\) and \(\lambda\) into the formula for \(N\):
\[ N = \frac{2.4 \times 10^{-2} \, \text{m}}{5.9 \times 10^{-7} \, \text{m}} \]

### Step 4: Perform the calculation
Calculating the above expression:
\[ N = \frac{2.4 \times 10^{-2}}{5.9 \times 10^{-7}} \approx 4.068 \times 10^{4} \]

### Step 5: Conclusion
Thus, the number of oscillations in the coherence length of sodium light is approximately \(4.068 \times 10^{4}\).

### Final Answer
The answer is \(N \approx 4.068 \times 10^{4}\).