Refractive index of air is `1.0003.` The correct thickness of air column which will have one more wavelength of yellow light `(6000 Å)` than in the same thickness in vacuum is

Options

A. `2 mm`
B. `2 cm`
C. `2m`
D. `2 km`

Explanation

To solve the problem, we need to find the correct thickness of the air column that will have one more wavelength of yellow light (6000 Å) than in the same thickness in vacuum.

### Step-by-Step Solution:

1. **Understand the relationship between wavelength, refractive index, and thickness**:
- The number of wavelengths in a medium is given by the formula:
\[ N = \frac{T}{\lambda} \]
- Where \( N \) is the number of wavelengths, \( T \) is the thickness of the medium, and \( \lambda \) is the wavelength of light in that medium.

2. **Define the variables**:
- Let \( T \) be the thickness of the air column.
- The wavelength of yellow light in vacuum is given as \( \lambda_V = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} \).
- The refractive index of air is given as \( \mu = 1.0003 \).

3. **Calculate the wavelength of light in air**:
- The wavelength of light in air (\( \lambda_A \)) can be calculated using the formula:
\[ \lambda_A = \frac{\lambda_V}{\mu} \]
- Substituting the values:
\[ \lambda_A = \frac{6000 \times 10^{-10}}{1.0003} \]

4. **Calculate the number of wavelengths in air and vacuum**:
- The number of wavelengths in air (\( N_A \)) is:
\[ N_A = \frac{T}{\lambda_A} \]
- The number of wavelengths in vacuum (\( N_V \)) is:
\[ N_V = \frac{T}{\lambda_V} \]

5. **Set up the equation based on the problem statement**:
- According to the problem, the difference in the number of wavelengths is 1:
\[ N_A - N_V = 1 \]
- Substituting the expressions for \( N_A \) and \( N_V \):
\[ \frac{T}{\lambda_A} - \frac{T}{\lambda_V} = 1 \]

6. **Factor out \( T \)**:
- Rearranging gives:
\[ T \left( \frac{1}{\lambda_A} - \frac{1}{\lambda_V} \right) = 1 \]
- Thus:
\[ T = \frac{1}{\frac{1}{\lambda_A} - \frac{1}{\lambda_V}} \]

7. **Substitute \( \lambda_A \) into the equation**:
- We know \( \lambda_A = \frac{\lambda_V}{\mu} \):
\[ T = \frac{1}{\frac{1}{\frac{\lambda_V}{\mu}} - \frac{1}{\lambda_V}} \]
- Simplifying this gives:
\[ T = \frac{\lambda_V \cdot \mu}{1 - \mu} \]

8. **Substituting the values**:
- Substitute \( \lambda_V = 6000 \times 10^{-10} \) m and \( \mu = 1.0003 \):
\[ T = \frac{6000 \times 10^{-10} \cdot 1.0003}{1 - 1.0003} \]
- This simplifies to:
\[ T = \frac{6000 \times 10^{-10} \cdot 1.0003}{-0.0003} \]

9. **Calculate \( T \)**:
- Calculate the value:
\[ T \approx \frac{6000 \times 10^{-10} \cdot 1.0003}{-0.0003} \approx 2 \times 10^{-3} \, \text{m} = 2 \, \text{mm} \]

### Final Answer:
The correct thickness of the air column is \( 2 \, \text{mm} \).