1. Given data

• Refractive index of air: $\mu = 1.0003$
• Wavelength in vacuum: $\lambda_0 = 6000\,\mathring{A} = 6000 \times 10^{-10}\,\text{m} = 6 \times 10^{-7}\,\text{m}$
• Let thickness of the column be $t$.

2. Number of waves in a medium

If light travels a distance $t$, the number of wavelengths contained is: $N = \frac{t}{\lambda}$

• In vacuum: $N_v = \frac{t}{\lambda_0}$

• In air, wavelength becomes: $\lambda_{air} = \frac{\lambda_0}{\mu}$ so number of waves is $N_{air} = \frac{t}{\lambda_{air}} = \frac{t}{\lambda_0/\mu} = \frac{\mu t}{\lambda_0}$

3. Use the given condition

The difference in number of waves is 1: $N_{air} - N_v = 1$

Substitute: $\frac{\mu t}{\lambda_0} - \frac{t}{\lambda_0} = 1$ $\frac{t}{\lambda_0}(\mu - 1) = 1$

Hence, $t = \frac{\lambda_0}{\mu - 1}$

4. Substitute values

$t = \frac{6 \times 10^{-7}}{1.0003 - 1}$ $t = \frac{6 \times 10^{-7}}{3 \times 10^{-4}}$ $t = 2 \times 10^{-3}\,\text{m}$

5. Convert to mm

$2 \times 10^{-3}\,\text{m} = 2\,\text{mm}$

6. Final answer

The thickness of the air column is: $\boxed{2\,\text{mm}}$