The difference in the number of waves when yellow light propagates through air and vacuum columns of the same thickness is one. The thickness of the air column is _________ mm. [Refractive index of air = 1.0003, the wavelength of yellow light in vacuum = 6000 Å]

Let the common thickness of the two columns be $$t$$.

For a light beam, the number of complete waves contained in a distance is obtained by dividing the distance by the wavelength in that medium. Hence, in a column of thickness $$t$$:

$$\text{Number of waves in vacuum} = N_{\text{vac}} = \dfrac{t}{\lambda_0}$$

where $$\lambda_0$$ is the wavelength of yellow light in vacuum. We are given

$$\lambda_0 = 6000\;\text{\AA} = 6000 \times 10^{-10}\;\text{m} = 6.0 \times 10^{-7}\;\text{m}.$$

Inside air the speed (and hence the wavelength) changes according to the refractive index. First we recall the basic relationship

$$n = \dfrac{\text{speed in vacuum}}{\text{speed in medium}} = \dfrac{\lambda_0}{\lambda_{\text{air}}}.$$

Re-arranging for the wavelength in air, we have

$$\lambda_{\text{air}} = \dfrac{\lambda_0}{n}.$$

The refractive index of air is supplied as $$n = 1.0003$$, so the wavelength in air becomes

$$\lambda_{\text{air}} = \dfrac{6.0 \times 10^{-7}\;\text{m}}{1.0003}.$$

Next we write the number of waves in the air column:

$$\text{Number of waves in air} = N_{\text{air}} = \dfrac{t}{\lambda_{\text{air}}} = \dfrac{t}{\lambda_0/n} = \dfrac{n\,t}{\lambda_0}.$$

The statement in the question says that the difference between the two wave counts equals one:

$$|N_{\text{air}} - N_{\text{vac}}| = 1.$$

Because air’s wavelength is slightly shorter, $$N_{\text{air}} > N_{\text{vac}}$$. Therefore we set

$$N_{\text{air}} - N_{\text{vac}} = 1.$$

Substituting the expressions we have derived:

$$\dfrac{n\,t}{\lambda_0} - \dfrac{t}{\lambda_0} = 1.$$

Factorising $$t/\lambda_0$$ gives

$$\dfrac{t}{\lambda_0}\,(n - 1) = 1.$$

Now we solve for the thickness $$t$$:

$$t = \dfrac{\lambda_0}{\,n - 1\,}.$$

Putting the numerical values,

$$t = \dfrac{6.0 \times 10^{-7}\;\text{m}}{1.0003 - 1} = \dfrac{6.0 \times 10^{-7}\;\text{m}}{0.0003}.$$

The division gives

$$t = 2.0 \times 10^{-3}\;\text{m}.$$

Converting metres to millimetres (since $$1\;\text{mm} = 10^{-3}\;\text{m}$$) we have

$$t = 2.0\;\text{mm}.$$

So, the answer is $$2\;\text{mm}$$.