Question

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, wavelength of yellow light in vacuum = 6000 Å]

Hint:

The concept of optical path length is useful here. The optical path is $n \times t$. The difference in the number of waves is related to the difference in optical path length: $\Delta N = \frac{\Delta(\text{Optical Path})}{\lambda_{vac}} = \frac{t(n_{air}-n_{vac})}{\lambda_{vac}}$. Since $n_{vac}=1$, this simplifies to the formula used.

Answer 1

Correct Answer: 2

Let the thickness of the column be $t$.
The number of waves ($N$) in a medium of thickness $t$ is given by $N = \frac{t}{\lambda}$, where $\lambda$ is the wavelength in that medium.
In vacuum, the number of waves is $N_{vac} = \frac{t}{\lambda_{vac}}$.
In air, the wavelength changes to $\lambda_{air} = \frac{\lambda_{vac}}{n_{air}}$, where $n_{air}$ is the refractive index of air.
The number of waves in air is $N_{air} = \frac{t}{\lambda_{air}} = \frac{t \cdot n_{air}}{\lambda_{vac}}$.
The problem states that the difference in the number of waves is one.
$N_{air} - N_{vac} = 1$.
Substitute the expressions for $N_{air}$ and $N_{vac}$:
$\frac{t \cdot n_{air}}{\lambda_{vac}} - \frac{t}{\lambda_{vac}} = 1$.
Factor out $\frac{t}{\lambda_{vac}}$:
$\frac{t}{\lambda_{vac}} (n_{air} - 1) = 1$.
Solve for the thickness $t$:
$t = \frac{\lambda_{vac}}{n_{air} - 1}$.
Now, substitute the given values, ensuring consistent units.
$\lambda_{vac} = 6000$ Å = $6000 \times 10^{-10}$ m.
$n_{air} = 1.0003$.
$t = \frac{6000 \times 10^{-10} \text{ m}}{1.0003 - 1} = \frac{6 \times 10^3 \times 10^{-10}}{0.0003} = \frac{6 \times 10^{-7}}{3 \times 10^{-4}}$.
$t = 2 \times 10^{-7 - (-4)} = 2 \times 10^{-3}$ m.
The question asks for the thickness in millimeters (mm).
$1 \text{ mm} = 10^{-3} \text{ m}$.
Therefore, $t = 2$ mm.