The difference in the number of waves when yellow light propagates through air and vacuum columns of

Q: Why does the number of waves change if the thickness is the same?

When light enters a denser medium like air from a vacuum, it slows down. Because its frequency must remain constant, its wavelength compresses. A shorter wavelength means more waves can fit into the exact same physical thickness!

Q: Does the frequency of light change when it enters the air?

No! Frequency is a fundamental property of the source emitting the light. It never changes when moving between mediums. Only the speed and wavelength change.

Q: Why do we use $10^{-10}$ for Angstroms?

An Angstrom ($\text{\AA}$) is a unit of length equal to $10^{-10}$ meters. It is commonly used to express the size of atoms and the wavelengths of visible light, which are extremely small.

Q: What if the light was travelling from water to glass?

The exact same principle applies! You would calculate the wavelength in water and the wavelength in glass using their respective refractive indices, and then find the difference in the number of waves based on those new wavelengths.

Q: Can the difference in the number of waves be a fraction?

Absolutely! In real-world interferometry, the path difference often results in a fractional number of waves, which manifests as a phase shift and creates beautiful interference patterns. In this specific problem, it was given as exactly one for simplicity.

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.
[Take, refractive index of air $=1.0003$, wavelength of yellow light in vacuum $=6000	ext{\AA}$]

Sigma Percentile · Accepted Answer

Have you ever wondered how light experiences the world? To us, a millimeter is just a tiny mark on a ruler. But to a wave of light, a millimeter is a vast journey, and the medium it travels through changes everything. Let's dive into a fascinating problem that explores exactly this!

### The Setup: Racing Through Different Worlds

Imagine a beam of yellow light. We split this beam and send it racing through two identical tunnels, both having the exact same thickness, $t$.

One tunnel is a perfect vacuum—empty space where light travels at its absolute maximum speed. The other tunnel is filled with air. Now, air might seem invisible to us, but to light, it's an obstacle course. The refractive index of air is $\mu = 1.0003$. This tiny decimal means light slows down just a tiny bit when it enters the air.

### The Mathematics of Waves

When light slows down, its frequency remains constant, but its wavelength compresses. 
The number of waves $n$ that can fit into a given thickness $t$ is simply the thickness divided by the length of a single wave:
$$n = \frac{t}{\lambda}$$

For the vacuum column, the number of waves is:
$$n_{vac} = \frac{t}{\lambda_{vac}}$$

For the air column, the wavelength is shorter ($\lambda_{air} = \frac{\lambda_{vac}}{\mu_{air}}$). Because the waves are squished together, more of them can fit into the same space!
$$n_{air} = \frac{t}{\lambda_{air}} = \frac{t \cdot \mu_{air}}{\lambda_{vac}}$$

### The Master Equation

The problem gives us a beautiful constraint: the air column fits exactly **one** more wave than the vacuum column. 
$$n_{air} - n_{vac} = 1$$

Let's substitute our expressions into this master equation:
$$\frac{t \cdot \mu_{air}}{\lambda_{vac}} - \frac{t}{\lambda_{vac}} = 1$$

Factoring out the common terms, we get:
$$\frac{t}{\lambda_{vac}} (\mu_{air} - 1) = 1$$

Now, we can easily isolate the thickness $t$:
$$t = \frac{\lambda_{vac}}{\mu_{air} - 1}$$

### The Final Calculation

It's time to plug in the numbers! We are given the vacuum wavelength $\lambda_{vac} = 6000 	ext{ \AA} = 6000 	imes 10^{-10} 	ext{ m}$, and the refractive index $\mu_{air} = 1.0003$.

$$t = \frac{6000 	imes 10^{-10}}{1.0003 - 1}$$
$$t = \frac{6000 	imes 10^{-10}}{0.0003}$$

Let's simplify the denominator to $3 	imes 10^{-4}$:
$$t = \frac{6000 	imes 10^{-10}}{3 	imes 10^{-4}}$$
$$t = 2000 	imes 10^{-6} 	ext{ m}$$

Converting this to millimeters, we get our final answer:
$$t = 2 	imes 10^{-3} 	ext{ m} = \mathbf{2 	ext{ mm}}$$

Isn't it amazing? A difference of just one single wave over a distance of 2 millimeters allows us to measure the refractive index of air! This very principle is the heart of interferometry, a technique used to measure gravitational waves and the expansion of the universe.

Enter Numerical Value:

Visualized Solution

$t$ Thickness Columns

Number of Waves

Wavelength in Medium

Given Condition

Substitution

Factoring

Isolating $t$

Calculation

Final Answer

Conclusion

The Sigma Insight: Refraction and Total Internal Reflection

The Setup

The Mathematics of Waves

The Master Equation

The Final Calculation

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