100 Nano-Stories: Refractive Index!

Episode #81: Transparency x Trigonometry x Density!

Carlos Manuel Jarquín Sánchez
7 min readApr 25, 2021

--

Preface! ✨

It’s your favorite material science & nanotechnology enthusiast! Today, I want to talk more about geometrical optics and the refractive index of aerogels!

There is no article this time to prepare you, reader! We will directly go into today’s topic! 😄

Refractive Index Explained! 💡

The Refractive Index is the reduction of speed of the light and the bending of the light as it passes from one material to another.

To explain the refractive index, let’s take a look at this photo below:

When light enters the aerogel, it will come at an angle. This is known as the angle of incidence (sin θ₁).

The incident light is the light that is going to enter a material.

But because the light is about to enter a different material, the light will bend at an angle.

The angle of the bending of the light after it enters the new material is the angle of refraction (sin θ₂).

Concepts of Air x Aerogel! 🔑

The Refractive Index of Air is about 1.0029, while the Refractive Index of Aerogel is approximately 1.0026 (thanks to low-density hydrophobic aerogels of 0.01 g/cm³.)

Authors Note:

1.0026 for a refractive index in aerogels is quite low, but that is not the normal range for common silica aerogel. The normal range for the refraction index in an aerogel is between 1.005 - 1.25. For today’s purposes, we will use the lowest refractve idnex value calculated (1.0026). 😃

So in this case, the refractive index is going from a “higher value” (air) to a “lower value” (aerogel).

If the refractive index moves from a higher value to a lower angle, the ray of light is going to bend farther away from the normal line. If the refractive index moves from a lower value to a higher angle, the ray of light is going to bend closer to the normal line.

For example, if the angle of the incident ray of light is 15°, then the angle of the refracted ray of light will be greater than 15°.

If it helps, the normal line in the refractive index is basically where the light from the air comes into contact with the surface of the aerogel. This “normal line” is also perpendicular to the contact surface between the air and the aerogel.

But how do we calculate the index of refraction in an aerogel? 🤔

Snell’s Law x Prism Method! 🔑

We can use Snell’s Law to calculate the Refractive Index, but in the case of the air and aerogels, we can solve the refractive index using the Prism method.

I will explain Snell’s Law for you to understand how to solve the refractive index with The Prism Method, reader!

The equation is listed below as follows:

Before I discuss what all the symbols mean, I want to talk about what arcsin means. The term arcsin is the inverse of the sine (sin) function/number. arcsin can also be labeled as sin-1.

The domain in arcsin is -1 to +1, and the range of arcsin is -π/2 to π/2.

But now that we know what arcsin means, we can label the refractive index symbols:

θ = The Angle of deflection.

ϕ = The angle of incidence (the light) to the surface of the aerogel.

n = Refractive Index of The Silica Aerogel.

∝ = The angle between the two adjacent sides of the aerogel. Monolithic aerogels mainly have 90° angles, so we can assume the angle of the adjacent sides to be 90°.

Thankfully, we have some numbers to work with to try and solve the angle of deflection in an aerogel. In this case, ϕ will have to be made up for today’s purposes.

ϕ → 10°

∝ → 90°

n ?

So it looks like we need to find the refractive index of the aerogel! But we cannot use Shell’s Law because the aerogel’s density contains the answer we are looking for! So how can we calculate the refractive index?

Calculating Refractive Index In Aerogel! 🔑

From defining the equation, we are lacking the number for the refractive index of aerogels.

k → A coefficient that depends on the light wavelength.

ρ → The density of the aerogel.

The “1” is a bit complicated to explain, but I will tell you as plainly as I can, reader. 😄

1 is the refractive index when light travels in a vacuum. When light travels in a vacuum, it is traveling at the speed of light, which is around 299, 792, 458 (300 million) meters per second (m/s).

But in an aerogel, there are more molecules and particles (despite the aerogel being around 99.8% air by volume), so this means we do have to calculate the speed of the light as it travels through the aerogel and the refractive index will be greater than 1.

Since the low density of the aerogel is 0.01 g/cm³ and the wavelengths of silica aerogel are normally blue light, we can calculate the refractive index of low-density aerogels to be 1.0026.

n → 1.0026

Density is equal to the mass per unit volume of a material/substance. (m / v)

Calculating The Speed Of Light In An Aerogel! 🔑

Now that we have our refractive index for aerogels (1.0026), we can calculate how fast the light travels in an aerogel!

We can find out by dividing the speed of light by the refractive index.

299,792,458 m /s ÷ 1.0026 = 299,015,018.951 m/s.

The speed of light in a silica aerogel with a refractive index of 1.0026 is 299,015,018.951 m/s.

This speed of light and the wavelength will decrease as the refractive index number increases.

But hold up, what do you mean by the wavelength? 🤔

Calculating Wavelength! 🔑

To calculate the wavelength that is in the aerogel, we can calculate the incident wavelength divided by the refractive index of the aerogel/material.

In this example, λ₀ will be the wavelength of sunlight, which is about 500 nanometers.

The reason we will use sunlight is that sunlight is the biggest source of light that can hit an aerogel, especially if the aerogel sample is outside. So ideally, the wavelength of sunlight is the best way to go. 😃

λₙ = λ₀ ÷ n

500 nm ÷ 1.0026 ≈ 498.7 nm.

The wavelength of light that passes through an aerogel with a refractive index of 1.0026 is around 498.7 nm, which is the highest range of blue light. The wavelength of blue light is around 450 nm - 495 nm.

Bonus:

However, for aerogels with a refractive index of 1.26, the wavelength will be around 396 nm, which means that this wavelength of light is in the Ultraviolet (UV) Spectrum. This means that we have UV light inside of the aerogel.

Final Points! 🔑

Since the incident angle was 10°, and because the refractive index of aerogel was 1.0026, I was able to calculate the results on my calculator, and in the end, I was able to get the result for the deflection angle in the aerogel (measured in degrees, of course!)

This is the original equation, and below it is the calculator result:

This means that θ (The Angle of deflection) is equal to 0.912307. This makes sense because most aerogels have a higher refraction index than air. When light travels from a low refractive index to a high refractive index, the angle will decrease and tend to be closer to the normal line (line of contact between the air & aerogel.)

Closing Thoughts! 💭

However, even with such low refraction indexes as 1.0026, there is another way to decrease the refraction index in aerogels.

We can further decrease the refraction index by decreasing the density in the aerogels to low density as 0.03 g/cm³ or lower.

But with such low density and refraction indexes, there come the optical properties and how they all connect!

See you tomorrow for the finalization on Optical Properties of Aerogel! 👋🏽

Vocabulary! 📓

Refractive IndexThe reduction of speed of the light and the bending of the light as it passes from one material to another.

Density Mass per unit volume of a material/substance. (m / v)

Incident Light → The light that is going to enter a material.

Speed Of Light → 299,015,018.951 m/s

arcsin → The inverse of sine (sin); labeled as sin-1.

Normal LineThe light from the air comes into contact with the surface of the aerogel.

Connect →🔗

LinkedIn

Twitter

Personal Newsletter

cjarquin0005@gmail.com

© 2021 by Carlos Manuel Jarquin Sanchez. All Rights Reserved.

--

--