Electromagnetic Waves in a Medium

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Send light of one fixed frequency from vacuum into a block of glass. Inside the glass, the light slows down. So, what pays for this slowing down? Does the frequency change, or the wavelength, or both? Hmm, think about it. Today we will see exactly what happens to an electromagnetic wave inside a medium, and why. Questions on refractive index and wavelength in a medium come up often in JEE, so let's begin.

0:35Speed of the wave in a medium

So, first recall the vacuum. In vacuum, the speed c equals one over the square root of mu naught times epsilon naught; the tick shaped sign on screen is the square root. Mu naught, the Greek letter mu with a small zero below it, is the permeability of free space, and epsilon naught, the curly e with a small zero, is the permittivity of free space. Okay? Now put the wave inside a material, a medium. The permittivity becomes epsilon, which equals epsilon r times epsilon naught. The small r written below stands for relative, so epsilon r is the relative permittivity of the medium. In the same way, the permeability mu equals mu r times mu naught, where mu r is the relative permeability. Right? Now let me ask you this. In vacuum we had mu naught and epsilon naught under the root. Inside the medium, what do you think the speed v will be? Yes, exactly, the same form! The speed v equals one over the square root of mu times epsilon. Put in mu r times mu naught, and epsilon r times epsilon naught, and the vacuum part gives back c. So v equals c over the square root of mu r times epsilon r. Clear? Now the refractive index, n. It is defined as c over v, the speed in vacuum divided by the speed in the medium. And from the last line, n equals the square root of mu r times epsilon r. Got it? Now a special case, a non-magnetic medium. For such a medium, mu r equals one, so the refractive index is simply the square root of epsilon r. Remember that square root; we will meet a trap on it very soon. With me?

2:16Entering a medium: what changes?

Now back to our puzzle. The wave goes from vacuum into glass, and it slows down. So tell me, which one changes, the frequency or the wavelength? The frequency does not change at all! The frequency nu, the Greek letter that looks like a v, is set by the source that makes the wave, and it stays the same when the wave enters the medium. Okay? So what changes? The speed equals nu times lambda, where lambda, the Greek letter like an upside down y, is the wavelength. The speed falls and nu stays the same, so lambda must fall. The double arrow on screen means this gives: lambda equals lambda naught over n, where lambda naught is the wavelength in vacuum. Right? One more thing. In vacuum, E naught over B naught, the ratio of the electric and the magnetic field amplitudes, equals c. The small zero below each letter marks the amplitude. So inside a medium, is this ratio still c? No, and this is exactly the trap many students fall into! Inside the medium, E naught over B naught equals v, the speed in that medium, and not c. So to find the magnetic field amplitude in a medium, divide E naught by the speed v. Clear? So here is a simple way to hold these two results together. In lambda equals v over nu, and in E naught over B naught, the vacuum version had c; inside the medium, we put v, which is c over n, in its place. Both results come straight from that. With me?

3:51Example 1: relative permittivity 4, find n and speed

Now our first example. Read it with me. Here, a non-magnetic medium has relative permittivity four. Find the speed of electromagnetic waves in it, and its refractive index. Pause the video and try it first. Done? Okay, let's check. The relative permittivity is four, so the refractive index is also four, right? No, that is the trap we talked about, so be careful here! The medium is non-magnetic, so mu r equals one, and n is the square root of epsilon r. The square root of four is two, so the refractive index is two. Got it? Now the speed. The speed v equals c over n, which is three times ten to the power eight, over two. That gives one point five times ten to the power eight metres per second, the m s to the power minus one on screen. Easy, isn't it?

4:46Example 2: ferrite, 100 MHz wavelength shrinks 3 m to 3 cm

Now a harder one. Here, a ferrite has epsilon r equal to ten, and mu r equal to one thousand. By what factor is the wavelength of a one hundred megahertz wave reduced inside it, and what is the wavelength there? The capital M, capital H and small z on screen stand for megahertz. Pause the video and try it first. Done? Okay, let's check. Refractive index first. Many students write n equals the square root of epsilon r, so the square root of ten. Is that right here? No, not here, and this is a very common mistake! That shortcut works only for a non-magnetic medium, where mu r is one. Here mu r is one thousand, so we need the full formula: n equals the square root of mu r times epsilon r, the square root of one thousand times ten. That is the square root of ten to the power four, which is one hundred. With me? Now, the frequency stays the same, and lambda equals lambda naught over n. So the wavelength is reduced one hundred times. That answers the first part. Right? Now the wavelength in vacuum. Mega means ten to the power six, so one hundred megahertz is ten to the power eight hertz. Lambda naught equals c over nu, which is three times ten to the power eight, over ten to the power eight. That gives three metres; the small m on screen is metres. Okay? And inside the ferrite, lambda equals lambda naught over n, so three metres over one hundred. That is three centimetres, the c m on screen. From three metres down to three centimetres, just because of the medium! If you got both parts, very good!

6:33Quick revision: same frequency, wavelength over n, E0/B0 = v

So, quick revision. In a medium, v equals c over the square root of mu r times epsilon r, and n equals the square root of mu r times epsilon r; for a non-magnetic medium, it is just the square root of epsilon r. On entering a medium, the frequency stays the same, and the wavelength becomes lambda naught over n. And inside the medium, E naught over B naught equals v, not c. So the answer to our puzzle: the frequency stays, and the wavelength shrinks. Now go and try the practice problems on this page. All the best!