JEE Advanced, AIEEE and Earlier JEE Main Questions: Energy

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Light has no rest mass. Yet when a short pulse of light is absorbed by a small object at rest, the object starts to move. So where does that push come from? Today we solve past questions on the energy of electromagnetic waves, from JEE Advanced and the earlier JEE Main papers. They come up often, so let's begin.

0:27Energy density: an equal share

So, the letter u is the energy density, the energy per unit volume. The u with a small E below it is the electric part, one half epsilon naught E squared, where epsilon naught, the Greek epsilon with a small zero, is the permittivity of free space. The u with a small B is the magnetic part, B squared over two mu naught, with mu naught, the Greek mu with a small zero, the permeability of free space. Okay? Now, JEE Main twenty fourteen asked this. When an electromagnetic wave travels in a medium, how do its electric and magnetic energy densities compare? They are equal, yes, exactly! u E equals u B: even in a medium, the energy is shared equally between the two fields. Let us see why. Right? In a medium, epsilon naught becomes epsilon, and mu naught becomes mu. So u E is one half epsilon E squared, and u B is B squared over two mu, with E equals v times B, where v is the speed of the wave in the medium. With me? Now put B equals E over v into the magnetic part. We get E squared, over two mu v squared. But v squared is one over mu epsilon, so this becomes one half epsilon E squared, exactly the electric part. Clear?

1:48Intensity and the field amplitude

Now let us take the average. E naught, the E with a small zero, is the amplitude of the electric field. Over one cycle, the average energy density, u inside angle brackets, is one half epsilon naught E naught squared, half the peak value, epsilon naught E naught squared. Got it? Next, intensity, capital I, the average energy crossing unit area, held perpendicular to the direction of travel, per unit time. Each second, the energy in a column of length c, the speed of light, crosses that area. So I equals the average energy density times c, which is one half epsilon naught E naught squared c. Okay? Sometimes the question gives E rms, the root mean square value, written with the small letters r m s. Then is there still a half in front? No, there is no half this time! E rms squared is already E naught squared over two, so I equals epsilon naught c E rms squared. Keep this in mind for our second example. Right? Now a point source of power P, shining uniformly all around. At distance r, the power spreads over a sphere of area four pi r squared, with pi the Greek letter pi. So I equals P over four pi r squared. Clear? Turn the intensity formula around, and E naught equals the square root of, two I over epsilon naught c; the tick shaped sign is the square root. Remember the two inside the root. With me?

3:22Light carries momentum

Now back to our puzzle. An electromagnetic wave that carries energy, capital U, also carries momentum, p, equal to U over c, in its direction of travel. So no rest mass is needed; the fields themselves carry the momentum. Okay? So tell me, when the light is absorbed by an object, where does this momentum go? Into the object, yes, exactly! Momentum is conserved, so when the light is absorbed completely, the object gains all of it, U over c. That is the push in our puzzle. Got it? But a perfect reflector reverses the momentum, so the push doubles. For a beam of power P falling normally, the force F is P over c on a perfect absorber, and two P over c on a perfect reflector. Clear?

4:12Example 1: momentum from a 30 mW, 100 ns absorbed pulse

Now Example one, from JEE Advanced twenty thirteen, our puzzle itself. Here, a pulse of light lasting one hundred nanoseconds, the n s on screen, is absorbed completely by a small object at rest. Its power is thirty milliwatts, the m W on screen. Find the final momentum. Pause the video and try it first. Done? Okay, let's check. Milli means ten to the power minus three, so the power P is thirty times ten to the power minus three watt. Nano means ten to the power minus nine, so the time t is one hundred times ten to the power minus nine second. Right? The energy U equals P times t. Thirty times one hundred is three thousand, and the powers of ten give ten to the power minus twelve. So U is three times ten to the power minus nine joule, the capital J on screen. Okay? Now, what is the momentum of the object: is it U over c, or two U over c? It is U over c, because the pulse is absorbed completely, not reflected. Two U over c would need a perfect reflector, and that is the common slip here. Got it? So p equals three times ten to the power minus nine, over three times ten to the power eight, which is one point zero times ten to the power minus seventeen kilogram metre per second, the k g m s to the power minus one on screen. It is tiny, but it is real, and that solves our puzzle perfectly.

5:50Example 2: field amplitude 1 m from a 0.1 W red LED

Now Example two, the harder one, from JEE Main twenty fifteen. Here, a red LED gives out light of power zero point one watt, the capital W on screen, uniformly all around it. Find the electric field amplitude one metre away. Pause the video and try it first. Done? Okay, let's check. The LED is a point source, so I equals zero point one, over four pi times one squared. With the wavy equals sign for approximately, I is about seven point nine six times ten to the power minus three watt per square metre, the W m to the power minus two on screen. With me? Now the amplitude. Many students forget the two inside the root and take the square root of I over epsilon naught c, about one point seven three. So is that the amplitude E naught? No, no, careful, that is the trap! Without the two, this root gives E rms, about one point seven three volt per metre, the V m to the power minus one on screen. That is not the amplitude E naught. Clear? So use the correct formula, the square root of, two I over epsilon naught c. On top, two times seven point nine six times ten to the power minus three. Below, epsilon naught, eight point eight five times ten to the power minus twelve, times c, three times ten to the power eight. Okay? The top is one point five nine two times ten to the power minus two, and the bottom is two point six five five times ten to the power minus three. Their ratio is about six, and its square root gives E naught, about two point four five volt per metre. Very well done if you got that right!

7:39Quick revision: uE = uB, intensity, p = U/c, P/c and 2P/c

So, quick revision. The electric and magnetic energy densities are equal, in vacuum and in a medium. Intensity is one half epsilon naught E naught squared c, or epsilon naught c E rms squared, and P over four pi r squared for a point source. Light carries momentum U over c: a beam of power P pushes a perfect absorber with force P over c, and a perfect reflector with two P over c. Now try the practice problems on this page. All the best!