Set 3

Transcript

0:00Introduction

Hello everyone! One small puzzle first. Picture a free mirror of five grams, at rest. We fire one laser pulse at it, with an energy of ten millijoules, and the mirror reflects it completely. Does the mirror move? If so, how fast? Keep your guess; we will check it at the end. Today, Set three of the mixed problems on electromagnetic waves, in the style of JEE Main. Four quick tools first, then two full problems. Let's go.

0:33Tool 1: from k to wavelength and frequency

So, tool one: reading a wave. Here k is the wave number, lambda, the first Greek letter on screen, is the wavelength, and nu, the Greek letter that looks like a v, is the frequency. Lambda equals two pi over k, and nu equals c over lambda, with c the speed of light. Okay? Now a problem from this set. Here, a wave in vacuum has k equal to four pi times ten to the power six radians per metre; that is the rad m to the power minus one on screen. What is its wavelength? In two pi over four pi times ten to the power six, the pi cancels, and two over four is one half, so lambda is zero point five times ten to the power minus six metre. That is zero point five micrometre; the Greek letter mu before the m means micro. Right? Three times ten to the power eight, over zero point five times ten to the power minus six, gives nu, six times ten to the power fourteen hertz, the H z on screen. Clear?

1:40Tool 2: intensity and energy density

Now, tool two: intensity and energy density. Intensity, capital I, is the average energy crossing one square metre, held perpendicular to the wave, every second. It equals half epsilon naught E naught squared c. Epsilon naught, the curly e with a small zero, is the permittivity of free space, and E naught is the peak electric field. With the r m s field, the root mean square value written small below the E, it is epsilon naught c E r m s squared. With me? Now, a wave has an r m s electric field of fifty volts per metre, the V m to the power minus one on screen. Find its intensity. Is there a half in front this time? There is no half, that is correct! The half goes only with the peak value, E naught. Fifty squared is two thousand five hundred. So epsilon naught, eight point eight five times ten to the power minus twelve, times three times ten to the power eight, times two thousand five hundred, gives six point six watts per square metre, the W m to the power minus two. Got it? Now, the average energy density, u inside the angle brackets, is I over c. So where the intensity is zero point three watts per square metre, zero point three over three times ten to the power eight gives ten to the power minus nine joules per cubic metre, the J m to the power minus three. Okay?

3:09Tool 3: a point source

Tool three: a point source. The electric field amplitude is twelve volts per metre at three metres from a point source. What is it at nine metres? Three times farther, so one ninth of the field, twelve over nine, right? No, no, careful, that is the classic trap here. The intensity falls as one over r squared, with r the distance from the source; this sign, like an open eight on its side, means proportional to. But the intensity goes as the square of the field, so E naught falls only as one over r. Got it? So E naught equals twelve times three over nine, which is four volts per metre. Three times farther, one third of the field. Clear?

3:54Tool 4: light pushes

Tool four, the one our mirror needs. Look, a wave that carries energy, capital U, also carries momentum, small p, equal to U over c, along its direction of travel. When the wave is absorbed or reflected, this momentum goes into the surface, which feels a pressure. With me? In the figure, on the left, light falls straight on an absorbing surface; this is normal incidence. The radiation pressure, p with a small rad below, is I over c; the figure writes it simply as small p. Okay? Now the middle picture: a perfect mirror sends the light straight back. Is the push the same, or more? More, twice as much, yes, exactly! The momentum arrives as plus p and leaves as minus p, a change of two p. So p rad is two I over c. On the right, light strikes the mirror at an angle theta to the normal, and the pressure becomes two I cos squared theta over c; we skip that today. Got it? And when a surface reflects only a fraction r and absorbs the rest, p rad is one plus r, in brackets, times I, over c. Put r equal to zero for the absorber, and r equal to one for the mirror. Clear?

5:12Example 1: pressure on a surface reflecting 40 percent

Now, example one. Light of intensity six hundred watts per square metre falls normally on a surface. The surface reflects forty percent of it and absorbs the rest. Find the radiation pressure. Pause the video and try it first. Done? Okay, let's check. Sixty percent is absorbed, so we put r equal to zero point six, right? Oh no, careful, that is the trap! Here r is the fraction reflected, not absorbed. So r is zero point four. Okay? One plus zero point four is one point four. One point four times six hundred gives eight hundred and forty, and we divide by three times ten to the power eight. With me? Eight hundred and forty over three times ten to the power eight gives two point eight times ten to the power minus six pascal; the P a on screen means pascal. Right?

6:12Example 2: speed of a 5 g mirror hit by a 10 mJ laser pulse

Now, example two: our mirror from the start! So, a laser pulse of energy ten millijoules, the m J on screen, is reflected completely by a free mirror of mass five grams, initially at rest. Find the speed the mirror acquires. Pause the video and try it first. Done? Okay, let's check. The pulse carries momentum U over c, so the mirror gets U over c, right? No, no, wait, the pulse is reflected completely! It arrives with U over c and leaves with U over c the other way. So the mirror gets two U over c. Got it? Now convert the units: ten millijoules is zero point zero one joule, and five grams is five times ten to the power minus three kilogram. Okay? The mirror's momentum, m times v, equals two U over c. So v is two times zero point zero one, divided by five times ten to the power minus three, times three times ten to the power eight. Clear? The top is zero point zero two; the bottom, one point five times ten to the power six. So v is one point three times ten to the power minus eight metres per second, the m s to the power minus one on screen. So yes, the mirror moves, very slowly, and our puzzle is solved. Right?

7:38Quick revision: wavelength, intensity, 1/r field, pressure

So, quick revision. For a wave, lambda is two pi over k and nu is c over lambda; intensity is epsilon naught c E r m s squared, and the average energy density is I over c. From a point source, the intensity falls as one over r squared, but the field only as one over r. And at normal incidence, the pressure is I over c for an absorber, two I over c for a mirror, and one plus r, times I, over c, when a fraction r is reflected. Now go and try the other problems of this set on this page. All the best!