Overview

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Light carries momentum, so when sunlight falls on your hand, it really does push on it. So why have you never felt that push, not even a little bit? Hmm, keep your guess ready. Today we will see how to use this unit's problem bank, settle this puzzle with one of its concept checks, and solve two of its problems. Radiation pressure is part of the JEE syllabus, and these ideas are tested, so let's begin.

0:32How to use this chapter

So, first, what is in this chapter? Part A has twelve solved problems, at JEE Main level. And they mix the whole unit: displacement current, the waves, intensity, pressure and the spectrum, just like the exam. Right? Then Part B has ten harder problems, at JEE Advanced level. With me? And Part C has eight concept checks. They test the ideas behind the formulae, the why questions. After that come eight practice problems on the page, for you to solve on your own. Got it? Now tell me: when should you read a solution? Straight away, or only after you have tried the problem yourself? Only after you try, yes, exactly! The course says it clearly: attempt each part as a test, and only then read the solutions. Okay?

1:26Part C: why does a mirror feel double?

Now, a taste of Part C, on radiation pressure. When a surface absorbs light, the momentum of the light goes into the surface, so the surface is pushed. For a perfect absorber at normal incidence, light hitting straight on, the radiation pressure, p with the small word rad below, for radiation, equals I over c. Capital I is the intensity, and c is the speed of light. That is the left panel of the figure. Right? Now the middle panel: a perfect reflector, which sends the light straight back. Its pressure is two I over c, double the absorber's. Clear? So here is the Part C question itself. The same light falls on both surfaces. Why should the mirror feel twice the pressure? Because the momentum of the light is reversed, that is correct! The absorber only stops the light. The mirror stops it and sends it back with the same momentum, so the change is doubled, and so is the push. On the right, the light hits the reflector at an angle theta, the Greek letter, measured from the normal, and the pressure becomes two I cos squared theta over c. But today we stay straight on. Got it?

2:35Part C: why is sunlight's push never noticed?

Now back to our puzzle, because Part C asks it too: why is the radiation pressure of sunlight not noticed in daily life? Look at I over c once more. What makes it so small? It is the speed of light, c, yes, exactly! We divide the intensity by a very large number. For sunlight, I over c is about five times ten to the power minus six pascal. The wavy equals sign means approximately, the cross means times, and capital P with a small a is the pascal. Okay? Now compare it with the air around you: the atmosphere presses with about ten to the power five pascal. Right? Divide one by the other: five times ten to the power minus six, over ten to the power five, gives five times ten to the power minus eleven. The push of sunlight is only that tiny fraction of atmospheric pressure. So that solves our puzzle: the push is real, but far too small to feel. Clear?

3:35Example 1: intensity of a wave with E0 = 30 V/m

Now, Example one, an easy one from Part A, okay? Find the intensity of a wave with E naught equal to thirty volts per metre. E naught, the E with a small zero below, is the peak value of the electric field, and V m to the power minus one means volts per metre. Pause the video and try it first. Done? Okay, let's check. Many students write the intensity as epsilon naught, times c, times E naught squared, with no half. Epsilon naught, the Greek epsilon with a small zero, is the permittivity of free space. That gives about two point four. So, is that right? Oh no, careful, that is the trap, because that formula needs the rms value, E with a small r m s below, the root mean square value. With the peak value E naught, there is a half, because sine squared averages to one half over a cycle. So I equals one half epsilon naught E naught squared, times c. Okay? Now the numbers. E naught squared is thirty squared, which is nine hundred. Right? So I is one half, times epsilon naught, eight point eight five times ten to the power minus twelve, times nine hundred, times c, three times ten to the power eight. Clear? Work it out, and the intensity comes to about one point two watt per square metre; W m to the power minus two means watt per square metre. Exactly half of the wrong answer, so never forget that half, okay?

5:02Example 2: force and acceleration of a 1 g absorbing plate

Now Example two, a hard one from Part B, okay? Here, a plane wave with E naught equal to one hundred volts per metre falls normally on a free, perfectly absorbing plate. The plate has an area of zero point five square metre, the small two up on the m, and a mass of one gram, the small g. Find the force on the plate, and its acceleration. Pause the video and try it first. Done? Okay, let's check. First, the plate is perfectly absorbing, and the light falls normally. So, is the pressure I over c, or two I over c? It is I over c, very good, because nothing comes back! And the course gives a neat shortcut: I over c equals one half epsilon naught E naught squared. So the force, capital F, is this pressure times capital A, the area: one half epsilon naught E naught squared, times the area. Okay? Now the numbers. E naught squared is one hundred squared, ten to the power four. One half, times eight point eight five times ten to the power minus twelve, times ten to the power four, gives four point four three times ten to the power minus eight pascal. Right? Multiply by the area, zero point five, and the force is about two point two times ten to the power minus eight newton, the capital N on screen. Tiny, just as Part C warned us. With me? Now the acceleration is force over mass. The mass is one, so the acceleration is just the force, two point two times ten to the power minus eight, right? No, no, wait, the mass is one gram, not one kilogram! One gram is ten to the power minus three kilogram, the k g on screen. Clear? So the acceleration, the small a, is F over m: two point two times ten to the power minus eight, over ten to the power minus three. That gives two point two times ten to the power minus five metres per second squared, which is what m s to the power minus two means. Still small, but a thousand times bigger than the wrong answer. Got it?

7:10Quick revision: I from E0, I/c and 2I/c, light's tiny push

So, quick revision. Part A is at JEE Main level, Part B at JEE Advanced level, and Part C checks the ideas; attempt each as a test, then read the solutions. With the peak field, I equals one half epsilon naught E naught squared, times c; the pressure is I over c on an absorber, and two I over c on a reflector. And light's push is tiny because of the huge c; always use kilograms in F over m. Now open Part A, solve it like a test, and then try the practice problems on this page. All the best!