Counting Photons

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Two lamps give out exactly the same power as light. One glows red, and the other glows blue. Which lamp sends out more photons every second? Many students say the blue one, because blue light feels stronger. Keep your answer in mind; we will test it soon. Counting photons comes up often in JEE problems. So, let's begin.

0:26Photons per second from a source

First, recall the last lesson. One photon of frequency nu, the Greek letter like a small v, carries energy E equals h nu, which equals h c over lambda. Here h is the Planck constant, c is the speed of light, and lambda, the last Greek letter on the line, is the wavelength. Okay? Now, every source has a power, capital P: the energy it gives out as light every second. Right? So let me ask you this. Each second the source gives out P joules, and each photon carries away h nu joules. How many photons leave per second? Just divide, yes, exactly! N, the number of photons per second, equals P over h nu, and with nu equal to c over lambda, it is P times lambda, over h c. This line is the heart of today's lesson. Clear? Now, back to our puzzle. Same power P, but red light has the longer wavelength. Using the formula, which lamp sends out more photons? The red lamp, that is correct! For the same power, N is proportional to lambda; the sign like an open loop means proportional to. Each red photon carries less energy, so more of them are needed to carry the same power. Got it?

1:49Intensity and photon flux

Now, a detector catches only the light falling on it. So we use intensity, capital I: the energy falling on unit area every second. Its unit, in the brackets, is watt metre to the power minus two, that is, watts per square metre. Okay? Divide by the energy of one photon, and the photon flux, the number of photons crossing unit area per second, is I over h nu. With me? Now take a tiny point source, shining equally in all directions. At a distance r, over what surface is its power spread? Over a sphere of radius r, yes! Its area is four pi r squared; pi is the Greek letter, about three point one four, and the small two means squared. So the intensity is P over four pi r squared, and the photon flux is N over four pi r squared. Clear? Here is one more result. In a beam, the energy in each cubic metre is I over c; the slanted line means divide. Divide that by h nu, and the photons per unit volume are I over h nu c. Right?

2:58Force due to photons

Now something surprising: photons carry momentum too, so light can push! Each photon carries momentum, small p, equal to h over lambda. Okay? So how much momentum does the beam bring every second? N photons, each with h over lambda; the dot means times. Put in N equals P times lambda over h c, and both h and lambda cancel. We are left with just P over c, nothing else. Beautiful, isn't it? And momentum delivered per second is a force. So if the surface absorbs the light completely, the force F on it is P over c, okay? Now make the surface a perfect mirror, with the light falling straight on it. Each photon bounces straight back. By how much does its momentum change? By two times h over lambda, exactly! It arrives with h over lambda and leaves with h over lambda the other way. So on a perfect reflector, at normal incidence, the force is two P over c. Writing P over c here is a common slip. Got it?

4:05Example 1: photons per second from a 2 mW 633 nm laser

Let's try the first example. Read it with me: a laser emits two milliwatts, m W on screen, of light of wavelength six hundred and thirty three nanometres, n m on screen. How many photons does it emit per second? Pause the video and try it first. Done? Okay, let's check. First, let us convert the units. Milli means ten to the power minus three, so the power P is two times ten to the power minus three watts. Okay? Now for the wavelength: six hundred and thirty three nanometres, in metres, is how much? Nano means ten to the power minus nine, so lambda is six point three three times ten to the power minus seven metres. Forget this conversion, and the answer is badly wrong. Right? Now a topper's trick: keep h c ready. Six point six three times ten to the power minus thirty four, times three times ten to the power eight, is one point nine eight nine times ten to the power minus twenty five joule metres. Clear? So N equals P times lambda, over h c. On top, two times ten to the power minus three, times six point three three times ten to the power minus seven, and below, one point nine eight nine times ten to the power minus twenty five. With me? Work it out, and N is about six point four times ten to the power fifteen per second. The wavy equals sign means approximately, and s to the power minus one means per second. Over a million billion photons a second, from just two milliwatts! Got it?

5:49Example 2: photons hitting a 1 cm2 detector 0.5 m away

Now a harder one, JEE level. Read it with me: a point source of power ten watts emits light of wavelength five hundred nanometres, uniformly. Find the number of photons striking, per second, a detector of area one square centimetre, c m with the small two, held zero point five metre away, facing the source. Pause the video and try it first. Here is a hint: think of a sphere. Done? Okay, let's check. First, find the photons from the whole source. N equals P times lambda, over h c: ten times five times ten to the power minus seven, over one point nine eight nine times ten to the power minus twenty five. That gives two point five one times ten to the power nineteen per second. Okay? Now look at the detector. One centimetre is ten to the power minus two metre. So is one square centimetre equal to ten to the power minus two square metre? No, no, careful, that is the trap! The area is squared, so the power of ten doubles too: one square centimetre is ten to the power minus four square metre. Right? Next, the photons spread equally over a sphere of radius zero point five metre. Its area is four pi times zero point five squared, which is exactly pi square metres. With me? So what fraction does our detector catch? Its own area over the area of the sphere. Ten to the power minus four over pi, about three point one eight times ten to the power minus five. Clear? Finally, we multiply. Two point five one times ten to the power nineteen, times three point one eight times ten to the power minus five, gives about eight point zero times ten to the power fourteen photons per second. So it was just N over four pi r squared, times the detector area. That's it!

7:46Quick revision: N = P lambda/hc, flux, force P/c and 2P/c

So, quick revision. Photons per second: N equals P over h nu, or P times lambda over h c; same power, longer wavelength, more photons. Photon flux is I over h nu; from a point source, divide N by four pi r squared. Force of light: P over c when absorbed, two P over c on a perfect reflector, with the light falling straight on it. Now go and try the practice problems on this page. All the best!