AIEEE and Earlier JEE Main Questions: Interference

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Mix red light and green light, and send both through the same two narrow slits. Each colour makes its own bright fringes, and red, with the longer wavelength, makes wider ones. So, away from the centre, where does a red bright fringe first land exactly on a green one? Hold that thought. Today we answer it with past questions from AIEEE and JEE Main. Questions like these come up often, so let's begin.

0:31What interference needs

First, a warm-up from AIEEE two thousand three: what must be true of two sources if they are to show interference? Let me make it concrete. You point two separate bulbs at one wall. Do you see bright and dark fringes? No, the wall just gets evenly brighter. The phase difference between separate lamps keeps changing randomly, so any pattern washes out at once. With me? So the sources must be coherent. Coherent means the phase difference between them stays constant in time. Okay? So how does Young get coherent sources? In the figure, one source, S, the dot on the left, lights both slits, S one and S two, that is S with a small one and a small two below. Same source, so their phase difference stays fixed. Right? So the full answer to that AIEEE question: the same frequency, and a definite, constant phase relationship. In fact, if the frequencies differed even slightly, the phase difference would keep drifting, so the two conditions go together. Clear?

1:37Bright fringes, and two colours together

Now, the bright fringes. In the figure, the slits are small d apart, the screen is capital D away, and P is at height y above O, the centre of the screen. Bright fringe number n sits at y n, y with a small n below, equal to n times lambda times capital D, over d. Lambda, the Greek letter, is the wavelength, and n is zero, one, two, and so on, as the dots say. Okay? The gap between neighbouring bright fringes is the fringe width, beta, the Greek letter on the right of the figure: lambda times capital D, over d. So a longer wavelength gives wider fringes, which is why the red fringes are wider. Got it? One condition: these formulas need small angles. Capital D is much greater than small d, that is what the double greater than sign means, and y is much smaller than capital D, the double less than sign. Both our questions assume this. With me? Now back to the puzzle. Is there any bright fringe that red and green already share? Yes, exactly, the central one! At the centre, n is zero, and the path difference is zero for every wavelength, so both colours are bright at O, the centre. Right? Away from the centre, two bright fringes coincide only where their positions y are equal. Capital D and small d are the same for both colours, so they cancel, and we get n one lambda one equals n two lambda two, with a small one and a small two below. This one line solves both our questions. Clear?

3:09Example 1: 3rd fringe of 590 nm on 4th of unknown wavelength

Now example one, from AIEEE two thousand nine. Light of five hundred and ninety nanometres, written n m, is mixed with light of an unknown wavelength, and lights a double slit. The third bright fringe of the known light coincides with the fourth bright fringe of the unknown light. Find the unknown wavelength. Pause the video and try it first. Done? Okay, let's check. Coinciding means the same spot, with the same capital D and small d, so n one lambda one equals n two lambda two. Okay? Now, many students write four times five hundred and ninety equals three lambda; the cross sign means times. Is that correct? No, no, careful, that is exactly the trap! Each order goes with its own wavelength: three goes with five hundred and ninety, and four goes with the unknown lambda. So three times five hundred and ninety equals four lambda. Right? Three times five hundred and ninety is one thousand seven hundred and seventy. Then one thousand seven hundred and seventy over four gives four hundred and forty two point five nanometres. Got it? And a quick check. The unknown light needs the higher order, four, to reach the same spot, so its wavelength must be shorter. Four hundred and forty two point five is less than five hundred and ninety, as the less than sign shows, so it checks out. With me?

4:31Example 2: least distance where 650 and 520 nm fringes meet

Now example two, from JEE Main twenty seventeen, and it is exactly our two colour puzzle. Slits zero point five millimetres apart, written m m, form fringes on a screen one hundred and fifty centimetres away, written c m. The light has wavelengths six hundred and fifty nanometres and five hundred and twenty nanometres. Find the least distance from the central maximum at which bright fringes of both coincide. Pause the video and try it first. Done? Okay, let's check. Both colours have to put a bright fringe on the same spot. Let me ask you this: which wavelength needs the bigger order to get there? The shorter one, five hundred and twenty, very good! Its fringes are narrower, so it needs more of them. So put six hundred and fifty at order n one, and five hundred and twenty at order n two: n one times six hundred and fifty equals n two times five hundred and twenty. The double arrow means which gives: n one over n two is five hundred and twenty over six hundred and fifty, which is four over five. Okay? But wait. Eight and ten also give four over five, right? So which pair do we take, four and five, or eight and ten? We want the least distance, so the smallest whole numbers, four and five; eight and ten coincide farther away. So four times six hundred and fifty equals five times five hundred and twenty, which is two thousand six hundred nanometres. Clear? Now, many students slip here, so be careful with the units, and change everything to metres. Two thousand six hundred nanometres is two point six times ten to the power minus six metres. Capital D, one hundred and fifty centimetres, is one point five metres, and small d is zero point five times ten to the power minus three metres. Got it? Now the bright fringe formula: y equals n one lambda one, times capital D, over d. That is two point six times ten to the power minus six, times one point five, over zero point five times ten to the power minus three. With me? Two point six times one point five is three point nine, and three point nine over zero point five gives seven point eight. So y is seven point eight times ten to the power minus three metres, that is, seven point eight millimetres. Clear? And a check with the fringe widths, which settles our puzzle. The fringe width is one point nine five millimetres for six hundred and fifty nanometres, and one point five six millimetres for five hundred and twenty. Four times one point nine five, and five times one point five six, both give seven point eight. So the fourth bright fringe of the longer wavelength sits exactly on the fifth of the shorter one. Okay?

7:15Quick revision: coherence, n1 lambda1 = n2 lambda2, least n

So, quick revision. Interference needs coherent sources, with a constant phase difference, of the same frequency. Bright fringes sit at n lambda capital D over d, and two wavelengths meet where n one lambda one equals n two lambda two. For the least distance, take the smallest whole numbers, and then work in metres. Now try the practice problems on this page. All the best!