Young's Double-Slit Experiment

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Two bulbs just make a wall brighter, everywhere. But send light from one source through two tiny slits, and a screen shows bright stripes, with dark stripes in between. So, light plus light, giving darkness? Today we find exactly where those dark stripes sit. Young's double slit experiment comes up often in JEE questions. So, let's begin.

0:28Path difference at a point on the screen

So, look at the figure. Light from one source, S, the dot on the left, reaches two narrow 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 constant. They are coherent sources, which interference needs. Our two bulbs were not coherent, so no stripes. With me? The slits are a distance small d apart, and the screen is capital D away from them. That double pointed sign means much greater than, so capital D is much bigger than small d, okay? Now take a point P on the screen, at a height y above the centre O, at an angle theta, the Greek letter, from the axis. Let me ask you this: which wave travels farther to reach P, from S one or from S two? From S two, yes, exactly! The extra path is the small piece marked d sine theta. So the path difference, capital delta, the triangle on screen, is d sine theta, approximately y times d, over capital D; the wavy equals sign means approximately. This holds for small angles, near the centre. Right? Now, for what path difference do the two waves arrive in step and make a bright fringe? When it is a whole number of wavelengths, very good! Lambda, the Greek letter, is the wavelength, and n is zero, one, two, and so on. So capital delta equals n lambda, and y n, that is y with a small n below, equals n times lambda times capital D, over small d, measured from the centre. And n equal to zero is the central bright fringe, right at O, the centre. Got it? And the dark fringes, where the waves arrive exactly out of step, sit at n minus one half, times lambda times capital D, over d, with n equal to one, two, three, and so on. Clear?

2:22Fringe width, and the dark-fringe trap

Now, the wavy curve on the right of the figure shows how bright the screen is at each height. So, how far apart are two neighbouring bright fringes? Subtract one bright fringe's position from the next, and you get lambda times capital D, over small d, every single time. This is the fringe width, beta, the Greek letter on the figure. It does not depend on n, so all fringes, bright and dark, are equally spaced, with the same width. Okay? So, the third dark fringe is at three beta, right? No, no, careful, that is the trap! Dark fringe number n is at n minus one half, times beta, not n times beta. So the third dark fringe is at two and a half beta, halfway between the second and third bright fringes. Got it? Next, the angle between neighbouring fringes, seen from the slits, is the angular fringe width. It is simply lambda over small d; the screen distance does not appear. With me? Now put the apparatus in a liquid of refractive index mu, the Greek letter on screen. The wavelength becomes lambda over mu, so the fringe width becomes beta over mu. Right? And one condition: all of this needs y much smaller than capital D, the double sign the other way round. For large angles, use d sine theta equals n lambda, and y equals capital D times tan theta. Clear?

3:49Intensity, and the highest order

Now let us look at how bright each point is. Let I zero, I with a small zero below, be the intensity from each slit alone. Then the intensity is four I zero, times cos squared of phi over two, where phi, the Greek letter, is the phase difference: two pi over lambda, times y times d, over capital D; the dot means times. Okay? So at the centre, two slits, so two I zero, right? Oh no, careful, that is how two bulbs add, not two coherent slits! At the centre phi is zero, and cos of zero is one, so the intensity is four I zero. Got it? There is also a limit on the number of fringes. Sine theta can never be more than one; that sign means less than or equal to. So, since d sine theta equals n lambda, the highest order is the integer part of small d over lambda. One small catch: if d over lambda is a whole number, that last order would sit at ninety degrees, so it never reaches the screen. With me?

4:53Two wavelengths, and white light

Now take light with two wavelengths, lambda one and lambda two, with a small one and a small two below. Each makes its own pattern, and their bright fringes fall together where n one lambda one equals n two lambda two. Okay? And with white light, every colour has zero path difference at the centre, so the central fringe is white. Right? Now, beta is proportional to lambda. So tell me, which colour has its fringes nearest the centre? Violet, yes, exactly! It has the shortest wavelength, so the smallest beta, and red is farthest. After a few coloured fringes, the colours overlap and you see uniform illumination. Clear?

5:34Example 1: 5th bright and 3rd dark fringe with 600 nm light

Now example one, an easy one. The wavelength is six hundred nanometres, small d is one millimetre, and capital D is one metre. Find the distances from the centre of the fifth bright fringe, and of the third dark fringe. Pause the video and try it first. Done? Okay, let's check. Beta equals lambda times capital D, over small d: six hundred times ten to the power minus nine, times one, over ten to the power minus three, in metres. That gives zero point six millimetres. With me? The fifth bright fringe is at five beta. Five times zero point six gives three point zero millimetres. Clear? Now the third dark fringe. Three beta, so one point eight millimetres, right? No, no, wait, remember the trap! The third dark fringe is at three minus one half, times beta, which is two point five beta. Two point five times zero point six gives one point five millimetres. Got it?

6:40Example 2: where 480 nm and 600 nm bright fringes coincide

Now example two, a little harder. Light with two wavelengths, four hundred and eighty nanometres and six hundred nanometres, is used with the same small d and capital D, one millimetre and one metre. Find the least distance from the centre at which bright fringes of both coincide. Pause the video and try it first. Done? Okay, let's check. Use our rule, n one lambda one equals n two lambda two. So n one times four hundred and eighty must equal n two times six hundred, with whole numbers n one and n two. Okay? Not zero and zero; that is the centre itself. So, what are the smallest whole numbers that make the two sides equal? Five and four, very good! Five times four hundred and eighty is two thousand four hundred, and four times six hundred is also two thousand four hundred, in nanometres. Right? So the distance, y, equals two thousand four hundred times ten to the power minus nine, times one, over ten to the power minus three, in metres. That gives two point four millimetres. Clear?

7:49Quick revision: beta, dark fringes, intensity, white light

So, quick revision. Beta is lambda times capital D over small d, with bright fringes at n beta, and dark ones at n minus one half, times beta. Intensity is four I zero times cos squared of phi over two, and in a liquid the fringe width is beta over mu. For two wavelengths, n one lambda one equals n two lambda two, and white light gives a white central fringe. Now try the practice problems on this page. All the best!