IIT-JEE, AIEEE and Earlier JEE Main Questions: Diffraction and Polarisation

Transcript

0:00Introduction

Hello everyone! Light travels in straight lines, right? Now send it through a slit just one micrometre wide, which is a millionth of a metre. On the screen, the central bright band spreads over a full sixty degrees! From just these two numbers, can you find the wavelength of the light? Hold that thought. Today we solve past questions on diffraction and polarisation from JEE Main, and that sixty degrees is the key to one of them. Questions like these do appear in JEE Main, so let's begin.

0:35Single slit: where the central maximum ends

So, first, diffraction at a single slit. In the figure, light comes from the left to a slit of width small a, and theta, the Greek letter, is the angle from the centre line. The dark bands, the minima, are where small a times sine theta equals n times lambda. Lambda, the other Greek letter, is the wavelength, and n is plus or minus one, plus or minus two, and so on, as the dots say. Okay? Now look at the big bright band in the middle of the pattern, the central maximum. On each side, where does it end? At the first minima, yes, exactly! That is n equal to plus or minus one, so small a times sine theta equals lambda, with one dark band on each side of the centre. Right? So the central maximum lies between the two first minima: from theta below the centre line to theta above it, between the two slanted lines in the figure. Got it? Now, a question gives you the angular width of the central maximum. Is that theta, or two theta? It is two theta, the full spread from one first minimum to the other. So theta, the angle that goes into the formula, is half the angular width. We need this in example two. With me? And for small angles, sine theta is about theta in radians, so the angular width is about two lambda over small a; the wavy equals sign means approximately. On a screen capital D away, that is the width two lambda capital D, over small a, the arrow with two heads in the figure. This shortcut needs small angles; for a large angle, use the sine itself. Clear?

2:19Polaroids and Malus's law

Now, polarisation. Light is a transverse wave, and its electric field is perpendicular to the direction of travel. In unpolarised light, the field takes all directions in that perpendicular plane, changing at random; that is the star on the left of the figure. Now, a polaroid lets through only the part of the field along its pass axis, the dark line drawn on it. So unpolarised light of intensity I nought, that is I with a small zero below, comes out with intensity I nought over two. Okay? Now, let us slowly rotate this polaroid. Does the light coming out change? No, it stays I nought over two, for any orientation, because the field of unpolarised light takes all directions across the beam, with no direction preferred. But what comes out is plane polarised: its field stays along one line, the small double arrow in the figure. Got it? Now send polarised light of intensity I to a polaroid whose axis makes an angle theta with the field. Only I times cos squared theta gets through. This is Malus's law, and the small two up there means squared. With me? So, unpolarised light through two polaroids at angle theta, the polariser and then the analyser, gives I nought over two, times cos squared theta, just as the figure shows on the right. First halve, then multiply by cos squared theta. Right? Now turn the analyser until theta is ninety degrees. These are called crossed polaroids. So how much light comes out? Nothing comes out at all, very good! The small circle on screen means degrees, and cos ninety degrees is zero, so crossed polaroids transmit nothing. Clear?

4:08Example 1: unpolarised I0 through polaroids at 45 degrees

Now example one, from JEE Main two thousand thirteen. Unpolarised light of intensity I nought passes through a polaroid A, and then through a polaroid B, whose axis is at forty five degrees to the axis of A, okay? Find the emergent intensity. Pause the video and try it first. Done? Okay, let's check. Polaroid A receives unpolarised light, so it passes half: I nought over two, polarised along the axis of A, right? Now, some students write I nought times cos squared forty five degrees. Is that correct? No, and that is exactly the trap, so be careful here. Malus's law is for polarised light, and the light reaching A is unpolarised. So we halve at A first, and only then use cos squared at B: I nought over two, times cos squared forty five degrees. Okay? Cos squared forty five degrees is one half. Then I nought over two, times one half, gives I nought over four; the cross sign means times. So one quarter of the light comes out.

5:20Example 2: slit separation from a 60 degree central maximum

Clear? Now example two, from JEE Main two thousand eighteen, and this is our puzzle from the start. The angular width of the central maximum in a single slit pattern is sixty degrees, and the slit is one micrometre wide; on screen that is mu m, with mu the Greek letter. Then, a second identical slit is made near it, and Young's fringes of width one centimetre are seen on a screen fifty centimetres away. Find the separation of the slits. Pause the video and try it, and find the wavelength first. Done? Okay, let's check. The wavelength comes from the single slit, where the first minimum is at small a times sine theta equals lambda. So we put theta equal to sixty degrees, right? Oh no, careful, that is the trap! Sixty degrees is the full angular width, from one first minimum to the other. So the first minimum is at half of it, and theta is thirty degrees. Got it? So small a times sine thirty degrees equals lambda. Small a is ten to the power minus six metres, and sine thirty degrees is one half. So lambda is five times ten to the power minus seven metres, which is zero point five micrometres. Thirty degrees is not small, so we use the sine itself. With me? Now the second slit. Two slits give Young's fringes, and for small angles the fringe width beta, the Greek letter, is lambda times capital D, over small d. Capital D is the screen distance, and small d is the separation of the slits. The double arrow means which gives: small d equals lambda times capital D, over beta. Okay? Now the units, so put everything in metres. Capital D, fifty centimetres, is zero point five metres, and beta, one centimetre, is ten to the power minus two metres. Right? So small d is five times ten to the power minus seven, times zero point five, over ten to the power minus two. Five times zero point five is two point five, so the top is two point five times ten to the power minus seven. Dividing by ten to the power minus two gives two point five times ten to the power minus five metres, which is twenty five micrometres. So, our puzzle: the sixty degrees, halved to thirty, gave the wavelength, and the wavelength gave the slit separation. Clear?

7:49Quick revision: single slit minima, Malus law, fringe width

So, quick revision. Single slit: the first minima are at small a times sine theta equals lambda, and the angular width of the central maximum is two theta, so halve it first. Unpolarised light through a polaroid: half comes out, and after that, Malus's law, I times cos squared theta. Young's fringes: beta equals lambda capital D, over small d, with all lengths in metres. Now go and try the practice problems on this page. All the best!