Overview

Transcript

0:00Introduction

Hello everyone! One small puzzle first. An object moves towards a concave mirror at four centimetres per second. How fast does its image move? Also four, surely, right? Hmm, hold that thought; Example two will settle it. In this chapter, all of optics comes together in mixed JEE Main-pattern problems. So, let's begin.

0:25How this chapter works

So, how does this chapter work? It has three timed sets, each with fifteen original problems, in the style of recent JEE Main papers. Okay? And every set mixes all the chapters of this unit. Right? Give yourself thirty minutes per set, and only then check the solutions. Now tell me: how much time is that for each problem? Two minutes each, very good! Thirty times sixty gives one thousand eight hundred seconds, the small s on screen. And one thousand eight hundred over fifteen gives one hundred and twenty seconds per problem. Now, the tools. With me?

1:06Mirrors and lenses: paraxial rays, signs first

The first tool is the sign convention. In the figure, P is the pole of the mirror, F the focus, and C the centre of curvature. All distances are measured from the pole: along the incident light they are plus, and against it, minus. Okay? With light from the left, a real object has a negative u, the object distance, less than zero on screen. For mirrors, a concave one has a negative f, the focal length, and a convex one a positive f. Clear? The mirror formula: one over v, plus one over u, equals one over f, where v is the image distance. It holds for paraxial rays, the rays close to the axis. And the magnification m is minus v over u, which is also f over, f minus u. Right? Now the lens. Tell me, is the lens formula also one over v, plus one over u? Not at all, the lens formula has a minus sign! One over v, minus one over u, equals one over f, and m is plus v over u. Here f is positive for a converging lens, and negative for a diverging one. Mixing up these signs is a very common trap. Got it? And read the sign of m: negative m means a real, inverted image, and positive m a virtual, erect one. Okay? One more tool: along the axis, and measured relative to the mirror or lens, the image speed is m squared, times the object speed. Clear?

2:38Refraction, total reflection and the prism

Now refraction. Snell's law: n one times sine i equals n two times sine r, where n one and n two, the n with small subscripts, are the refractive indices. The angles i and r are measured from the normal, not from the surface. Okay? Let me ask you this: light goes from air into glass at a very large angle of incidence. Can it be totally reflected? It cannot, because air is the rarer medium! Total internal reflection needs light going from the denser medium, n one, to the rarer one, and an angle of incidence greater than theta c, the critical angle, the Greek theta with a small c. And sine theta c equals n two over n one. Right? For a prism, capital A is the prism angle, and delta m, the Greek delta with a small m, is the angle of minimum deviation. Then n, the refractive index of the prism, equals the sine of, A plus delta m over two, divided by the sine of A over two. With me? And only for a thin prism, the deviation delta is n minus one, times the prism angle. Got it?

3:45Waves: fringes, slits and polaroids

Now the wave chapters. Interference needs coherent sources. The fringe width beta, the Greek letter, is lambda times capital D, over small d: lambda is the wavelength, capital D the screen distance, and small d the slit separation. It holds when capital D is much greater than small d, the double greater-than sign, and for small angles. Okay? At a distance y from the centre, the bright fringes sit at y equals n times beta, and the dark fringes at n minus a half, times beta. Right? Now a single slit of width small a, with theta the angle from the axis. In Young's experiment, an equation of exactly this shape, with small d, gave the bright fringes. So here, does a times sine theta equals n lambda give the bright ones too? Careful, it gives the minima, the dark ones, for n equal to plus or minus one, plus or minus two, and so on, never zero. It is the same-looking equation with the opposite meaning, a classic trap. Clear? Polarisation next. Unpolarised light of intensity capital I naught, the I with a small zero, comes out of one polaroid as I naught over two, whatever its orientation. And Malus's law, for light already polarised: capital I times cos squared theta, with theta the angle between the field and the pass axis. Got it?

5:07Example 1: fringe width and 4th dark fringe, d = 0.2 mm

Now Example one, from Set one. In a double-slit experiment, small d is zero point two millimetres, the m m on screen, capital D is one point two metres, and lambda is five hundred nanometres, the n m. Find the fringe width, and the distance of the fourth dark fringe from the centre. Pause the video and try it first. Done? Okay, let's check. First, everything in metres: five hundred nanometres is five times ten to the power minus seven metres, and zero point two millimetres is two times ten to the power minus four metres. Okay? In the figure, S one and S two are the slits, small d apart, with the screen capital D away. So beta is five times ten to the power minus seven, times one point two, over two times ten to the power minus four. That gives three times ten to the power minus three metres, which is three millimetres. Right? Now the fourth dark fringe. It is four times beta, so twelve millimetres, right? Oh no, careful, that is the trap! Dark fringes sit at n minus a half, times beta. So four minus a half gives three point five, and three point five times three gives ten point five millimetres.

6:19Example 2: image speed for object at 4 cm/s, concave mirror

Got it? And now Example two, a hard one from Set two: our puzzle. An object fifty centimetres from a concave mirror of focal length thirty centimetres moves towards it along the axis, at four centimetres per second, the c m s to the power minus one on screen. Find the speed of the image. Pause the video and try it first. Done? Okay, let's check. Signs first: the object is real, so u is minus fifty centimetres, and the mirror is concave, so f is minus thirty centimetres. Okay? Now the one-line trick: m equals f over, f minus u. That is minus thirty over, minus thirty plus fifty, which is minus thirty over twenty, so minus one point five: a real, inverted image. Right? So the image moves at one point five times four, which is six centimetres per second, right? No, no, wait, that is the trap! Along the axis, the speed goes as m squared. Minus one point five, squared, is two point two five, and two point two five times four gives nine centimetres per second. Clear? And which way does the image move? Towards the mirror, just like the object? It moves away from the mirror, because for a mirror the object and the image always move in opposite directions along the axis. Puzzle solved: the object moves at four, but its image moves at nine, the other way! Got it?

7:51Quick revision: mirror and lens signs, m squared, fringes

So, quick revision. Thirty minutes a set, about two minutes a problem. Signs first: plus in the mirror formula, minus in the lens formula, and along the axis the image speed goes as m squared. Dark fringes sit at n minus a half, times beta, and total reflection needs denser to rarer. Now open Set one, start your timer, and try it. All the best!