Set 1

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Picture a man, one point seven metres tall, who wants to see his full height in a plane mirror on the wall. Does he need a mirror one point seven metres tall? And if he steps back, does he need a bigger mirror, or a smaller one? Hold that thought, because we will answer it today. This is our first mixed problem set on optics, and today we take its mirror problems, their tools, and their traps. Questions like these come up often in JEE Main. So, let's begin.

0:34Tool 1: the plane mirror

So, first tool: the plane mirror. On the left of the figure, a ray is reflected at the mirror. The dashed line is the normal, and i and r, the angles of incidence and reflection, are both measured from the normal, not from the mirror. And i equals r; that is the law of reflection. Okay? Now look at the right side. The point marked O is the object, and its rays bounce off the mirror. Extend them backwards along the grey dashed lines, and they all seem to come from the point marked I, which is the image. No light really reaches that point, so the image is virtual. It is also erect, and of the same size. Right? Now tell me this. The object is a distance d in front of the mirror, where d is the length you see below the figure. So how far behind the mirror is the image? The same distance d, yes, exactly! Look at the two equal arrows marked d. The image is as far behind the mirror as the object is in front. Clear?

1:35Tool 2: sign convention and the mirror formula

Second tool, for curved mirrors. The figure shows a concave mirror and a convex mirror, with light coming from the left; the black arrow, A B on the left, is the object, and the dark red arrow is the image. P is the pole, the centre of the mirror's surface, C is the centre of curvature, and F is the focus, midway between them. Our sign convention: measure every distance from the pole, positive along the incident light, and negative against it. With me? So, a real object sits on the left, in front of the mirror. Is u, its distance from the pole, positive or negative? Negative, yes, exactly! From the pole, you walk back towards the object, against the light. So for a real object, u is negative. Okay? Same idea for the focus. The focus of a concave mirror is in front of it, so f, the focal length, is negative. For a convex mirror the focus is behind, so f is positive. Right? Now the mirror formula: one over v, plus one over u, equals one over f, which equals two over capital R, okay? Here v is the image distance, and capital R is the radius of curvature, so f is half of it. It holds for paraxial rays, the rays close to the axis. Clear? And the magnification, m, is h prime over h: the image height, h with a small dash, over the object height. That equals minus v over u, and also f over f minus u. This last form is the topper's trick: it answers many mirror questions in one line, with the signs put in. Got it? Then read the sign of m. When m is less than zero, the less than sign you see, the image is real and inverted. When m is greater than zero, it is virtual and erect, like every image a convex mirror forms of a real object. With me?

3:29Problem 3: one line with the trick

Let us try the trick on problem three of the set. Here, a convex mirror forms an image one fifth the size of an object forty centimetres in front of it; c m is centimetres. We need its radius of curvature. So tell me first: is m plus one fifth, or minus one fifth? Plus one fifth, that is correct, because for a real object, a convex mirror always gives a virtual, erect, smaller image. And the object is real, so u is minus forty centimetres. Okay? Now the trick: one fifth equals f over f minus u. Minus u is plus forty, so we get f over f plus forty. Cross multiply, and f plus forty equals five times f; the double arrow means this follows. Right? So four times f equals forty, and f is ten centimetres, positive, as a convex mirror should be. Then capital R is two times f, which is twenty centimetres. Clear?

4:30Example 1: shortest mirror for a 1.7 m man to see himself

Now, example one, and it is our opening puzzle: problem one of the set. The man is one point seven metres tall, and he stands two metres from a vertical plane mirror; the m after each number means metres. Find his distance from his image, and the shortest mirror in which he sees himself fully. Pause the video and try it first. Done? Okay, let's check. He stands two metres from the mirror, so he is two metres from his image, right? No, no, wait, that is the trap, because two metres is only the distance to the mirror. The image is as far behind the mirror as he is in front, so it is two metres behind the mirror, like the two equal distances marked d in the figure. Okay? So from the man to his image, it is two plus two, which gives four metres. Got it? Now let us size the mirror. To see his full height, a person needs a mirror of only half his height, and this holds whatever his distance from the mirror. Right? One half times one point seven gives zero point eight five metres. Puzzle solved: not one point seven metres, and stepping back changes nothing. Clear?

5:42Example 2: concave mirror f = 18 cm, real image 1.5 times

Now the harder one: problem two of the set. Here, a concave mirror of focal length eighteen centimetres forms a real image, one point five times the size of the object. Find the object and image distances. Pause the video and try it first. Done? Okay, let's check. The image is one point five times the size of the object, so m is plus one point five, right? No, no, careful, that is the trap, because the image is real. For a mirror, a real image is inverted, so m is minus one point five. And the mirror is concave, so f is minus eighteen centimetres. With me? Now we use the trick again. Minus one point five equals f over f minus u, which is minus eighteen, over minus eighteen minus u. Okay? Now cross multiply: minus one point five, times the whole of minus eighteen minus u, gives twenty seven plus one point five times u, and that equals minus eighteen. So one point five times u is minus forty five, and u is minus thirty centimetres. Right? Now let us find the image distance. From m equals minus v over u, we get v equals minus m times u. The brackets just hold the two negative numbers. Minus one point five times minus thirty is plus forty five, and the minus sign in front makes it minus forty five centimetres. Clear? Both are negative, so both lie in front of the mirror, and the image is real, as the question said. So the object is thirty centimetres away, and the image forty five centimetres. Got it?

7:18Quick revision: half-height rule, signs, m = f/(f - u)

So, quick revision. Plane mirror: the image is as far behind as in front, and a full view needs half the height. Measure from the pole: a real object has u negative, and a concave mirror has f negative. One over v plus one over u equals one over f, and m equals f over f minus u; for a real image, m is negative. Now try all fifteen problems of this set on this page, in thirty minutes, then check the solutions. All the best!