Size and Speed of the Image

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Picture a small object sliding along the axis towards a concave mirror at five centimetres per second. At one moment, its image moves at twenty centimetres per second, four times faster, and in the opposite direction! So how can an image move faster than its own object? By the end of this video, you will calculate that twenty yourself. This idea comes up often in JEE, so let's begin.

0:32Quick recap: signs and the mirror formula

First, a quick recap of the sign convention. In the course figure, light comes in from the left onto a concave and a convex mirror. All distances are measured from the pole, capital P, where the principal axis meets the mirror. Okay? Distances along the incident light are positive, the plus sign, and distances against it are negative, the minus sign. Now, u is the object distance, and the object sits on the left. So is u positive or negative? It is negative, yes, exactly! So, with the less-than sign on screen, u is less than zero for a real object. And f, the focal length, is less than zero for a concave mirror, and, with the greater-than sign, greater than zero for a convex one. Right? Now the two formulas, for paraxial rays, the rays close to the axis. One over v plus one over u equals one over f, with v the image distance, and the magnification m equals minus v over u, which is also f over, f minus u. Clear? When m is less than zero, the image is real and inverted; when m is greater than zero, it is virtual and erect. Got it?

1:47Differentiate the mirror formula

Now comes the new idea. If the object moves a tiny bit along the axis, the image moves too. The mirror does not change, so f is fixed, and we differentiate both sides. So tell me, d of one over u, what is it? It is minus d u over u squared, yes! In the same way, d of one over v is minus d v over v squared, and one over f is a constant, so its change is zero. So, minus d v over v squared, minus d u over u squared, equals zero. With me? Rearrange, and d v over d u equals minus v squared over u squared. Since m is minus v over u, that is just minus m squared. This line is the heart of today's lesson, so write it down. Okay?

2:36Size: along the axis and across it

So, the first use is size. For a short object of length b lying along the axis, the small change d u is just b, so the image length is m squared times b. Clear? Across the axis, a height h becomes h prime, the h with a small dash, and h prime equals m times h. So across the axis the factor is m, and along the axis it is m squared. Right? Now let me ask you this. Take a small flat area perpendicular to the axis. How many times is its area magnified? It is m squared times, very good! Its length and its breadth both lie across the axis, so each picks up a factor of m, and the area picks up m times m. Got it? Now here comes a trap question. Suppose a long rod lies along the axis. Its image length is simply m squared times its length, right? No, no, wait, that is the trap! The m squared rule came from a derivative, so it works only for a short object; along a long rod, m changes from one end to the other. So find the images of the two ends separately, and subtract. With me?

3:49Speed: along the axis and across it

Now let us look at speed. v O, the v with a small capital O below it, is the velocity of the object along the axis, relative to the mirror. Careful, v alone is still the image distance. Okay? The velocity of the image is v I, the v with a small capital I below it. In each small step, the image moves d v over d u times as far as the object. So v I equals d v over d u, times v O, which is minus m squared times v O, clear? Now look at the sign. m squared is never negative. So what does this minus sign tell you about the directions? Yes, exactly, the object and the image always move in opposite directions along the axis! If the object moves to the right, the image moves to the left. Right? And what if the object moves across the axis instead? Remember, heights pick up a factor of m. So what is the velocity of the image now? That is correct, it is just m times v O, with no square, because heights pick up only one factor of m. Okay?

4:57Example 1: length of the image of a 2 mm pin on the axis

Now, our first example. Here is a pin, two millimetres long, the m m on screen, lying along the axis of a concave mirror of focal length twenty centimetres, the c m on screen. The pin is thirty centimetres from the mirror. Find the length of its image. Pause the video and try it first. Done? Okay, let's check. The object is real, so u is minus thirty centimetres, and the mirror is concave, so f is minus twenty centimetres. With me? Now, m equals f over, f minus u, which is minus twenty, over minus twenty plus thirty, that is, minus twenty over ten. So m is minus two, which means a real, inverted image. Right? So the size of m is two, and the image of the pin is two times two millimetres, which is four millimetres long, right? Oh no, careful, the pin lies along the axis! Along the axis we use m squared, which is four. So the image length is four times two millimetres, which is eight millimetres. Clear?

6:03Example 2: image speed with the object at 60 cm and 30 cm

Now our second example, and it settles our puzzle. An object moves along the axis towards a concave mirror of focal length twenty centimetres, at five centimetres per second; that is the c m s to the power minus one on screen. Find the speed of the image when the object is, part A, sixty centimetres, and part B, thirty centimetres from the mirror. Pause the video and try it first. Done? Okay, let's check part A first. With u minus sixty and f minus twenty, m equals minus twenty, over minus twenty plus sixty, which is minus twenty over forty. So m is minus one half. With me? Now tell me, which one multiplies the five here, m or m squared? It is m squared, because the motion is along the axis, very good! So the speed is one quarter times five, which is one point two five centimetres per second. Right? Now part B, at thirty centimetres. It is the same mirror and distance as in our first example, so m is minus two. m squared is four, so the speed is four times five, which is twenty centimetres per second. There it is, the twenty from our puzzle! Okay? And now, the object moves towards the mirror. So which way does the image move? It moves away from the mirror, that is correct, because the minus sign said opposite directions. Got it?

7:31Quick revision: m squared along the axis, m across, long rod

So, quick revision; the cross on screen is the times sign. Along the axis, a length picks up a factor of m squared, and v I equals minus m squared times v O, so the object and the image move in opposite directions. Across the axis, height and velocity pick up a factor of m, and a small area a factor of m squared. And for a long object, image the two ends separately. Now go and try the practice problems on this page. All the best!