Parallax and the Concave Mirror

Transcript

0:00Introduction

Hello everyone, try this right now. Hold up one finger, with a far wall behind it, and move your head a little to one side. The finger seems to slide across the wall, right? Now here is the puzzle for you. When a concave mirror forms a real image of a pin, that image hangs in the air, and you cannot touch it. So how do you find exactly where it is? We will do it with a second pin, and that same sideways movement of your eye. This is part of experimental skills in JEE, and questions on it do come up. So, let's begin.

0:37Parallax: when the eye moves sideways

So, what you just saw with your finger has a name, parallax. It is the apparent shift of one object against another object at a different distance, when your eye moves sideways. Okay? Now think about the finger and the wall again. When you move your head, one seems to move with your eye, and the other against it. So which one moves with your eye, the nearer or the farther? If you said the farther one, that is correct. The farther object appears to move with the eye, and the nearer object moves against it, like your finger. Right? Now the key idea of this whole experiment. When two objects show no relative shift as you move your eye, they are at the same place. Remember this one line, because it is the heart of today's lesson. Clear?

1:25Locating the image with two pins

Now look at the left part of the figure. The curved line is the concave mirror, and capital P marks its pole. The arrow with the orange head, pointing up, is the object pin, capital O, placed beyond the focus. But why beyond the focus? What goes wrong if it is closer than that? Beyond the focus, the mirror forms a real image of the pin, and it is inverted, upside down. That real image is what our second pin can reach. Okay? But between the pole and the focus, the image is virtual, and it lies behind the mirror. No pin can be placed back there to coincide with it, so the method fails. With me? Now the image pin, capital I, at the dark arrow pointing down. We move it until its tip and the tip of the image of the object pin show no parallax, tip to tip, because the image is inverted. Then this pin marks the image, as the note under the figure says. Clear? Now measure both distances from the pole, the two double arrows at the bottom: small u for the object pin, and small v for the image pin. We take them as magnitudes, with no signs. Right? Then one over f equals one over u, plus one over v, where f is the focal length. Or in one step, f equals u times v, over u plus v. Got it?

2:46Two graphs that give the focal length

With several pairs of readings, we draw graphs. In the middle graph, small v goes up against small u, and the orange curve is a rectangular hyperbola. The dashed line through the origin is the line v equals u. So tell me, at which point of the curve is v equal to u? Put v equal to u in our formula, and one over f becomes two over u. So u equals two f, and the line meets the curve at the point two f, two f, as the dashed lines show. Read it, halve it, and you have the focal length. Okay? Now the graph on the right: one over v going up, against one over u going across. This one is a straight line, with a slope of minus one. Why exactly minus one? Because our formula can be written as one over v equals one over f, minus one over u. So the slope is minus one, and the line cuts both axes at one over f, as the figure shows. Invert an intercept, and you get f. Clear?

3:51Index correction with a knitting needle

Now here is a practical point that students often miss. The bench scale gives the separation of the uprights that hold the mirror and the pins. But that is not the distance between the pole and the pin tip, right? So we find the index correction with a knitting needle held between them. Index correction equals the true length of the needle, minus the separation read on the bench. Okay? And this correction is added, with its sign, to every observed distance, both u and v. If it comes out negative, the distances get smaller. Clear?

4:28The error in the focal length

One last idea: the error in the focal length. The little triangle is capital delta, and delta f is the error in f. Delta f over f squared equals delta u over u squared, plus delta v over v squared. With me? And when you write the answer, quote the error to one significant figure, and give f to the same decimal place as the error. Right?

4:51Example 1: f from u = 30.0 cm and v = 60.0 cm

Now the examples, the easier one first. With a concave mirror, the object pin is thirty point zero centimetres, and the image pin sixty point zero centimetres from the pole, when parallax is removed. Find the focal length. Pause the video and try it first. Done? Okay, let's check. One over f equals one over thirty, plus one over sixty, which is three over sixty. That gives zero point zero five zero zero per centimetre, and the small minus one written above the c m on screen means per centimetre. So the focal length is zero point zero five zero zero centimetre, right? No, and this is the trap, because that is one over f, not f itself. Invert it, and one over zero point zero five zero zero gives twenty point zero centimetres. Clear? Or use the faster form, f equals u times v, over u plus v. Thirty point zero times sixty point zero, over ninety point zero, gives twenty point zero centimetres in one line. Got it?

5:56Example 2: error in f when u and v are known to 0.1 cm

Now the harder one, with the same readings. Each distance in the last example is known to zero point one centimetre. Find the error in the focal length. Pause the video and try it first. Done? Okay, let's check. Multiply our error formula by f squared. So delta f equals f squared, times the bracket, delta u over u squared, plus delta v over v squared. Okay? Now f is twenty, so f squared is four hundred. Delta u and delta v are both zero point one. And u squared is nine hundred, while v squared is three thousand six hundred. Right? Before you add, look at the two fractions inside the bracket. Which one is bigger, the term from u or the term from v? The term from u is bigger, four times bigger. Over a common denominator of three thousand six hundred, the u term becomes zero point four, and the v term is zero point one. So four hundred times zero point five, over three thousand six hundred, gives two hundred over three thousand six hundred. That is about zero point zero five six centimetre, and the wavy equals sign means approximately. With me? To one significant figure, the error is zero point zero six centimetre. So f goes to the second decimal place, and f equals twenty point zero zero, plus or minus zero point zero six centimetre, with the plus or minus sign on screen. If you got this on your own, very well done!

7:32Quick revision: no parallax, f = uv/(u+v), index correction

So, quick revision. First, no parallax between the tips means the image pin marks the image. Second, f equals u times v, over u plus v, v equals u at two f, and the straight line graph has intercepts of one over f. Third, add the index correction to every distance, and remember the error formula. Now you can catch an image hanging in the air. Go and try the practice problems on this page, and all the best!