The Convex Mirror

Transcript

0:00Introduction

Hello everyone, here is a small puzzle before we start. With a concave mirror, we can catch the real image with a pin, and remove the parallax. Now someone hands you a convex mirror, and asks for its focal length. But the image in a convex mirror is always virtual, and it sits behind the mirror. So where will you put your pin, behind the mirror? That does not work here, so we need another way. Today we will borrow a convex lens to help. This belongs to experimental skills in JEE, and questions on it do come up. So, let's begin.

0:38Why the convex mirror needs a lens

So, look at the figure. The left part, labelled small a, shows a convex lens alone, and the right part, labelled small b, shows our convex mirror with a helper, an auxiliary lens. Now, the convex mirror alone gives only virtual images, okay? So let me ask you this. Why can we not locate this image with a pin, the way we do for a concave mirror? Yes, exactly, because the virtual image lies behind the mirror, and no pin can be placed there to coincide with it. So a pin cannot locate a virtual image, right? So we bring in a helper, a convex lens, the shaded shape marked capital L in the figure. The lens gives a converging beam, and that lets us find the mirror's centre of curvature from a real image. Clear?

1:28Step 1: the lens alone

Now step one, in the left part of the figure. The orange arrow standing on the axis is capital O, the object pin. The lens, capital L, forms a real and inverted image of it at capital I, the dark arrow pointing down. Got it? We find the exact place of capital I with a second pin, the image pin, using parallax. We move this pin until its tip and the tip of the image show no shift against each other when the eye moves sideways. Then we note its mark on the bench. With me?

2:00Step 2: bring in the convex mirror

Now step two, in the right part of the figure. The curved line marked capital M is the convex mirror, and we place it between the lens and the point capital I, okay? Now the light from the pin goes through the lens, reflects from the mirror, and comes back through the lens again. We slide the mirror along the bench until this final image of capital O lands on capital O itself. Right? And we check it with parallax again. There must be no parallax between the pin, capital O, and its inverted image. Clear?

2:33Why the point I is the centre of curvature

Now the big question is why the image lands exactly on the pin. Think about one ray. When does a ray, after reflection from a mirror, come back along its own path? Yes, exactly, it happens when the ray strikes the mirror along the normal. In this position the rays strike the mirror normally and retrace their paths, back through the lens to the pin. That is why the pin meets its own image, okay? Now, every normal to a spherical mirror passes through one special point. Which point is that? It is the centre of curvature, and that is correct. Now look at the dashed line behind the mirror: produced, the rays would meet at capital I, so capital I is the centre of curvature, capital C, as the figure shows in brackets. Got it? So the radius of curvature, capital R, equals M I, the distance from capital M to capital I, shown by the double arrow at the bottom of the right part. And the focal length, small f, is capital R over two. Clear?

3:36The condition: the lens image must lie beyond R

Now one condition you must not forget. The mirror sits between the lens and capital I, a distance capital R in front of it. So the lens must form its image farther away than capital R, okay? That is why we choose a lens of suitable focal length, and a suitable object distance. Right? Now try this one. The lens forms the image of the pin eighteen centimetres away, and the convex mirror has a radius of twenty four centimetres. Can we use this method? No, we cannot, and here is why. The mirror would have to be twenty four centimetres in front of the image point. But the lens is only eighteen centimetres from that point, so the mirror would land on the wrong side of the lens. With me? So the fix is to change the object distance, so that the lens forms its image more than twenty four centimetres away. Got it?

4:30Example 1: image at 80.0 cm, mirror at 56.0 cm, find f

Now Example one, an easy start. The lens alone forms the image at the eighty point zero centimetre mark. With the convex mirror at the fifty six point zero centimetre mark, the object pin coincides with its own image. Find the focal length of the mirror. Pause the video and try it first. Are you done? Okay, let us check it together. Capital I is at the eighty point zero mark, and the mirror, capital M, is at the fifty six point zero mark, okay? Notice that the question never tells us where the lens is. Do we need it? We do not need it, because capital R is just the distance from capital I to the mirror. So eighty point zero minus fifty six point zero gives twenty four point zero centimetres. Right? And small f is capital R over two, so twenty four point zero over two gives twelve point zero centimetres. If you wrote twenty four as the focal length, be careful, that is the radius. Clear?

5:34Example 2: lens f 15 cm, pin at 20 cm, mirror 25 cm away

Now Example two, a little harder. An object pin is twenty centimetres from a convex lens of focal length fifteen centimetres. Then a convex mirror twenty five centimetres behind the lens sends the light back, so that the final image coincides with the pin. Find the focal length of the mirror. Pause the video and try it first. Are you done? Okay, let us check it together. The mirror is twenty five centimetres from the lens, and the pin meets its own image. So is capital R simply twenty five centimetres? That would be a mistake, because capital R is measured from the mirror to capital I, so first we find capital I with the lens alone. With magnitudes, one over f equals one over u, plus one over v, where small f here is the focal length of the lens, small u the object distance and small v the image distance. So one over fifteen equals one over twenty, plus one over small v, okay? So one over small v equals one over fifteen, minus one over twenty, which gives one over sixty. So small v is sixty centimetres. With me? Now the mirror is twenty five centimetres from the lens. So the distance from the mirror to capital I is sixty minus twenty five, which gives thirty five centimetres, and that is capital R, the radius. Right? And small f is capital R over two, so thirty five over two gives seventeen point five centimetres. Very good if you found capital I first, that is the whole trick.

7:12Quick revision: lens image I is C, R = MI, f = R/2

So, let us do a quick revision. The convex mirror alone gives a virtual image, so a convex lens helps. The lens forms its image at capital I, and we move the mirror until the pin meets its own image. Then capital I is the centre of curvature, capital R is the distance from the mirror to capital I, and small f is capital R over two, as long as the lens image lies beyond capital R, the radius. Now go and try the practice problems on this page, and all the very best to you.