The Convex Lens

Transcript

0:00Introduction

Hello everyone, a small puzzle first. Fix an object pin and a screen one metre apart, and slide a convex lens between them. At one place you get a sharp, big image. Keep sliding, and there is a second sharp image, now a small one! Same pin, same screen, and two sharp images? How is that possible? Today we find the focal length of a convex lens in three ways. Questions on them do come up in JEE, so let's begin.

0:31Method 1: the u–v method

So, method one, the u v method. In part small a of the figure, the shaded shape, capital L, is the convex lens, and part small b is the convex mirror, for another lesson. The orange arrow, capital O, is the object pin, with rays from its tip. Now tell me, must this pin be inside the focus, or beyond it? Yes, exactly, the pin goes beyond the focus. Inside the focus the image is virtual, on the object's side, and no pin on the far side can meet it. Beyond it, the lens forms a real, inverted image at capital I, the dark arrow pointing down. Okay? To locate this image, we use a second pin, the image pin. We move the eye sideways and slide the image pin until its tip and the image show no parallax, no shift against each other. With me? Now we read small u, from the pin to the lens, and small v, from the lens to the image, the double arrows at the bottom. With magnitudes, so both positive, one over f equals one over u, plus one over v, where small f is the focal length. So f equals u times v, over u plus v, the product over the sum. Clear?

1:44The two graphs: same as for the concave mirror

Now the graphs. They are the same as for the concave mirror, whose set-up is on the left of this figure. In the middle graph, small v is plotted against small u: the orange curve, with a dashed line where v equals u. So tell me, for which object distance is the image distance the same? Yes, exactly, it happens when small u is two f. So the curve meets the dashed line at the point two f, two f, in brackets. Take half of that reading, and you have the focal length. Got it? Now the right graph: one over v against one over u. It is a straight line, and it cuts both axes at one over f. So one over the intercept gives the focal length. Right?

2:25Least distance between an object and its real image

Now remember this fact. The least distance between an object and its real image is four f, four times the focal length. Okay? Try this one from the course. The focal length is fifteen centimetres, and the two pins are fifty centimetres apart. Can any lens position give a real image on the second pin? Let us check it. The least distance is four times fifteen, which gives sixty centimetres. Right? But the pins are only fifty centimetres apart, less than sixty. So no position of the lens gives a real image on the second pin. Many students say, just keep sliding the lens and the image will come. That is the trap, because below sixty centimetres it never comes, okay?

3:08Method 2: the displacement method

Now, back to our puzzle. Fix the object pin and the screen a distance capital D apart, with capital D more than four f, the least distance we just saw, right? Now slide the lens between them. Tell me, how many positions of the lens will give a sharp image on the screen? Yes, exactly, there are two, and that answers our puzzle. The distance between these two lens positions is small d. With me? Then f equals capital D squared minus small d squared, all over four capital D, okay? The little two means squared. See, no small u and no small v, only capital D and the shift of the lens. So the position of the optical centre does not enter, and that is the big advantage. Clear? The two images have heights h one and h two, with little numbers written below the h. Then the object size is the square root of h one times h two, the square root sign on screen. Got it?

4:05Method 3: the plane mirror method

Now method three, the plane mirror method. We lay the lens on a plane mirror, and move a pin above it, along the axis of the lens. Okay? At one height, the pin coincides with its own image, with no parallax. Which special point of the lens is that? Yes, exactly, it is the focus. So the height of the pin above the lens is the focal length. Right? The reason is this. Light from the focus leaves the lens parallel, meets the mirror normally, and returns along the same path to the pin. Simple, isn't it?

4:40Example 1: f from u = 24 cm and v = 72 cm

Now Example one, an easy start. Here, a convex lens forms the image of a pin twenty four centimetres away, at seventy two centimetres on the other side. Find the focal length. Pause the video and try it first. Are you done? Okay, let us check. The pin is the object, so small u is twenty four centimetres, and small v is seventy two, the two double arrows in the figure, both magnitudes, okay? Use the product over the sum. Twenty four times seventy two, over their sum, ninety six, gives eighteen. So small f is eighteen centimetres. Right? Always check the conditions. Twenty four is more than eighteen, so the object is beyond the focus. And small u plus small v is ninety six, more than four f, which is seventy two. Everything fits, well done.

5:33Example 2: D = 100 cm, d = 20 cm, find f, u and object size

Now Example two, harder: our puzzle with real numbers. Capital D is one hundred centimetres, the lens positions are twenty centimetres apart, and the image heights are four point five and two point zero centimetres. Find the focal length, the two object distances and the height of the object. Pause the video and try it first. Are you done? Okay, let us check. Capital D squared is ten thousand, small d squared is four hundred, and four capital D is also four hundred. So small f is ten thousand minus four hundred, over four hundred, which gives twenty four centimetres. With me? Now the object distances. They are capital D minus or plus small d, over two, and the minus or plus sign gives two answers. One hundred minus twenty, over two, gives forty, and one hundred plus twenty, over two, gives sixty centimetres. Okay? Let us check with method one. When small u is forty, small v is one hundred minus forty, which is sixty. Forty times sixty, over forty plus sixty, gives twenty four centimetres again. Both methods agree, right? Many students take the average of four point five and two point zero, which is three point two five centimetres. So, is that the height of the object? No, it is not, and this is the trap. The height is the square root of h one times h two, not the average. Four point five times two point zero is nine, and its square root gives three point zero centimetres. Very good if you got that.

7:08Quick revision: u-v, displacement and plane mirror methods

So, quick revision. In the u v method, f equals u times v, over u plus v, and the graphs are the same as for the concave mirror. The least distance from an object to its real image is four f. In the displacement method, f equals capital D squared minus small d squared, over four capital D, with capital D more than four f. And in the plane mirror method, the pin meets its own image at the focus. Now go and try the practice problems on this page. All the very best!