The Thin Lens Formula and Magnification

Transcript

0:00Introduction

Hello everyone! Put a candle and a screen ninety centimetres apart, and slide a convex lens from the candle towards the screen. At one spot, the flame's image on the screen is sharp. Keep sliding, and at a second spot, it is sharp again, but smaller! So how can one lens give a sharp image in two places? Today, you will find that lens's focal length from these two spots. The lens formula comes up often in JEE, so let's begin.

0:31Sign convention and the lens formula

First, let us fix the signs. In the figure, light comes from the left. All distances are measured from the centre of the lens. Along the incident light is positive, the plus sign, and against it is negative, the minus sign. Now, u is the object distance, and the object, the black arrow, sits on the left. So, is u positive or negative? It is negative, yes, exactly! We measure back to the object, against the light. So for a real object, u is less than zero; the pointed sign you see there is read as less than. Okay? Next comes f, the focal length. It is greater than zero for a converging lens, like the convex one, and less than zero for a diverging lens, like the concave one. Right? And now the lens formula: one over v, minus one over u, equals one over f, with v the image distance. It holds for a thin lens and paraxial rays, the rays close to the axis. Careful with that minus sign; the mirror formula had a plus there. Clear? Next, look at what the sign of v tells you. When v is greater than zero, the image is real, on the far side, like the dark red arrow in the left figure. When v is less than zero, it is virtual, on the object's side, like the small dark red arrow in the right figure. Got it?

1:55Magnification: size and nature of the image

Now let us look at magnification. m equals h prime over h, image height over object height; the small dash on h is called prime. Heights above the axis are positive, and for a lens, m is just v over u. So, for a real image, v is positive and u negative. Is m positive or negative? It is negative, that is correct, and that means an inverted image. For a virtual image, v and u are both negative, so m is positive, and the image is erect. With me? And two shortcuts: m equals f over, f plus u, and also f minus v, over f. But be careful here: for a mirror, it was f over, f minus u, so do not mix them up, okay?

2:41Convex lens: where is the image?

Now, where does the image form? In the left figure, F is the focus, and two F is at twice the focal length. With the object beyond two F, the image, the dark red arrow, forms between F and two F, real, inverted and smaller. Right? Bring the object closer, to two F, and the image is at two F on the other side, real, inverted and the same size. Clear? Now move it closer still, between two F and F, and tell me, where does the image go, and how big is it? It goes beyond two F, and it is larger, yes, exactly! Still real, still inverted, and this is our first example. Okay? And between F and the lens? Now the image is virtual, erect and larger, on the same side as the object. With me? Now the right figure, a concave lens. Wherever the object is, its image is always virtual, erect and smaller, between the lens and its focus. Okay?

3:40Least distance and the displacement method

Remember the table from a moment ago. Object at two F, image at two F on the other side. How far apart are the object and the image? Two f plus two f, four times f, correct! That is the least possible distance between a real object and its real image, with the object at two f. Bring them any closer, and the lens cannot form a real image on a screen. Right? Now comes our puzzle. Fix the object and the screen a distance capital D apart, with D greater than four times f. Then two lens positions give a sharp image, a distance small d apart, just like our candle! Clear? The focal length is capital D squared minus small d squared, over four capital D; the small two up top means squared. Measure two distances, and you have f. Got it? In m one and m two, the small numbers below are labels, and the straight bars mean the size only, without the sign. The size of m one is capital D plus small d, over capital D minus small d, which is one over the size of m two. And the object height is the square root of h one times h two, the image heights; that tick is the square root sign. With me?

4:56Example 1: convex lens f = 20 cm, object at 30 cm, find v, m

Now example one, an easy one. An object is thirty centimetres from a convex lens of focal length twenty centimetres; the c m on screen means centimetres. Find the image and its magnification. Pause the video and try it first. Done? Okay, let's check. The object is thirty centimetres away, so u is plus thirty, right? No, no, careful, that is the trap! The object is on the left, against the light, so u is minus thirty. The lens is convex, so f is plus twenty. Okay? From the lens formula, v equals u times f, over u plus f. That is minus thirty times twenty, over minus thirty plus twenty, so minus six hundred over minus ten, which gives sixty centimetres. Since v is positive, the image is real. Right? Now m equals v over u, and sixty over minus thirty gives minus two. So the image is real, sixty centimetres beyond the lens, inverted, and twice the size. Just as our table said, the image is beyond two F, and larger.

6:06Example 2: displacement method, D = 90 cm, d = 30 cm, find f

Clear? Now example two, our puzzle from the start. An object and a screen are ninety centimetres apart. Here, a convex lens gives a sharp image in two positions, thirty centimetres apart. Find the focal length and the two magnifications. Pause the video and try it first. Done? Okay, let's check. Capital D, object to screen, is ninety centimetres, and small d, the gap between the lens positions, is thirty. With me? So f equals ninety squared minus thirty squared, over four times ninety; the cross means times. Eight thousand one hundred minus nine hundred is seven thousand two hundred, and seven thousand two hundred over three hundred and sixty gives twenty centimetres. Okay? Now the size of m one is capital D plus small d, over capital D minus small d, so one hundred and twenty over sixty gives two. And the size of m two is one over that, which is one half. Got it? Now, two and one half are both positive numbers, so both images are erect, right? No, no, wait, that is the trap! These formulas give only the sizes. Both images fall on a screen, so both are real, and inverted. With signs, m one is minus two, and m two is minus one half. Clear? See, in the first position, u is minus thirty and v is sixty, exactly example one. In the second, they swap, so u is minus sixty and v is thirty. Both add up to ninety, and the lens moved thirty centimetres, which solves our puzzle.

7:47Quick revision: lens formula, m = v/u, displacement method

Got it? So, quick revision. Along the light is positive, a real object has a negative u, and one over v, minus one over u, equals one over f. Then m equals v over u: negative means real and inverted, positive means virtual and erect. And the displacement method: f equals capital D squared minus small d squared, over four capital D, when D is greater than four times f. Now go and try the practice problems on this page. All the best!