JEE Advanced Questions (2010–2018): Lenses

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. An object stands thirty centimetres in front of a convex lens, and just behind the lens there is a plane mirror. The light goes through the lens, hits the mirror, and comes back. So where does the final image form? Many students find where the mirror sends the light, and stop there. But they forget one thing. In JEE, lens and mirror combinations like this come up often. Today we solve two past lens questions from IITJEE, and the second one is this very puzzle. So, let's begin.

0:38Thin lens: sign convention and formula

So, first our tools. The figure shows a convex lens on the left, and a concave lens on the right; capital F marks a focus, and two F the point at twice the focal length. Our sign convention: measure every distance from the lens, and distances along the incident light are positive. Okay? Now, u is the object distance. The light travels to the right, and the object stands on the left. So what sign does u take? Negative, yes, exactly! The object is against the light, so a real object has u less than zero; that little sign means less than. Right? Now the lens formula, with v the image distance and f the focal length: one over v, minus one over u, equals one over f. It holds for a thin lens and paraxial rays, the rays close to the axis. With me? And f is greater than zero for a converging lens, like the convex one, and less than zero for a diverging lens, like the concave one. Clear? Positive v means a real image, on the far side; negative v means a virtual image, on the object side, like the one on the right, where the dashed line goes back to the focus. Got it? Last, the magnification, m, equals v over u. Negative m means inverted, like the real image on the left; positive m means erect, like the virtual one. Okay?

1:59Lens maker's formula and lenses in contact

Next, where does f come from? The lens maker's formula. For a thin lens of refractive index n in air, one over f equals n minus one, times, in brackets, one over R one, minus one over R two. R one and R two, with the small one and two below, are the radii of curvature of the two faces, R one for the face the light meets first, signed by the same convention. Got it? Now let me ask you this. What is the radius of curvature of a flat face? It is infinity, very good; that is the sideways eight on screen. So one over R is zero, and a flat face adds nothing. Right? So for a plano-convex lens, flat on one side and curved on the other, one over f equals n minus one, all over R, okay? That means f equals R over n minus one. With me? And for thin lenses in contact, one over capital F equals one over f one, plus one over f two. f one and f two are the two focal lengths, and capital F is that of the combination. Clear?

3:03Lens with a mirror: one element at a time

Now the third tool, for a lens with a mirror, or any system of elements. Work step by step: the image formed by one element is the object for the next. Okay? Measure the new u from the new element, along the direction the light travels now. After a mirror, the light turns back, so the plus direction turns back too. Got it? Now suppose a beam converges towards a point, but the next element stands in the way, before that point. Is that point still an object? Yes, it is, a virtual object! The next element receives a beam converging towards that point, which lies ahead of the light, so its u is greater than zero. Clear?

3:45Example 1: plano-convex lenses of n 1.5 and 1.2 in contact

Now example one, from IITJEE twenty twelve. Two thin plano-convex lenses, of indices one point five and one point two, are placed with their plane faces in contact, making a biconvex lens. Each curved face has radius fourteen centimetres; c m means centimetres. An object is forty centimetres from the lens. Find the image distance. Pause the video and try it first. Done? Okay, let's check. Two different glasses, so we treat it as two thin plano-convex lenses in contact. But first, the two flat faces pressed together in the middle: what do they add to one over f? They add nothing at all, that is correct. So each half gives n minus one, all over R, and these add: one point five minus one is zero point five, and one point two minus one is zero point two. So zero point five over fourteen, plus zero point two over fourteen, gives zero point seven over fourteen, which is one over twenty, and f is twenty centimetres. Okay? Now, the object is forty centimetres away. So u is plus forty, right? No, no, careful, that is the trap! The object is against the light, so u is minus forty centimetres. With plus forty, every number after this goes wrong. With me? Now rearrange the lens formula: v equals u times f, over u plus f. That is minus forty times twenty, over minus forty plus twenty, so minus eight hundred over minus twenty. Right? So v is plus forty centimetres: positive, so the image is real, on the far side. And see, the object at two f, twice the focal length, gives its image at two f. Clear?

5:30Example 2: convex lens 10 cm in front of a plane mirror

Now example two, our puzzle, from IITJEE twenty ten. Here, a biconvex lens of focal length fifteen centimetres stands in front of a plane mirror, ten centimetres from it. Then a point object is thirty centimetres in front of the lens. Find the final image. Pause the video and try it first. Done? Okay, let's check. Step one, the lens alone: u is minus thirty, f is plus fifteen. So v equals minus thirty times fifteen, over minus thirty plus fifteen, which gives thirty centimetres. Okay? But wait. The mirror is only ten centimetres behind the lens. So can the light really reach that point, thirty centimetres away? No, it cannot, the mirror is in the way! The beam converges towards a point thirty minus ten, that is twenty centimetres, behind the mirror: a virtual object for the mirror. Right? Now a plane mirror keeps the distance and flips the side. So the reflected beam converges towards a point twenty centimetres in front of the mirror. With me? So the final image is twenty centimetres in front of the mirror, and we are done, right? No, no, wait, this is the trap from our puzzle! That point is ten centimetres past the lens, on the object's side. So the returning light meets the lens first. Clear? So the lens acts again. Along the returning light, the point is ten centimetres ahead of the lens, so u is plus ten, a virtual object, and f is still plus fifteen. One over v equals one over fifteen, plus one over ten, which is one over six. Got it? So v is plus six centimetres. The final image is real, six centimetres from the lens on the object's side, and sixteen centimetres from the mirror. Puzzle solved: the forgotten thing was the lens, meeting the light a second time! Okay?

7:27Quick revision: lens formula, f = R/(n-1), virtual objects

So, quick revision. The lens formula: one over v, minus one over u, equals one over f, and a real object has u less than zero. Plano-convex, f equals R over n minus one; in contact, one over capital F equals one over f one, plus one over f two. And with several elements, go one at a time, and watch for virtual objects. Now try the practice problems on this page. All the best!