Combinations of Lenses and Mirrors

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take a plane mirror. On its own, for a real object, it only gives a virtual image, behind the mirror, right? Now put it right behind a convex lens of focal length twenty centimetres, touching it, and place an object thirty centimetres in front. Suddenly, a real image appears, in front of the lens! So how can a plane mirror help to make a real image? By the end of this video, you will know exactly where that image sits. Combinations of lenses and mirrors come up often in JEE, so let's begin.

0:39Thin lenses in contact: the powers add

First, the ground rules. Everything here is for thin lenses and paraxial rays, that is, rays close to the axis. Distances are measured from the lens, and they are positive along the direction of the incident light, okay? So, a real object sits before the lens, against the light, and that makes u, the object distance, less than zero. And the focal length f is greater than zero for a converging lens, and less than zero for a diverging lens. Remember this sign. Clear? Now, let me ask you this. Two thin lenses touch each other. How do you get the focal length of the pair? You simply add the reciprocals! One over capital F equals one over f one, plus one over f two, and so on. Capital F is the focal length of the pair, the small one and two are subscripts, and the three dots mean more lenses can follow. Got it? Now, one over f is the power, written as P, and with f in metres, it is measured in dioptres. So for lenses in contact, the powers simply add: P equals P one plus P two, and so on, okay? And the magnification of the combination is the product, so m equals m one times m two, and so on. With me?

1:56Two lenses a distance d apart

Now, pull the two lenses apart, with a distance d between them. One over capital F equals one over f one, plus one over f two, minus d over f one times f two. Put d equal to zero, and we are back to lenses in contact. Nice check, right? Here is a question for you. Suppose a parallel beam goes in. For what separation does it come out still parallel? Yes, exactly, when one over capital F is zero, so the pair has no focusing power at all! Put that into the formula, and you get d equals f one plus f two. Okay? But in any other case, do not hunt for a magic formula. Just go step by step. Clear? The image formed by one element becomes the object for the next one. Right? And here is the trap. Measure the new u from the new element, not from the first lens. Many students keep the old distance and lose easy marks, so be careful, okay?

2:54Silvered lens: a lens that acts as a mirror

Now the most interesting one. On the left of the figure is a lens whose back face is silvered, the thick grey layer. The orange ray enters, reflects from the silvered face, and comes back out, so the whole thing acts as a mirror. Now tell me, how many times does the light pass through the lens? Yes, exactly, the light passes through the lens twice! It crosses the lens, reflects, and crosses the lens again. That is why, after the three-line sign, which means is equivalent to, the figure shows lens, then mirror, then lens. Right? So we add the powers along the way. P equals two times P L, plus P M, where the small L and M are subscripts for the lens and the mirror. Okay? Here, P L is one over f L, the focal length of the lens, and P M is minus one over f M, with a minus sign in front. Why the minus sign? Because a concave mirror has a negative focal length in our convention, and yet it converges light, so its power must be positive. Clear? And f M is the focal length of the silvered face acting as a mirror, taken with its sign. Got it? Finally, capital F, the focal length of the whole system as a mirror, equals minus one over capital P, okay? So if P comes out greater than zero, the system is just a concave mirror, with focal length one over P, as simple as that!

4:15Example 1: convex 20 cm and concave 30 cm lenses in contact

Alright, now two questions from the course. Here is the first one. We have a convex lens of focal length twenty centimetres, in contact with a concave lens of focal length thirty centimetres. Find the focal length of the combination. Pause the video, and try it first. Done? Okay, let's check. The lenses are in contact, so add the reciprocals. One over twenty, plus one over thirty, gives one over twelve, so twelve centimetres, right? No, no, wait, that is exactly the trap we talked about! The concave lens is diverging, so its focal length is negative. So f one is plus twenty centimetres, and f two is minus thirty centimetres. Okay? Now, one over capital F equals one over twenty, minus one over thirty. Take sixty as the common denominator: three minus two, over sixty, which gives one over sixty. Right? So capital F is plus sixty centimetres. It is positive, so the combination is converging, because the convex lens, with the shorter focal length, is the stronger lens.

5:28Example 2: convex lens with plane mirror behind, find image

Clear? Now, remember our puzzle from the start? Let's finish it. We have a convex lens of focal length twenty centimetres, in contact with a plane mirror behind it, and an object thirty centimetres in front. Find the image. Pause the video and try it first. Here is a hint: this is just a silvered lens. Done? Okay, let's check. So first, tell me, what is the power of a plane mirror? It is zero, that is correct, because a plane mirror has an infinite focal length. So P M is zero. Okay? Then P equals two times P L, which is two over twenty, that is, one over ten per centimetre. The centimetre to the power minus one means per centimetre, since f was in centimetres. Right? So capital F equals minus one over P, which is minus ten centimetres. P is positive, so the system is a concave mirror of focal length ten centimetres. And look, ten is exactly half of twenty, the focal length of the lens. That is a topper's trick: with a plane mirror, P M is zero, so the equivalent mirror has exactly half the focal length of the lens. Got it? Now we use the mirror formula, one over v plus one over u equals one over f, where v is the image distance. The object is real, so u is minus thirty centimetres, and for our concave mirror, f is minus ten centimetres. With me? Rearranging, v equals u times f, over u minus f. Minus thirty times minus ten gives three hundred, and u minus f is minus thirty plus ten, which is minus twenty. Three hundred over minus twenty gives minus fifteen centimetres. Now, v is negative. So is the image in front of the mirror, or behind it? That is right, the image is in front of the mirror! For a mirror, a negative v means the image is in front of it, so it is real. So the image is real, fifteen centimetres in front of the lens. For a real object, the plane mirror alone could never do this, but with the lens, it can, and that solves our puzzle!

7:38Quick revision: lenses in contact, apart, silvered lens

So, quick revision. Lenses in contact: the powers add, and a concave lens goes in with a negative focal length. Lenses apart: subtract d over f one times f two, or go step by step. And a silvered lens: P equals two times P L, plus P M, and then capital F equals minus one over P, okay? Now go and try the practice problems on this page. You will find them easy. All the best!