Microscopes

Transcript

0:00Introduction

Hello everyone! Look through a school microscope and gently push the slide to the left. The cell you are watching moves to the right! So why does it move the wrong way? Today you will find out why, and how two small lenses make it look sixty times bigger. Microscopes come up often in JEE, so let's begin.

0:21Simple microscope: one convex lens

Let us start with the simplest microscope, one convex lens close to your eye, the magnifying glass. Without any lens, where do you hold a tiny thing to see it as big as possible? As close as you can while it still looks sharp, yes, exactly! That is the near point, at a distance capital D, twenty five centimetres for a normal eye; c m means centimetres. With me? How big a thing looks depends on the angle it makes at your eye. So the magnifying power, capital M, is the angle of the image at the eye, over the angle of the object placed at the near point. Okay? With the image at your near point, M equals one plus D over f, where f is the focal length. With the image at infinity, the eye is relaxed, and M is just D over f, one less. Right?

1:14The signs we use for a lens

Now our signs, the course's convention for lenses. Every distance is measured from the lens, and along the incident light is plus, against it is minus. Keep this in mind for every lens problem. So u, the object distance, is negative for a real object; the less-than sign means it is below zero. And v, the image distance, is negative for a virtual image on the object's side, as with a magnifying glass. Got it? The lens formula is one over v, minus one over u, equals one over f, for a thin lens and paraxial rays, the rays close to the axis. Okay? And small m, the magnification, is v over u. Negative m means real and inverted; positive m means virtual and erect. With me?

2:03Compound microscope: two lenses

Now the compound microscope in the figure. The black arrow on the left, B, is the object, placed just beyond the focus of the objective, the first lens. Its focal length f o, f with a small o below, is short. Okay? So the objective forms a real, inverted, magnified image, the dark red arrow, B prime; the small dash is called prime. The two coloured lines are rays from the tip of B, crossing exactly at B prime. Right? The tall lens on the right is the eyepiece, with focal length f e, f with a small e below. It is just a magnifying glass for B prime. And the formula at the top right, we will come to in a minute. Got it? Now, a magnifying glass gives an erect image of what it looks at, and B prime is already upside down. So is the final image erect or inverted, compared with the object? Inverted, yes, exactly! The eyepiece does not flip B prime back, so the final image stays upside down. And that solves our puzzle. The image is turned right around, so push the slide left, and the cell seems to move right. Clear?

3:14Magnifying power of a compound microscope

Each lens magnifies in turn, so M equals m o times m e, the magnifications of the objective and the eyepiece. Here m o is v o over u o, the objective's image and object distances, and with the final image at D, m e is one plus D over f e, in brackets. Since u o is negative, M comes out negative, which means inverted. With me? With the final image at infinity, m e is just D over f e, and the dot means times. This is normal adjustment, exactly as in the figure, with the relaxed eye. At the bottom, capital L is the tube length, from F o, the objective's second focus, to F e, the eyepiece's first focus. Then m o equals L over f o in size, and the size of M is approximately L over f o, times D over f e, the vertical bars meaning size. It is the same formula as at the top right of the figure, which shows sizes only; the wavy equals sign means approximately. Got it? So, for a really large magnifying power, should the two focal lengths be long or short? Both should be short, yes, exactly, because f o and f e sit in the denominators. Right?

4:32Two traps: object distance and lens separation

Now two traps. First, the object is just outside the focus, so u o is about f o in size, but never equal to it. Okay? Second, the lenses are v o plus the modulus of u e apart, where u e is the eyepiece's object distance, and the vertical bars mean its size. That is v o plus f e only for the final image at infinity, with B prime at the eyepiece's focus. Clear?

5:00Example 1: magnifying glass, f 10 cm, image at 25 cm

Now our first example. Here, a magnifying glass of focal length ten centimetres forms the image twenty five centimetres from the eye, which is close to the lens. Where is the object, and what is the magnification? Pause the video and try it first. Are you done? Okay, let's check. Many students put v equal to plus twenty five centimetres. Is that right? Oh no, careful, that is exactly the trap! The image is virtual, on the object's side, so v is minus twenty five centimetres, and f is plus ten for this convex lens. Okay? So one over u equals one over v, minus one over f, which is minus one over twenty five, minus one over ten. With fifty at the bottom, they add up to minus seven over fifty. Right? Flip it, and u is minus fifty over seven centimetres, so the object is seven point one four centimetres from the lens. Got it? And the image is at the near point, so M equals one plus twenty five over ten, which gives three point five. With me?

6:09Example 2: compound microscope power and lens separation

Now a harder one. Here, a compound microscope has an objective of focal length one centimetre and an eyepiece of focal length five centimetres. An object is one point one centimetres from the objective, and the final image is at twenty five centimetres. Find the magnifying power and the separation of the lenses. Pause the video and try it first. Are you done? Okay, let's check. For the objective, u o is minus one point one and f o is plus one centimetre. Rearranging the lens formula, v o equals u o times f o, over u o plus f o, with brackets around each number. That is minus one point one over minus zero point one, which gives eleven centimetres. Right? So m o is eleven over minus one point one, which is minus ten, so the image is real and inverted. Clear? With the final image at twenty five centimetres, m e is one plus twenty five over five, which is six. So M equals minus ten times six, which is minus sixty, the sixty from our puzzle, and inverted! Okay? Now, are the lenses eleven plus five, sixteen centimetres apart? Oh no, careful, sixteen is right only when the final image is at infinity! Here the final image is at minus twenty five centimetres, so for the eyepiece, one over u e equals minus one over twenty five, minus one over five. The double arrow means this gives, and the size of u e is four point one seven centimetres. Got it? So the separation is v o plus the modulus of u e, eleven plus four point one seven, which gives about fifteen point two centimetres, not sixteen. With me?

7:59Quick revision: magnifier, compound microscope, separation

So, quick revision. First, a magnifier gives one plus D over f with the image at D, or D over f at infinity. Second, a compound microscope gives m o times m e, about L over f o, times D over f e, and an inverted final image. Third, the lenses are v o plus the modulus of u e apart, which is v o plus f e only at infinity. You have done really well today. Now try the practice problems on this page. All the best!