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# Lens Equation

###### Jonathan Osbourne

###### Jonathan Osbourne

**PhD., University of Maryland**

Published author

Jonathan is a published author and recently completed a book on physics and applied mathematics.

The **lens equation** allows us to understand geometric optic in a quantitative way where *1/d0 + 1/di = 1/f*. The lens equation essentially states that the *magnification of the object = - distance of the image over distance of the object*.

So let's talk about the lens equation, the lens equation allows us to do Geometric optics but in a quantitative way now it will turn out that it works exactly the same way for lenses as it does for mirrors just like Geometric optics does. So I'm going to spend most of my time talking about lenses because they're easier to draw the diagrams for, alright so we've got a diagram like this for a converging lens. Now remember the principal rays we've got coming in parallel out through focus. Remember that the focus is on the proper side of a lens when it's converging. Alright then we've got straight through the vertex no problem then we've got in through the kind of wrong focus and out parallel. The guys all meet there's my image everything is wonderful. Now the way that we're going to name things on here so we can have an equation is do is the distance that the object is along the principal axis from the vertex do the distance, the object from the lens. Alright what about di? That's the distance that the image takes from the lens again along the principal axis from the vertex. And then f is the focal length so that's the distance from the vertex that the focus is on the principal axis.

Alright so just like before we'll draw the little diagrams and put the axis at the focus on either side now the lens equation says that if you do 1 over do plus 1 over di you get 1 over f. You have to be very careful with this equation because it really means 1 over you can't do di plus do equals f because it doesn't okay that's never true. Alright so you need to be careful with that but once you are it's not that difficult to use this equation. The only kind of strange things are what does it mean when di is positive versus di is negative. If di is positive out of this equation then what it means is the image appears on the correct side of the lens. So what that means is that the object is on one side here's the lens, the image is on the other side. If I got di was negative and it's a lens, then the image and the object are on the same side. Now the reason why I say the right side here is that when it's a mirror it kind of goes the other way. Mirrors are supposed to reflect light, so if I've got a mirror here and the object is here, positive di means that the image is on the same side as the object because that's where the light is supposed to go.

Negative di with a mirror would mean that the image is on the other side which is really in darkness there's a wall there. Alright so that's what you should think about di positive means the correct side, the side that you would expect light to go. Alright so with the lens it means you'd expect light to go through with the mirror you'll expect light to bounce off. Alright so that's with di same thing with the focus, focus is positive when it's a converging lens and that means that the focus is on the opposite side of the lens than the object. It's negative when it's a diverging lens and that means that light doesn't actually come from the focus or go through that focus it only kind of pretends as if it came from the focus. Alright so that's the way that we need to interpret it and those are the only tricky things. Now we can also pull out of this a quantitative description of the magnification of the object. Magnification is defined as the size of the image divided by the size of the object and it's also given by negative distance of the image over distance of the object. So what that indicates is that if the distance of the image is positive which means that the image is on the correct side, then the magnification would be negative.

And that means that real images must be inverted if you're only using one lens. If you're using 2 lenses then you might be able to put yourself in a situation where the distance to the object is negative also and then you've got a nice positive magnification. So 2 lenses you can get an upright real image but only 1 lens, if it's a real image it's inverted, if it's a virtual image it's erect upright. Alright so let's do some examples, so over here I've got an object that's 2 centimeters tall so this is the size of the object and it's placed 20 centimeters in front of a converging lens. So this is going to be my distance of the object with a focal length of 15 centimeters. So this is going to be my f, alright now I'm I going to take it as positive or negative? Well it's a converging lens and so that means that my focus should be positive. Alright so let's just use the lens equation, so we'll say 1 over do plus 1 over di equal 1 over f. Alright so 1 over do, well do is 20 so I'll do 1 over 20 I'm not going to convert to SI units here because everything is in centimeters. So I should just understand I'm looking for a length it's going to have to be in centimeters. Alright but you could convert if you wanted to but I think the Math looks easier like that plus 1 over I don't know that so I'm going to call it di equal 1 over f is 15 so I'm going to do 15. So that means 1 over di is equal to one fifteenth minus one twentieth.

Alright now if you go through that you'll find that this is 1 over 60 the easiest way to do this is just type in your calculator 15 to the negative 1 minus 20 to the negative 1 equals and then answer to the negative 1 and you'll get 60. So this means that di is 60 centimeters alright now [IB] describe the image not just write down di is 60 centimeters so what I would say is the image is a real image di's positive that is 60 centimeters on the other side of the lens from the object alright what about it's size? Well so is 2 centimeters alright so that means that the magnification which is minus di over do is negative 3 so that means it's an inverted image that's 3 times bigger so it's going to be 6 centimeters tall but up-side-down. Alright and so that's the answer to this guy not at that, the problems come in when we don't do this Math properly. So if you ended up with negative 5 that's a problem it's not 15 minus 20 it's 1 over 15 minus 1 over 20 and then we have to do 1 over that kind of annoying but I don't know what to tell you, that's the way the equation works.

Alright so let's look down here the next example is everything is exactly the same except it's a diverging lens this time not a converging lens. Alright well what's the difference between a converging lens and a diverging lens? The difference is that the focal length should now be taken to be negative. So we'll start off with the same equation but then when we plug in the numbers it's negative 1 over 15 and that changes everything because now it's 1 over di equals negative 1 over 15 minus 1 over 20. Alright so that if you type in your calculator you'll find that di is negative 8.57 centimeters not as nice as it was before, alright so now what we would say is that the image appears on the same side of the lens as the object it is a virtual image it is 8.57 centimeters away from the lens which is much closer to the lens than the object. Now nice easy thing to remember is that wherever is closer will be smaller. So since this image is closer to the lens it's going to be smaller and that means the magnification would be less than 1.

In fact we can calculate the magnification, magnification is 8.57 divided by 20 because it's negative distance of the image over distance of the object and this is 0.4286 alright the magnification is positive that means that it is an erect or upright image alright. And the size of the image will be equal to this times the size of the object, so it's going to be 0.8572 centimeters okay well just certainly less than the size of the object. And that's the lens equation most important thing to remember is 1 over do plus 1 over di equals 1 over f that and also what does negative distance of the image represent? Remember that, that always represents that the image appears on the wrong side whichever side you do not expect light to be on that's the side that the image is on. Always a virtual image when di is negative and that's the lens equation.

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###### Jonathan Osbourne

PhD., University of Maryland

He’s the most fun and energetic teacher you’ll ever meet. He makes every lesson sound like the most exciting topic ever so you never get bored when he's teaching.

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## aftab ahmed khan · 1 month, 2 weeks ago

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## Halley · 11 months, 3 weeks ago

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## haljabri · 1 year, 7 months ago

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