Practice: A biconvex lens has a focal length of 12 cm. If an object is placed 5 cm from the lens, where is the image formed? Is it real or virtual? Is it upright or inverted? What’s the height of the image if the object is 2 cm tall?

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Ray Nature Of Light | 11 mins | 0 completed | Learn |

Reflection Of Light | 12 mins | 0 completed | Learn Summary |

Refraction Of Light | 32 mins | 0 completed | Learn Summary |

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Ray Diagrams For Mirrors | 36 mins | 0 completed | Learn |

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Ray Diagrams For Lenses | 23 mins | 0 completed | Learn |

Thin Lens And Lens Maker Equations | 25 mins | 0 completed | Learn |

Concept #1: Thin Lens Equation

**Transcript**

Hey guys, in this video we're going to cover something called the thin lens equation. Just like for mirrors, we used ray diagrams to find qualitative information about the images but when it came down to actual numbers, we relied on the mirror equation. The thin lens equation is going to replace ray diagrams for lenses and allow us to find numeric answers to where the image is located, the height of the image, etc. Alright let's get to it. For images produced by lenses really all we're going to consider are thin lenses. What a thin lens is is it's a lens composed of two pieces of glass that are spherical and all that it means to be spherical is that it's part of some imaginary sphere with some radius of curvature. And to be thin, the radius of curvature has to be a lot larger than the thickness of the lens. For instance, maybe the radius of curvature is 10 centimeters for a particular piece of glass, but when you put 2 of them together the lens is actually only a couple millimeters thick, so it's very very thin. Thin lenses come in five basic types. We have our biconvex. We have our convex concave, convex from one side, concave from the other. We have our plano-convex. It's flat or plane from one side and it's convex from the other.We have our biconcave lenses which are concave from both sides and lastly we have our plano-concave lenses. Concave from one side, plane from the other. Now the rule of thumb for deciding which of the lenses are converging and which of the lenses are diverging are the lenses that are the thickest in the middle are converging lenses, and lenses that are the thinnest in the middle are diverging lenses. So you can look at this lens right here, it's thick in the middle, thin at the edges. That's a converging lens. This lens, thin in the middle, thick in the middle sorry, thin at the edges that's a converging lens. For the biconcave, thin in the middle, thick at the edges. That is a diverging lens. For the plano-concave, thin in the middle, thick at the edges. That is a diverging lens the only one of for grabs is the convex concave lens. That one could be either converging or diverging. It just depends on the radius or the radii of curvature for the two pieces of glass that make it up. For a lens with some focal length F, the image location is given by the thin lens equation and if you notice, this is the exact equation as the mirror equation because this is all based on geometry and it actually doesn't matter whether you have a spherical mirror or a spherical lens. That was why our rules for drawing rate diagrams were almost identical for lenses and for mirrors the equation too is identical. Now, the sign conventions that are important are if the lens is converging, remember the rule of thumb for a converging lens is it's thick in the middle, thick at the edges, the focal length is positive. If the lens is diverging, the focal length is negative. And just like always, if the image distance is positive it is a real image and it is an inverted image. You guys should have this memorized down packed by now. There have been so many videos that cover this exact concept and if the image distance is negative, it's virtual and it's upright. We also have a magnification equation for thin lenses that is identical to that of mirrors. So luckily you don't have to memorize a new set of equations. The thin lens equation is identical to the mirror equation and magnification for images produced by thin lenses is identical to the magnification for images produced by mirrors. Let's do a quick example. A biconcave lens has a focal length of 2 centimeters, if an object is placed 7 centimeters in front of it, where is the image located? Is this image real or virtual? Is it upright or inverted? Now this biconcave lens looks like this. Actually, I'm going to minimize myself I'm going to draw off the side where we don't really need it except to illustrate this point. Biconvex, sorry, biconcave means that the image is concave on sorry the lens is concave on both sides. So this is thin in the middle and thick at the edges, so this is a diverging lens. If it has a focal length of 2 centimeters, since it's diverging, that focal length has to be negative so the focal length is going to be -2 centimeters. Besides that now we can use the thin lens equation. The object is placed 7 centimeters in front of the lens so the object distance is 7 centimeters. So one over S I plus 1 over S O equals one over F. We're going to isolate the 1 over S I which is 1 over F minus 1 over S O. This is 1 over -2 right that's the sign for sorry yeah that's the sign for the focal length minus 1 over 7. This whole thing is going to be -0.64 but we still have to reciprocate our answer. This is only the solution for 1 over S I, it's not the solution for S I. So if I reciprocate the answer, we get -1.6 centimeters. So is this image real or is it virtual? Well the image distance is negative so you should know automatically that this is a virtual image. And since it's virtual automatically, it's upright. Alright guys that wraps up our discussion on the thin lens equation. Thanks for watching.

Practice: A biconvex lens has a focal length of 12 cm. If an object is placed 5 cm from the lens, where is the image formed? Is it real or virtual? Is it upright or inverted? What’s the height of the image if the object is 2 cm tall?

Concept #2: Lens Maker Equation

**Transcript**

Hey guys, in this video we're going to talk about something called the lens maker equation which is the equation that will tell us the focal length of a thin lens based on the shape of the two pieces of glass that make it up. Alright let's get to it. The focal length of the thin lens depends upon three things. It depends on the radius of curvature of the near glass, what I mean by near is if I have an object over here, this face is the nearer glass. So depends upon that radius of curvature, it depends upon the radius of curvature of the far glass, the glass on the opposite side, so that radius of curvature and it depends on the index of refraction of the glass itself whatever that index is. The lens maker equation is going to tell us what the focal length of this thin lens is going to be and it is N minus 1 times 1 over R1, where our one is the radius of the near glass minus 1 over R2 where R2 is the radius of the far glass. Now remember that this equation is 1 over the focal length it's not the focal length so don't forget to reciprocate your answer. There is an important sign convention that we need to know in order to apply this equation. If the center of curvature is in front of the lens like this guy right here, the near sorry the far glass has a center of curvature on the front side of the lens, then the radius is negative. If the center of curvature is behind the lens like this guy, the near glass, then the radius is positive. So this radius is positive this radius is negative, alright? Let's do a quick example to illustrate this point. The following lens is formed by glass with a refractive index of 1.52. What is the focal length of the following lens when an object is placed in front of the convex side? What if an object is placed in front of the concave side. So first, I'll apply the lens maker equation to find it if an object is placed here in from the convex side. That's going to be N minus 1, 1 over R1 minus 1 over R2. The index of refraction of the glass is 1.52 minus 1, what is the radius of the near glass? That's 10 centimeters. Is it positive or negative? It's positive because the center of curvature is behind the lens. So this is 1 over positive 10 minus 1 over positive 7. That radius of curvature is also behind the lens. Plugging this into a calculator, we're going to get -0.022 but don't forget we have to reciprocate our answer so F1 is -45 centimeters, that's what the focal length is if you place an object in front of the convex side, alright. Now for the second part I'm going to minimize myself so that I don't get in the way and what would happen if we were to place an object here? What would the focal length be? Well, we're going to use the same lens maker equation, 1 over F2, N minus 1, 1 over R1 minus 1 over R2. The N is the same, 1.52 minus 1. What about R1? Now what's the near glass? The near glass is the 7 centimeters. Is the center of curvature in front of the lens or behind the lens? Now it's in front of the lens and if it's in front of the lens it's negative so this is -7 minus 1 over -10 for that 10 centimeter piece of glass, the center of curvature is also in front of the lens. Plugging this into a calculator, we get -0.022. Well look at that we got the same answer regardless of which side of the lens we put our object on, and this is actually a fundamental result of the lens maker equation that no matter what side of the lens you put the object on, you're going to have the same focal length. When we were drawing ray diagrams for lenses, we assumed that the focus was at the same distance on either side of the lens. This is a fundamental result of the lens maker equation. Alright guys, that wraps up this talk on the lens maker equation. Thanks for watching.

Example #1: What Types of Images Can be Formed by Lenses?

**Transcript**

Hey guys, let's do an example. What types of images can be formed by converging lenses? What about diverging lenses? So let's start with converging lenses. For converging lenses, we know by convention the focal length is positive so let's take a look at the thin lens equation and see what types of images we can produce. I'm going to isolate one over S I and this becomes 1 over F minus 1 over S O, and by convention that object distance is always positive. For a converging lens, that focal length is always positive so this number could be either positive or negative depending on if 1 over F is larger than 1 over S or if it's smaller. So the image distance is positive if 1 over F is a larger than one over S O or if S O is larger than, not a capital F it's a lowercase F. If that is true, we have positive image distances so we have real images. But the image distance can also be negative. The image distance can be negative if 1 over F is less than 1 over S not or if S not is less than 1 over F, then we're going to have virtual images. So if your object is placed outside of the focus, if the object distance is greater than the focal length, you will always get real images but if your object is placed inside of the focus at a distance less than the focal length, you will always get virtual images. Now, what about diverging lenses? Before even doing any math, what do you guys think are going to be the images produced by diverging lenses? They should always be virtual because you can never have converging light from a diverging lens. By convention, the focal length of a diverging lens is negative. So applying the same equation and the same process, that image sorry that object distance is always going to be positive but now, the focal length is negative so you have a negative number minus a positive number that's always going to be negative. So the image distance is always negative. That means only virtual images can be formed by diverging lenses exactly as we would expect regardless of what the actual numbers are. That wraps up this problem. Thanks for watching guys.

Example #2: Image Formation by a Biconcave Lens

-.393 should be multiplied by .52, f=-4.9

**Transcript**

Hey guys, let's do an example. A biconcave lens has two different radii of curvature if the radius of curvature of one piece of glass with a refractive index of 1.52 is 4 centimeters and the radius of curvature of the other piece is 7 centimeters, what is the focal length of the lens? If an object is placed 5 centimeters from the lens where will the image be formed? And is this image real or virtual? And finally, if the object is 1 centimeter tall, what's the height of the image? So let's start all the way at the beginning. What's the focal length of this lens? Now the lens maker equation tells us as we know that it doesn't matter the orientation of this lens, we're going to get the same focal length. So I'm just going to choose an orientation so we can assign a near radius and a far radius. So I'll say the near radius is 4 centimeters and the far radius is 7 centimeters. Now the lens maker equation tells us that 1 over a F is N minus 1 times 1 over R1 minus 1 over R2. The index of refraction is 1.52, the near radius is 4 centimeters but the center of curvature appears in front of the lens so by convention it's negative. The far radius is 7 centimeters and the center of curvature appears and sorry behind the lens so by convention that's positive. Plugging this into a calculator we get -0.393 but that isn't our answer we have to reciprocate this because this is 1 over F. So the focal length is going to be -2.5 centimeters. One answer done. Now if we place an object 5 centimeters from the lens, once again, it doesn't matter the orientation of the lens because it's the same focal length on either side. If we place it 5 centimeters from the lens, where will the image be formed? Now we need to use the thin lens equation that 1 over S O plus 1 over S I equals 1 over F. 1 over S I is, therefore, going to be 1 over F sorry minus 1 over S O which is 1 over -2.5 minus 1 over 5 which is -0.6. So if I reciprocate this answer because, once again, -0.6 is not the answer, it's the reciprocal then I get an image distance of -1.7 centimeters. Two answers down, two more to go. Is this a real image or virtual image? You guys should know this instantly by now. This is a virtual image. Why? Because the image distance is negative and finally, we want to know if the object is 1 centimeter tall, what is the height of the image? So for that we need to use the magnification. S I over S O. So, once again, not going to mess with the negative sign because we know that since this is a virtual image it's going to be upright. That negative sign will just tell us whether it's upright or inverted and we don't need that information. So this is 1.7 centimeters over the object distance which was 5 centimeters and that's 0.34. So, the height of the image is the magnification, 0.34, times the height of the object which is 1 centimeter. So the height of our image is 0.34 centimeters. Alright, so we know our focal length of this lens, -2.5 centimeters, image distance, -1.7 centimeters, which means it has to be a virtual image. The question wasn't asked, but this image is therefore upright. The magnification's 0.34 which means that the image is roughly one third the height of the object or 0.34 centimeters. Alright, guys, that wraps up this problem. Thanks for watching.

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The cornea of the human eye (the outtermost layer) acts like a thin lens, which is why imperfections in the cornea can lead to poor eyesight. The front and rear surfaces of the cornea is roughly spherical, with a radius of curvature of about 7.8 mm for the front and a radius of curvature of about 7.3 mm for the rear. What would the focal length of a typical cornea be?

An object is to the left of a thin lens. The lens forms an image on a screen that is 2.60 m to the right of the object. The height of the image is 2.50 times the height of the object. What is the focal length of the lens?

An object is placed 16 cm from a double concave thin lens (the index of refraction is 1.5) with the magnitude of the radius of curvature 20 cm on both sides. What is the magnification of the image?
a) –0.56
b) +0.89
c) –0.89
d) 0.56
e) –1.10

The cross-section of a glass lens with an index of refraction 1.5, is shown below. Determine the focal length f using the small angle approximation. Include both the magnitude and sign, which indicates whether this is a divergent or convergent lens.
1. 0.514286
2. 1.03196
3. -0.902762
4. -3.99724
5. -3.02298
6. 0.992568
7. -3.60345
8. 0.904751
9. 1.08561
10. -0.703276

A slide projector creates a 90.0 cm high image of a 2.0 cm tall slide. The screen is 300 cm from the lens. (a) How far is the slide from the lens? (b) What is the focal length of the lens?

In order to form a real image with a converging lens that is twice as large as the object, how far away from the lens should the object be placed? Assume the lens has focal length f.
(a) f
(b) 3f / 2
(c) 2f
(d) 2f / 3
(e) f / 2
(f) None of the above

An object that is 2 mm tall is placed 10 cm to the left of a thin lens that has f = +30 cm. What is the height of the image?

When a 4.0 mm tall object is 40.0 cm to the left of a lens, an upright image is formed that is 2.0 mm tall. Is the lens converging (f > 0) or diverging (f < 0)?
A. Converging
B. Diverging

Sally places an object 3 cm from a thin convex (converging) lens along its axis. The lens has a focal length of 4.5 cm. What are the respective values of the image distance and magnification?
a. -9 cm and +3
b. 18 cm and -3
c. +7.2 cm and -0.6
d. -18 cm and -3

Consider a divergent lens with a focal length f. An upright object is placed within the interval f and 2 f to the left of the lens. The corresponding image is
1. real, upright, and the same size.
2. real, upright, and reduced.
3. virtual, upright, and the same size.
4. virtual, upright, and reduced.
5. virtual, inverted, and reduced.
6. non-existant.
7. virtual, inverted, and enlarged.
8. virtual, upright, and enlarged.
9. virtual, inverted, and the same size.
10. real, upright, and enlarged.

A concave mirror with a radius of curvature of 1.4 m is illuminated by a candle located on the symmetry axis 3.4 m from the mirror. Where is the image of the candle?

A 4 cm tall object is placed 60 cm away from a converging lens of focal length 30 cm. What is the nature and location of the image? The image isa. Real, 2.5 cm tall, and 30 cm from the lens on the same side as the object.b. Virtual, 2.5 cm tall, and 30 cm from the lens on the side opposite the object.c. Virtual, 2.0 cm tall, and 15 cm from the lens on the side opposite the object.d. Virtual, 4.0 cm tall, and 60 cm from the lens on the same side as the object.e. real, 4.0 cm tall, and 60 cm from the lens on the side opposite the object.

An object (bold arrow) is located at a distance of 3 f / 2 (i.e. at x = 3 f / 2) in front of a diverging lens as shown in the figure. Note that f is positive, so the focal length of this lens is - f.What is the location x i of the image?a) x i = -3f / 5b) x i = 3f / 2c) x i = 3f / 2d) x i = f / 2e) x i = f / 5

An object (bold arrow) is located at a distance of 3 f / 2 (i.e. at x = 3 f / 2) in front of a diverging lens as shown in the figure. Note that f is positive, so the focal length of this lens is - f.Choose the correct description of the image in the situation above.a) The image is real and inverted.b) The image is virtual and inverted.c) The image is real and upright.d) The image is virtual and upright.e) There is no image formed.

An object (bold arrow) is located at a distance of 3 f / 2 (i.e. at x = 3 f / 2) in front of a diverging lens as shown in the figure. Note that f is positive, so the focal length of this lens is - f.
Suppose the object is now placed at x = - f. What is the magnification M?
a) M = 0
b) M = 1/2
c) M = 1

A convex lens forms a virtual image 4 times the size of the object. The object distance is 2.43333 cm. Find the focal length of the lens.1. 2.43752. 10.63. 2.406254. 2.156255. 2.666676. 7.07. 2.933338. 3.244449. 0.79210. 1.77778

The magnitudes of the radii of curvature for the spherical surfaces A and B are, respectively, |RA| = a and |R B| = 2a. The material of which the lens is made has an index of refraction n = 1.5. Find the focal length of the thin lens.1. f = a2. f = 2a3. f = −2a4. f = 4a5. f = −3a6. f = 3a7. f = −a8. f = −4a

An object is placed 30 cm to the left of a converging lens that has a focal length of 15 cm. Describe what the resulting image will look like (i.e. image distance, magnification, upright or inverted images, real or virtual images)?

When using the Lens Equation, a real OBJECT has a a) positive object distance b) negative image distance c) positive image distance?

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