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 |

Total Internal Reflection | 13 mins | 0 completed | Learn Summary |

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

Mirror Equation | 20 mins | 0 completed | Learn |

Refraction At Spherical Surfaces | 10 mins | 0 completed | Learn |

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

Anna holds a 14 D magnifier directly in front of her eye to get a close look at a 19-mm-diameter penny. Assume a typical 25 cm near point and a distance of 1.7 cm between the lens and the retina.What is the closest possible distance that she can hold the coin to have it appear in focus? Express your answer with appropriate units.

An object is 30 cm in front of a converging lens with a focal length of 10 cm. (a) Use ray tracing to determine the location of the image. (b What are the image characteristics: (i) upright or inverted, (ii) magnification, and (iii) real or virtual?

A 3.0-cm-tall object is 23 cm in front of a diverging lens that has a -30 cm focal length.Part ACalculate the image position. Express your answer to two significant figures and include the appropriate units.Part BCalculate the image height. Type a positive value if the image is upright and a negative value if it is inverted. Express your answer to two significant figures and include the appropriate units.

A lens produces a real image of a real object.a) Is the image inverted or upright?b) Is the lens diverging or converging?c) Is the image enlarged or reduced in size?d) If two convex lenses identical in size and shape are manufactured from glass with two different indices of refraction, would the focal length of the lens with the greater index of refraction (lens 1) be larger or smaller than that of the other lens (lens 2)?

The ___________ of a lens or mirror is a rotational symmetry axis of the surfaces.a. light rayb. paraxial ray approximationc. optic axisd. focal pointe. focal plane

A student is working in an optics lab and has a light bulb which is 90 cm from a screen. She needs to make a real image of the bulb on the screen and she has a positive lens with a focal length of 12.5 cm. Find two places she can put the lens and give object and image distances to both setups.

An object is 16.0cm to the left of a lens. The lens forms an image 36.0cm to the right of the lens.Part AWhat is the focal length of the lens?Part BIs the lens converging or diverging?Part CIf the object is 8.00mm tall, how tall is the image?Part DIs it erect or inverted?

A farsighted person has a near point of 72 cm rather than the normal 25 cm. What strength lens, in diopters, should be prescribed to correct this vision problem?

A candle (ho = 0.39 m) is placed to the left of a diverging lens (f = -0.054 m). The candle is so = 0.34 m to the left of the lens.ho = 0.39 mf = -0.054 mso = 0.34 mPart (a) Write an expression for the image distance, si.Part (b) Numerically, what is the image distance in meters?Part (c) Is this real or virtual?Part (d) Numerically, what is the image height, hi?

A slide projector needs to create a 98-cm-high image of a 2.0-cm-tall slide. The screen is 300 cm from the slide.Part A: What focal length does the lens need? Assume that it is a thin lens.Part B: How far should you place the lens from the slide?

An object is located 30.0 cm from a certain lens. The lens forms a real image that is twice as high as the object. What is the focal length of this lens? a. 90 0 cm b. 10.0 cm c. 5.00 cm d. 20.0 cm

The sun is 150,000,000 km from earth; its diameter is 1,400,000 km. A student uses a 5.2-cm-diameter lens with f = 10 cm to cast an image of the sun on a piece of paper.Where should the paper be placed relative to the lens to get a sharp image?

1.) Find the position of the final image of the 1.0-cm-tall object2.) Find the size of the final image of the 1.0-cm-tall object3.) Is the object inverted or erect?

The sun is 150,000,000 km from earth; its diameter is 1,400,000 km. A student uses a 5.2-cm-diameter lens with f = 10 cm to cast an image of the sun on a piece of paper.What is the diameter of the image on the paper?

Anna holds an 11 D magnifier directly in front of her eye to get a close look at a 19-mm -diameter penny. Assume a typical 25 cm near point and a distance of 1.7 cm between the lens and the retina.(a) What is the closest possible distance that she can hold the coin to have it appear in focus?(b) At this distance, how large does the image of the coin appear on her retina?Express your answer with the appropriate units.

You are using a converging lens with a focal length of 10 cm. What will the image distance be and what will the magnification of that image be if you place an object:a) 30 cm to the left of the lensb) 20 cm to the left of the lensc) 15 cm to the left of the lensd) 5 cm to the left of the lens

A shopper standing 3.00 m from a convex security mirror sees his image with a magnification of 0.250.(a) Where is his image?(b) What is the focal length of the mirror?(c) What is its radius of curvature?

Rank the images on the basis of the magnitude of their magnification, from greatest to smallest.Rank the images on the basis of their size, assuming the object size is the same. (in cm)1. d=15; f=202. d=15; f=53. d=5; f=204. d=20; f=105. d=10; f=206. d=10; f=5

The figure shows a combination of two lensesa. Find the position of the final image of the 1.0 cm tall objectb. Find the size of the final image of the 1.0-cm-tall objectc. Find the orientation of the final image of the 1.0-cm-tall object

You are using a converging lens with a focal length of 10 cm. What will the image distance be and what will the magnification of that image be if you place an object: a) 15 cm to the left of the lens.b) 5.0 cm to the left of the lens.

You are using a converging lens with a focal length of 10 cm what will the image distance be and what will the magnification of that image be if you place an object: a) 30 cm to the left of the lens. b) 20 cm to the left of the lens.

A real image is four times as far from a lens as is the object.What is the object distance, measured in focal lengths? ____from the lens.

An object is 1.0 cm tall and its erect image is 3.0 cm tall. What is the exact magnification?

A converging lens creates the image shown in:Is the object distance less than the focal length f, between f and 2f, or greater than 2f?

You have lenses with the following focal lengths: f = 25 mm, 50 mm, 100 mm, and 200 mm. In what arrangement would you use these lenses to get the highest-power telescope? fo = 200 mm, fe= 25 mm fo = 25 mm, fe= 200 mm fo = 25 mm, fe= 50 mm fo = 50 mm, fe= 25 mm

An object is 1.0 cm tall and its inverted image is 2.0 cm tall. What is the exact magnification?

1. What can one say about the image produced by a thin lens that produces a positive magnification?a. It is real and invertedb. It is real and erect.c.It is virtual and inverted.d. It is virtual and erect.2. If the diameter of a lens is reduced, what happens to the magnification produced by the lens?a. It increasesb. It decreasesc. It is unchanged

The diameter of Mars is 6794 km, and its minimum distance from the earth is 5.58×107 km. When Mars is at this distance, find the diameter of the image of Mars formed by a spherical, concave telescope mirror with a focal length of 1.60 m.

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)?

Ellen wears eyeglasses with the prescription -1.0D. What is her far point without the glasses?

You have lenses with the following focal lengths: f = 25 mm, 50 mm, 100 mm, and 200 mm. Which lens would you use to get the highest-power magnifier?

You have lenses with the following focal lengths: f = 25 mm, 50 mm, 100 mm, and 200 mm. Which pair of lenses would you use to get the highest-power microscope? Check all that apply. 25 mm 50 mm 100 mm 200 mm

Image Formation by Lenses Your camera's zoom lens has an adjustable focal length ranging from 80.0 to 200 mm. What is its range of powers? (You do not need to enter any units.) Pmax = Pmin =

A severely myopic patient has a far point of 5.75 cm. By how many diopters should the power of his eye be reduced in laser vision correction to obtain normal distant vision for him? (Assume a lens-to-retina distance of 2.00 cm.)

1/S+ 1/S'=1/f(S is the object distance, S' is the image distance and f is the focal length of the mirror or lens)When we choose 1/S as the x axis and 1/S' as the y axis... Then we make a linear fit. Show that the slope is -1. And if intercept is C, calculate f as function of C only.

True or False: The quantities di and do are measured from the focal point of a lens or mirror.

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

A "biconvex" lens is one in which both surfaces of the lens bulge outwards. Suppose you had a biconvex lens with radii of curvature with magnitudes of |R1|=10cm and |R2|=15 cm. The lens is made of glass with index of refraction nglass=1.5. We will employ the convention that R1 refers to the radius of curvature of the surface through which light will enter the lens, and R2 refers to the radius of curvature of the surface from which light will exit the lens.A) What is the focal length of the lens if it is immersed in water (nwater = 1.3)f= ____________ cmB) What is the focal length f of this lens in air (index of refraction for air is nair=1)?

Explain why you cannot measure the focal length directly for a diverging lens. Using no more than four sentences, explain why the technique of placing the diverging lens touching another converging lens with the light reaching the converging lens first, allows you to determine the focal length of the diverging lens.

Ramon has contact lenses with the prescription +2.0 D.a. What eye condition does Ramon have?b. What is the near point without the lenses?

A lens creates an image as shown. In this situation, the object distance s isa) Larger than the focal length f. ?b) Smaller than focal length f. ?c) Equal to the focal length f. ?

The ___________ is a plane perpendicular to the optic axis through the focal point.a) optic axisb) focal pointc) light rayd) paraxial ray approximatione) focal plane

Suppose a 200 mm focal length telephoto lens is being used to photograph mountains 10.0 km away. (a) Where is the image? (b) What is the height of the image of a 1000 m high cliff on one of the mountains?

An amoeba is 0.305 cm away from the 0.300 cm focal length objective lens of a microscope. Randomized Variables: D = 19.93 cmfe = 2.14 cmPart (a) What is the image distance (in cm) for this configuration?Part (b) What is this image's magnification?Part (c) An eyepiece with a 2.14 cm focal length is placed 19.93 cm from the objective. What is the image distance for the eyepiece in cm?Part (d) What magnification is produced by the eyepiece?Part (e) What is the overall magnification?

When using the Lens Equation, a virtual image has a positive object distance, positive image distance, or negative image distance?

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