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Goal: To understand multiple lens systems. Objectives: 1)To understand how to calculate values for 2 lens combos 2)To understand the human eye

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2 lenses We have learned how to do the equations for the first lens, but what happens when you have 2 lenses? The image from the first lens becomes the object for the second lens. The separation distance between lenses is denoted as s. Everything is normal after that. So, 1/p2 + 1/q2 = 1/f2 And p2 = s – q1 The final image type is determined by the 2 nd lens.

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Sample Lets make sure we still know the lesson from yesterday… You have 2 lenses. We will do the 2 nd lens in the 2 nd sample. An object is 10 cm from lens one. The focal length is 3 cm. What is the image distance (i.e. q1)?

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Sample 2 1/p2 + 1/q2 = 1/f2 And p2 = s – q1 For the previous question, if the separation distance between lenses is 20 cm then what is p2?

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Sample 3 Now that we have p2 as 15.7 cm, if the focal length for the 2 nd lens is 10 cm then what is the image distance for the 2 nd lens?

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Magnification Each lens will have a magnification. How do you think this will work with 2 lenses? A) M = M1 + M2 B) M = M1 – M2 C) M = M1 * M2 D) M = M1 / M2

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Sample We will put all this together… We have 2 lenses. An object is placed 6 cm from the first lens. The focal length of the first lens is 2 cm. The 2 nd lens is 7 cm from the first lens. The focal length of the 2 nd lens is 3 cm. What is the total magnification of the system (to find this you will have to find q1, q2, M1, and M2)?

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HW wrinkles Remember that a diverging lens has a negative value for the focal length. For the projector question, you are supposed to know that the object is upside down (so the initial height is negative). Sometimes they give you measurements between either object to image or in the multi lens they ask for the magnitude of distance from the FIRST lens and not the 2 nd. Just be on the lookout for this trip ups.

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The Eye Okay time for some cool stuff and some concepts.

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But what happens… If the eye is not correctly shaped? Myopia (nearsightedness) – you can focus on nearby objects but not distant ones. This is because the eye’s lens forms the image just in front of the retina. You compensate for this by using a divergent lens. Can you start fires with a lens like this?

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Hyperopia This is farsightedness. This time you have exactly the opposite problem. The lens doesn’t bend the light enough so the focus is just beyond the retina. To fix this you use a convergent lens. Note you COULD start a fire with this kind of lens!

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Presbyopia At some point a person’s eyes can loose flexibility. You start to have trouble focusing on nearby objects. While you are not better at seeing far away ones as you would be with farsightedness, to read something near by you need reading glasses. The reading glasses would be similar to the glasses used for farsightedness.

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The math There is some math for using glasses. There are two values to worry about – the near point and the far point. The near point is the nearest you can clearly see something. The far point is the farthest you can clearly see something.

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Refractive power P is the refractive power of a lens. P = 1 / f and is in units of Diopters (D). D has units of 1/meters Note that P also = 1/q + 1/p And if you have multiple lenses close together: P = P1 + P2 + …

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The equations… P = 1 / f = 1/q + 1/p However q and p are now something a bit different. You have 2 limits to your vision – new point and far point. For one of those two you find the original value and the new value. The original value is q and is negative (you will be making a virtual image). The new value is p. Be sure that you choose either far points or near points for both, don’t mix and match.

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How to know what to use: Correcting nearsightedness: You will be fixing your far point – what do you want your new far point to be? You want p to be infinity (so that your new far point is infinity). Your original far point distance is what you use for q, but use a negative number (because it is a divergent lens, and divergent lenses have negative values of q – and yes that means they have negative values of focal length). Sample: if your far point was 40 cm then what power of lens would you need to correct it (note that you need to convert to meters)?

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Correcting farsightedness q is the original near point (and is negative). p is the distance you want to be able to see clearly (i.e. the new near point) NOTE: these last two examples are for contact lens. The distances used in the equations are measured from the lens, so if the lens is 2 cm from the eyes then you have to subtract 2 cm off of both p and from the magnitude of q (so, the magnitude of q goes down) because the lens will be closer to your near and far point than your eye.

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Sample A pair of eyeglasses are 2 cm from the eye. If the closest someone can see is 0.5 meters and they wish to be able to see things clearly at 0.2 meters then what power of glasses do they need?

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Conclusion We have learned how to solve for distances in multiple lens systems We learned that the magnifications for each lens multiply together. We learned about the eye and different eye conditions. We learned how to correct vision problems and how to calculate the power required to do so.

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Physics 1161: Lecture 19 Lenses and your EYE

Physics 1161: Lecture 19 Lenses and your EYE

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