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PES 2130 Fall 2014, Spendier Lecture 21/Page 1 Lecture today: Chapter 34 Images 1) Images formed by thin lenses 2) Combination of thin lenses 3) Eye...
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PES 2130 Fall 2014, Spendier

Lecture 21/Page 1

Lecture today: Chapter 34 Images 1) Images formed by thin lenses 2) Combination of thin lenses 3) Eye glasses Announcements: - Monday Nov 10, Prof. Tolya Pinchuk will fill in for me. I will hold office hours that Monday from 10am-11am but will leave around noon. Last lecture: Thin lenses Summary: Mirrors: 1 1 1 2    p i f r Thin lenses: 1 1 1 1 1     n  1    p i f  r1 r  Magnification formula for mirrors and thin lenses i h' m  p h h' ... image height h ... object height

PES 2130 Fall 2014, Spendier

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Example 1: A botanist holds a magnifying glass 3.00 cm from a ladybug that is 4.05 mm long. It apperas through the lens to be 1.70 cm and has the correct orientation. a) What is the focal length of the magnifying glass? b) Draw a ray diagram illustrating this situation.

PES 2130 Fall 2014, Spendier

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2) Combination of thin lenses: We can use both the ray diagram technique and the thin-lens equation to analyze optical systems consisting of two or more lenses. What is the advantage of using 2 lenses? Remember that for 1 lens: m = -i/p So to get 100x magnification, the image needs to be located 100 times further away than the object. But then it is so far away that we may not be able to see is so well (loss of intensity) Applications: - camera - microscope (magnifies object by very large amount) - refracting telescope (magnifies object by very large amount) - eye glasses: eye plus corrective lens Basic technique: • The image formed by the first lens is located as if the second lens were not present. • The image formed by the first lens serves as the object for the second lens. • If image formed by the first lens lies in front of the second lens, the “object” distance is positive; otherwise negative. • The image formed by the second lens serves as the object for the third lens. • and so on. • The same rules can be applied to the combination of thin lenses and mirrors. Example 2: An object sits 35 cm to the left of a converging lens with focal length of 20 cm. 75 cm to the right is a second converging lens with focal length of 15 cm. Locate and characterize the final image.

PES 2130 Fall 2014, Spendier

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A ray diagram is always a good way to start. There is usually more real understanding available in the ray diagram than in "merely" solving the problem numerically. In addition, a good diagram also provides a place to keep all the dimensions of the problem.

Standard optical microscope

Eye or Camera h0 hi hv

- object lens: closest to the image, forms intermediate image. - The object is outside the focal length of the objective lens - The intermediate image is inside the focal length of the second lens, the eyepiece lens - eyepiece lens: closest to eye, it takes intermediate image as the object and forms a second "virtual" image

PES 2130 Fall 2014, Spendier

M objective 

 hi h0

M eyepiece 

 hv hv   hi hi

M system 

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 h  h   h  h  hv    i  v    i  v   M objective  M eyepiece / tube lens h0  h0  hi   h0  hi 

Eyepieces and objectives both have magnification that each contribute to the overall system magnification.

Refracting Telescope focused at infinity: The astronomical telescope can be used to view objects at finite distances, although there is the inconvenience of having the image upside down. Objects are so far away that light coming from them may be considered parallel. F1 coincides with F2.

The final “virtual” image is more infinite away than the original image (what?!)  so there is magnification. Eye glasses: DEMO Typically, to focus on objects the eye lens can change shape slightly as the muscles around it tighten or relax. - For close objects lens needs to have a shorter focal length ==> smaller radius of curvature ==> lens must be more rounded But the lens can only be tuned so much, that is why we can suffer from nearsightedness and farsightedness.

PES 2130 Fall 2014, Spendier

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Near-sighted: Light focuses short of the retina b/c the eye is too long or lens may be defective. We can correct this by inserting a diverging lens

Far-sighted: Light focuses behind the retina. We can correct this by inserting a converging lens.

Lenses for vision correction are usually described in terms of the power defined as the reciprocal of the focal length expressed in meters. The unit of power is the diopter. Thus a lens with f = 0.50m has power P of P = 1/f = 1/0.50 = 2.0 diopters

(note: f must be in meters!)

LASIK An alternative approach for correcting many defects of vision is to reshape the cornea. This is often done using a procedure called laser-assisted in situ keratomileusis, or LASIK. An incision is made into the cornea and a flap of outer cornea; tissue is folded back. A pulsed ultraviolet laser with a beam only 50 µm wide (about 1/200 the width of a human hair) is then used to vaporize away microscopic areas of the underlying tissue. The flap is then folded back into position, where it conforms to the new shape “carved” by the laser.

PES 2130 Fall 2014, Spendier

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Example 3 (Correcting for farsightedness): The near point of a certain far-sighted eye is 100 cm in front of the eye. To see clearly an object that is 25 cm in front of the eye, what contact lens is required? In this problem, we want the lens to form a virtual image of the object at the near point of the eye, 100 cm from it. That is, when p = 25 cm, we want i = -100 cm (negative b/c image is virtual)

Note: for correcting nearsightedness, you need a diverging lens  diopters will be negative!