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Upcoming Schedule How many physicists does it take to change a light bulb? Eleven. One to do it and ten to co-author the paper. Dec. 1 22, 23.1, 23.4, 23.5 Dec. 3 boardwork Dec. 5 23.6 Quiz 10a Dec. 8 23.7-23.9 boardwork Dec. 10 23.2-23.3 Quiz 10b Dec. 12 review Dec. 15 Final Exam 4-6 pm Monday, Dec. 8 start at slide 12

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Upcoming Schedule. Monday, Dec. 8 start at slide 12. How many physicists does it take to change a light bulb? Eleven. One to do it and ten to co-author the paper. 23.6 Total Internal Reflection; Fiber Optics. Recall Snell’s law:. - PowerPoint PPT Presentation

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Page 1: Upcoming Schedule

Upcoming Schedule

How many physicists does it take to change a light bulb? Eleven. One to do it and ten to co-author the paper.

Dec. 122, 23.1,

23.4, 23.5

Dec. 3boardwork

Dec. 523.6

Quiz 10a

Dec. 823.7-23.9boardwork

Dec. 1023.2-23.3Quiz 10b

Dec. 12review

Dec. 15Final Exam

4-6 pm

Monday, Dec. 8start atslide 12

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Recall Snell’s law:

23.6 Total Internal Reflection; Fiber Optics

Suppose n2>n1. The largest possible value of sin(2) is 1 (when 2 = 90). The largest possible value of sin(1) is

1 1 2 2n sin θ = n sin θ

21 2

1

nsin θ = sin θ

n

21,max

1

nsin θ = .

n

This value of is called the critical angle, C. For any angle of incidence larger than C, all of the light incident at an interface is reflected, and none is transmitted.

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1 < C 1 close to C

1 > C

1

Visualization here.

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http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/totint.html

Total internal reflection explains the disappearing coin trick we saw during the last lecture.

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The picture below is for a prism, but it also explains why no light reaches your eyes from the coin under the glass.

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application: fiber optics

http://laser.physics.sunysb.edu/~wise/wise187/janfeb2001/reports/andrea/report.html

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from www.howstuffworks.com

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application: swimming underwater

If you are looking up from underwater, if your angle of sight (relative to the normal to the surface) is too large, you see an underwater reflection instead of what’s above the water.

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23.7 Thin Lenses; Ray Tracing

We investigated lenses qualitatively in class last week.

A convex surface has its middle part bent towards you as you look at it.

A double convex lens has two convex surfaces, and is called a “converging lens.”

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A concave surface has its middle part bent away from you as you look at it.

A double concave lens has two concave surfaces, and is called a “diverging lens.”

The axis of one of these lenses is a straight line through its center, perpendicular to both surfaces.

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Light rays striking a convex lens are bent towards the axis at both surfaces.

If the lens is infinitesimally thin, all light rays striking the lens parallel to the axis converge at the same point.

The point at which the rays converge is called the “focal point.”

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For a lens with spherical surfaces, the rays do not quite converge at the focal point, but if the lens is very thin and the object very far away compared to the diameter of the lens, then the rays very nearly converge.

We will consider thin lenses in this section. The focal point of a thin convex lens is the point where the image of an infinitely distant object is formed.

http://webphysics.ph.msstate.edu/javamirror/ipmj/java/clens/

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The focal length, f, is the distance along the lens axis of the focal point from the center of the lens. A an incoming ray parallel to the axis will cross the axis at the focal point F.

F

f

Light rays not parallel to the axis will all converge on the focal plane.

F

f

focalplane

A converging lens like the one shown has focal points on both sides of the lens.

F'

The two focal lengths may differ. The light rays from an object to the left actually converge at F, not at F'.

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To study the image formed by a thin lens qualitatively, we use ray tracing to form the image. Only two rays are needed to determine the location of the image from a point on the object. For redundancy, we use three rays.An incident ray parallel to the lens axis passes through F.A ray through the center of the lens.

A ray through F' which emerges from the lens parallel to the axis.Image points for other points on the object are found similarly; this lets us complete the image of the object.

http://www.physics.brocku.ca/faculty/sternin/120/applets/Lenses/

F

F'

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We follow the same procedure to find the image for a diverging lens.

F F'

The image “appears” to be on the same side of the lens as the object; that’s why F and F' have been switched compared to the double convex lens!

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23.8 The Lens Equation

In this section Giancoli derives the lens equation, which describes the position and size of an imaged formed by a lens.The derivation is simple geometry, and can be done for both converging and diverging lenses.

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The result isO i

1 1 1 + =

d d f

where f is the focal length of the lens, do is the distance from the object to the lens, and di is the distance from the lens to the object.

This equation uses these sign conventions:

F is + for converging lenses and – for diverging lenses

dO is + if object is on side of lens that light comes from (and – otherwise)

di is + if image is on opposite side of lens to light source (and - otherwise)

image height hi is + if image is upright and – if image is inverted (object height hO is always +)

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The magnification of a lens is the image height divided by the object height, and from the figures used to derive the lens equation,

i i

O O

h dm = = -

h d

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Brief litany for solving lens equations:

draw a picture

draw a ray diagram

define and label focal lengths, distances, and heights

OSE

solve algebraically, then numerically

23.9 Problem Solving for Lenses

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Images formed by convex lenses may be real…

…or they may be virtual.

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Images formed by convex lenses are always virtual.

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Example 23-11 What are the position and size of the image of a 7.6 cm high flower placed 1 m from a +50 mm focal length converging lens?

O i

1 1 1 + =

d d f

i

1 1 1 + =

100 cm d 5 cm

i d = 5.26 cm

i

O

d 5.26 cmm = - = - = -0.0526

d 100 cm

i 0h = mh = -0.0526 7.6 cm = -0.40 cm

Bad! Skipping many steps! Only way to get the lecture done!

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Example 23-12 What is the position and size of the image of an object placed 10 cm away from a +15 cm focal length converging lens?

O i

1 1 1 + =

d d f

i

1 1 1 + =

10 cm d 15 cm

i d = -30 cm

i

O

d -30 cmm = - = - = 3

d 10 cm

The image height is three times the object height.

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Example 23-13 Where must a small insect be placed if a 25 cm focal length diverging lens is to form a virtual image 20 cm in front of the lens?

O i

1 1 1 + =

d d f

O

1 1 1 + =

d -20 cm -25 cm

O d = 100 cm

Good links:http://www.physics.brocku.ca/faculty/sternin/120/applets/Lenses/ http://www.cs.unc.edu/~dorianm/academics/comp235/lenstrace/javaapplet/index.html http://webphysics.ph.msstate.edu/jc/library/22-6/ http://webphysics.ph.msstate.edu/javamirror/ipmj/java/clens/