physics 212 lecture 26, slide 1 physics 212 lecture 26: lenses

Physics 212 Lecture 26, Slide Physics 212 Lecture 26, Slide 11

Physics 212Physics 212Lecture 26: Lecture 26: LensesLenses


MusicWho is the Artist?Who is the Artist?

A)A) Ramblin’ Jack ElliottRamblin’ Jack ElliottB)B) Arlo GuthrieArlo GuthrieC)C) Pete SeegerPete SeegerD)D) Phil OchsPhil OchsE)E) U. Utah PhillipsU. Utah Phillips

Why?Why?

Last time we talked about the Last time we talked about the Brewster angle diagram with “circles Brewster angle diagram with “circles & arrows”& arrows”


Your CommentsYour Comments

““This prelecture could not have better timing (I’m actually doing a project about lenses and mirrors in another class)! I am wondering how a flat lens, such as a fresnel lens, works though.””

““would like to go over the diagrams much more slowly, I’ve never been very good at understanding them.””

““i really dont understand what a virtual image is can you explain that in detail please””

““It was good and going along at a nice pace, but the lens-maker’s formula must have been on a slide with a smaller index of refraction than the others because it went super fast.””

“The demonstraion will be very valuable for this lecture. Choose wisely, or a kitten will die ...”

““This prelecture reminded me of my childhood frying ants in the sunlight. Ahh, memories.””

Hour Exam 3: 1 week from tomorrow (Wed. Apr. 25) Hour Exam 3: 1 week from tomorrow (Wed. Apr. 25) • covers L19-26 inclusive (LC circuits to lenses) covers L19-26 inclusive (LC circuits to lenses) • Sign up for conflicts, etc. no later than Mon. Apr. 23 at 10:00 Sign up for conflicts, etc. no later than Mon. Apr. 23 at 10:00 p.m.p.m.

We’ll show the key points with We’ll show the key points with examplesexamples


Refraction

That’s all of the physics – That’s all of the physics – everything else is just geometry!everything else is just geometry!

Snell’s Lawn1sin(1) = n2sin(2)

11

22

nn11

nn22


airair waterwater

glassglass glassglass

CaseCase II CaseCase IIIIIn In Case ICase I light in light in airair heads toward a piece of glass with incident angle heads toward a piece of glass with incident angle iiIn In Case IICase II, light in , light in waterwater heads toward a piece of glass at the heads toward a piece of glass at the samesame angle. angle.

In which case is the light In which case is the light bentbent most as it enters the glass? most as it enters the glass?

II or or II II oror Same Same (A) (B) (C)(A) (B) (C)

ii ii

2222 1.51.5

1.3

The angle of refraction in BIGGER for the water – glass interface:n1sin(1) = n2sin(2) sin(2)/sin(1) =

n1/n2 Therefore the BEND ANGLE (1 – 2) is BIGGER for air – glass interface


Checkpoint 2What happens to the focal length of a converging lens when it is placed under water?A. increases B. decreases C. stays the same

“Under water, the index of refraction is greater and closer to the index of refraction of the lens; this makes the angle of refraction smaller by Snell's law so the focal length will increase.”

“The index of refraction will increase so the focal length will decrease.”

“The lens makers formula shows focal length only depends on N and R”


Checkpoint 2

The rays are bent more from air to glass than from water to glassTherefore, the focal length in air is less than the focal length in water

nnairairnnlenslens

We can see this also from Lensmaker’s Formula

What happens to the focal length of a converging lens when it is placed under water?A. increases B. decreases C. stays the same


Object Location• Light rays from sun bounce off object and go in all

directions– Some hits your eyes

We know object’s location by where rays come from.

We will discuss eyes in lecture 28…


Waves from object are focused by lens


Two Different Types of Lenses


Converging Lens:Converging Lens: Consider the case where the shape of the Consider the case where the shape of the lens is such that light rays parallel to the axis of the mirror lens is such that light rays parallel to the axis of the mirror are all “focused” to a common spot a distance are all “focused” to a common spot a distance ff behindbehind the the lens:lens:

ff

ff


objectobject

1) Draw ray parallel to axis1) Draw ray parallel to axisrefracted ray goes through focusrefracted ray goes through focus2) Draw ray through center2) Draw ray through centerrefracted ray is symmetricrefracted ray is symmetric

imageimage

You now know the position of the same point on the imageYou now know the position of the same point on the image

Recipe for finding image:Recipe for finding image:

ff


S > 0S > 0 S’ > 0S’ > 0

f > 0f > 0

1 1 1

S S f

SMS

SS > 2 > 2ffimage is:image is:

realrealinvertedinvertedsmallersmaller

objectobject

imageimage

ff

Example


S > 0S > 0

f > 0f > 0

SS = = ffimage is:image is:

at infinity…at infinity…

objectobjectff

1 1 1

S S f

SMS

Example


S > 0S > 0

S’ < 0S’ < 0

f > 0f > 0

objectobject

imageimageff

0 < S0 < S < < ff

image is:image is:virtualvirtual

uprightuprightbiggerbigger

1 1 1

S S f

SMS

Example


Checkpoint 1a

ABCD

“The object is real, therefore its image is real. Real things make real things.”

“The screen is to the downstream of the lens, so s prime must be positive and the image must be real. Since both s and s prime are positive, the magnification will be inverted.”“The image is virtual because it is behind the lens. It is upright because it is always upright. ”“converging lens always give the virtual image and if it can be projected on the screen, it is inverted”


Checkpoint 1a

0sImage on screen

MUST BE REAL

ABCD

0sMs

MUST BE INVERTED


Diverging Lens:Diverging Lens: Consider the case where the shape of the lens Consider the case where the shape of the lens is such that light rays parallel to the axis of the lens all diverge is such that light rays parallel to the axis of the lens all diverge but appear to come from a common spot a distance but appear to come from a common spot a distance ff in front ofin front of the lens:the lens:

ff


S>0S>0 S’<0S’<0

f<0f<0

image is:image is:virtualvirtual

uprightuprightsmallersmaller

objectobject

imageimageff

1 1 1

S S f

SMS

Example


Executive Summary - Lenses:Executive Summary - Lenses:

SS > > 2f2frealreal

invertedinvertedsmallersmaller

2f2f > > S > f S > f realreal

invertedinvertedbiggerbigger

ff > > S > 0S > 0 virtualvirtual

uprightuprightbiggerbigger

SS > > 00 virtualvirtual

uprightuprightsmallersmaller divergingdiverging ff

convergingconverging ff

1 1 1

S S f

SMS


It’s always the same:It’s always the same:

1 1 1

S S f

SMS

You just have to keep the signs straight:You just have to keep the signs straight:

The sign conventionsThe sign conventionsSS: positive if object is “upstream” of lens : positive if object is “upstream” of lens S’S’ : positive if image is “downstream” of lens : positive if image is “downstream” of lensf:f: positive if converging lens positive if converging lens


Checkpoint 1b

“The light from the bottom half only makes it through the lens.“

A converging lens is used to project the image of an arrow onto a screen as shown above.A piece of black tape is now placed over the upper half of the lens. Which of the following is true?A. Only the lower half of the object (i.e. the arrow tail) will show on the screenB. Only the upper half of the object (i. e. the arrow head) will show on the screenC. The whole object will show on the screen

“When the top half is covered, the image would cover the lower end of the screen. The image is inverted but real size.”

“The entire image will still be projected onto the screen but there will not be as many rays so the intensity will be lower.”


Light from top of objectLight from top of object

objectobject imageimage

Light from bottom of objectLight from bottom of object

objectobject imageimage

Cover top half of lensCover top half of lens

Cover top half of lensCover top half of lens

What’s the Point?

The rays from the bottom half still focusThe image is there, but it will be dimmer !!A converging lens is used to project the image of an arrow onto a screen as shown above.A piece of black tape is now placed over the upper half of the lens. Which of the following is true?A. Only the lower half of the object (i.e. the arrow tail) will show on the screenB. Only the upper half of the object (i. e. the arrow head) will show on the screenC. The whole object will show on the screen


CalculationCalculation

• Conceptual Analysis • Lens Equation: 1/s + 1/s’ = 1/f• Magnification: M = -s’/s

• Strategic Analysis• Consider nature of image (real or virtual?) to determine relation between object position and focal point• Use magnification to determine object position

A magnifying glass is used to read the fine print on a document. The focal length of the lens is 10mm.

At what distance from the lens must the document be placed in order to obtain an image magnified by a factor of 5 that is NOT inverted?


Is the image real or virtual? (A) REAL (B) VIRTUAL

fh

h’

A real image would be A real image would be invertedinverted

A virtual image will be uprightA virtual image will be upright

h

f

h’




Real objectReal object s > 0s > 0

How does the object distance compare to the focal length? (A) (B) (C)s f s f s f

Converging lensConverging lens f > 0f > 0

Lens Lens equationequation

1 1 1s f s

fsss f

0s f

Virtual ImageVirtual Image s’ < 0s’ < 0 h

f

h’

ss’




fMs f

What is the magnification M in terms of s and f? (A) (B) (C) (D)fM

s f

s fM

f

f sMf

h

f

h’

ss’

Magnification Magnification equation:equation:

sMs

Lens Lens equation:equation:1 1 1s f s

fsss f

fM

s f



fsss f


fMs f

(A) 1.7mm (B) 6mm (C) 8mm (D) 40 mm

(E) 60 mm

h’

40 s sM mm

5M 10 f mm

1Ms f

M

4 8 mm5

s f hf

fMs f




Follow UpFollow Up

(A) sdiv < sconv (B) sdiv = sconv (C) (C) sdiv > sconv (D) s(D) sdivdiv doesn’t doesn’t existexist

Suppose we replace the converging lens with a diverging lens with focal length of 10mm.

If we still want to get an image magnified by a factor of 5 that is NOT inverted, how does the object sdiv compare to the original object distance sconv?

s negative s negative not real object not real object

EQUATIONSEQUATIONSfM

s f

1Ms fM

5M 10 f mm

4 8 mm5

s f f

h

sh’

s’f

h

sh’

s’

Draw the rays: s’ will always be smaller than sDraw the rays: s’ will always be smaller than sMagnification will always be less than 1Magnification will always be less than 1

PICTURESPICTURES


Follow UpFollow UpSuppose we replace the converging lens with a diverging lens with focal length of 10mm.

What is the magnification if we place the object at s = 8mm?

EQUATIONSEQUATIONSfM

s f

8 mms 10 f mm

fMs f

(A) (B) (C)(C) (D) (E)(D) (E)5

9M 5M

45

M 12

M 38

M

10 10 58 ( 10) 18 9

M

h’f

hh’

PICTURESPICTURES

physics 212 lecture 26, slide 1 physics 212 lecture 26: lenses

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