1

• Building on concepts from the last chapter

– Now that we know how rays and images work we can understand how curved mirrors and lenses produce images

– Practical applications include magnifying mirrors, rear-view mirrors, glasses and contact lenses, and cameras.

• Spherical mirrors– Significance of focal point– Tracing reflected rays to find virtual

images in convex mirrors– Tracing reflected rays to find virtual

and real images in concave mirros

• Lenses– Significance of focal points,

focal lengths, focal planes and power of a lens

– Tracing rays to find images from thin convex and concave lenses.

– Fresnel lenses– Compound lenses and using

intermediate images to find final images

Physics 1230: Light and ColorCh. 3: Mirrors and Lenses Ivan Smalyukh

What is a spherical convex mirror? (Think of the rear view mirror on your car)

• A convex mirror bulges outwards towards you

Reflecting surface ofconvex spherical mirror

Axis of mirrorCenter of sphere

Mirrors curved like a cylinder can make you look fat or skinny

Looking at yourself in a convex mirror,the image is compressed vertically like

rear view mirror but in one dimension only.Your image looks fatLooking at yourself in a concave mirror, the image is

vertically expanded like a bathroom magnifying mirrorbut in one dimension only. Your image looks skinny

Virtual imageon other side of mirror iscompressedvertically

Virtual imageon other side of mirror isstretchedvertically

A concave mirror bulges away from you

• The center of the sphere (circle) is in front of the mirror

• The focal length is also in front of the mirror, halfway between the center and the mirror surface

Axis Center Focalpoint

Mirrorsurface

Why are the rays from a distant light source such as the sun or a star essentially parallel?

Here the rays from a distant light sourcesuch as the sun

By the time they arrive here(into a camera or mirror) only

the nearby almost parallel rays enter

Convex mirror

Convex mirror

Rays which arrive at the mirrorfrom a close source, like Alex’s

nose are not almost parallel

Whenever we speak of incoming parallel rays

you can always visualize rays from the sun or a star.

http://micro.magnet.fsu.edu/primer/java/mirrors/concavemirrors/index.htmlhttp://micro.magnet.fsu.edu/primer/java/mirrors/convexmirrors/index.html

http://micro.magnet.fsu.edu/primer/java/mirrors/convexmirrors3d/index.htmlhttp://micro.magnet.fsu.edu/primer/java/mirrors/concave.html

http://micro.magnet.fsu.edu/primer/java/mirrors/convex.html

Here is how we use those rays to find the image of an object in a concave mirror

• Where is Alex's image when he is between the center and it's focal point?

• Let's find the image of his nose– Here is a ray of type 1 from his nose reflecting

off the mirror– Here is a ray of type 3 from his nose reflecting

off the mirror– The image of the nose is at the intersection of

reflected rays of type 1 and type 3. Why?– Can we use a ray of type 2?

• Here is Alex's image– Is it (a) real or (b) virtual? – Magnified or reduced?

• From which points can your eye see his nose?• From all points (A, B, C)?• From A and C, but not B?• From B and C, but not A?• From A only?• From C only

Center

Focalpoint

Mirrorsurface

Axis

A

B

C

Clicker Question

Special rays 1, 2 and 33 are not necessary to find the image of an object but are easier to work with than others

Special rays 1, 2 and 33 are not necessary to find the image of an object but are easier to work with than others

• Here are rays 1 and 3 used to find the image Alex's nose

• Here are some other (less convenient) incident and reflected rays– They all go through the same

image point as rays 1 and 3 after satisfying the law of specular reflection!

– Any two rays from an object point will intersect at the image point but rays 1, 2 and 3 are the easiest to use to find the image point

What happens to his image if Alex is closer to the mirror than the focal point?

Center Focalpoint

Mirrorsurface

• Draw rays of type 1 and type 3 from his nose to the mirror

• The reflected rays will never intersect behind Alex since they diverge behind him

• However we can extend the reflected rays behind the mirror where they do intersect– The image of Alex's nose is at the

intersection of the backward extended reflected rays

12. Is it real (a) or virtual (b)?13. Reduced (a) or magnified (b)?14. Right side up (a) or inverted (b)?

• Here is the whole image–This is a magnifying mirror (for shaving or cosmetics)Clicker question

Informatiuon

• HW # 4 due on Thursday, Sept. 24

• Exam #1 on Tuesday, September 29 – Chapters 1-3 of the textbook;

• Exam preparation: http://www.colorado.edu/physics/phys1230/phys1230_fa09/Exams.htm

• On Thursday: questions/answers & exam #1 material overview + new material

HW#2

Concept question

• Is your make-up/shaving mirror

A.Flat B.ConcaveC.Convex

Lenses

Focal point of a converging lens is where image of a distant object (sun) lies

• The focal point of a mirror was at the image point of a distant star (or the sun) seen in that mirror

• This is true for both convex mirrors (virtual image of star) and for concave mirrors (real image of star).

• The reason for involving the sun or a distant star is to guarantee that the incoming rays are essentially parallel.

• The same is true for lenses– The converging lens of a magnifying glass

brings the sun’s (parallel) rays to a focus at the “hot spot” which can produce a fire. The location of the hot spot is at the focal point of the lens.

Focal length

Focal point

Converginglens

(Parallel) rays from the sun

Unlike a mirror there are two focal points of a lens, one on either side

• Here is a converging lens with its axis and BOTH focal points shown– Both sides of a converging

lens bulge outwards

• Parallel incoming rays are refracted so that they ALL pass through the focal point F'– This fact defines F' and can

be used to find its location!

• Rays diverging from F and passing through the lens ALL come out parallel

F'F

Incoming ray is benttowards nornal

Outging ray is bentaway from nornal

F'F

Can we be more precise with distances and magnifications?

• Do we have to draw a to-scale ray-tracing for each different object location to find the image location?

• No, you don't have to do any ray-tracing at all (Thank goodness!!)

• But you have to use some algebra

• The distance between the center of the lens and either F or F' is the same.– It is called the focal length, f

F F'

These distances are theSAME. Either one is

called the focal length, f

How to use rays of type 1, 2 and 3 to find the image produced by a thin converging lens

• The image produced by a thin converging lens whose axis and focal points are known can be found by replacing the lens by a plane through its center

• A ray of type 1 goes from Alex's nose parallel to the axis to the plane and then refracts through the focal point F'

• A ray of type 2 goes from his nose through the center of the lens (plane)

• The image of the nose is at the intersection of the two rays

• You can check your results with a ray of type 3 which goes from his nose through focal point F and then parallel to the axis

F F'

Is the image upside down or right side up?Use the axis to help determinethe image by locating its bottom!

Where is the image when Alex is close to the converging lens (closer than F)?

• We draw incident and transmitted rays of type 1 and 2.– Note that the transmitted (bent) rays

diverge on the other side of the lens so the image cannot be there.

• Try extending the transmitted rays backwards. – The image is where they intersect– Note rays of type 3 are in agreement

with the image point determined by rays 1 and 2 even though they don't pass through the lens!

– Is it inverted (a) or right-side-up (b)?– Is it magnified (a), reduced (b) or

unchanged (c) in size?

F F'

This is a magnifying glass!!

Compare the results of ray tracing to find an image of Alex for each of his two positions.

F F'F F'

Alex is closer to the lensthan the focal point F

Image is virtual

Alex is further from the lensthan the focal point F

Image is real

What happens if we put a translucent screen at the location of the real image?

• Each image point is now scattered from the screen in all directions

• Hence, you eye can see the image not only by looking into rays 1, 2 and 3, but also by looking at the screen into the new (scattered) rays

• We can even see the image from the other side of the screen since rays are scattered there too!

F F'

New rays scattered off translucentscreen particles at each point

Incoming ray is benttowards nornal

How does a diverging lens differ from a converging lens?

• A diverging lens has concave surfaces (one or both)

• The focal points on a diverging lens are reversed.

– Parallel incoming (type 1) rays come out the other side diverging as though they came from the first focal point which is called F'.

– Incoming (type 3) rays aimed at the focal point behind the lens (F) come out parallel

• Diverging lenses deflect rays so that they point further away from the axis

FF'

Outging ray is bentaway from nornal

FF'

http://micro.magnet.fsu.edu/primer/java/lenses/converginglenses/index.htmlhttp://micro.magnet.fsu.edu/primer/lightandcolor/lenseshome.html

http://micro.magnet.fsu.edu/primer/java/lenses/simplethinlens/index.htmlhttp://micro.magnet.fsu.edu/primer/java/lenses/diverginglenses/index.html

http://micro.magnet.fsu.edu/primer/java/components/perfectlens/index.html

Computer simulation of light focusing/defocusing by lenses:

As a pencil moves closer to a converging lens what happens to its image?

See the following online interactiveexperiment to learn the answer

http://www.colorado.edu/physics/phet/simulations/lens/lens.swf

Sept 24

The lens equation gives the same results as ray-tracing but without any rays!

• 1/x0 + 1/xi = 1/f• f is focal length of lens

• xo = positive distance from object to center of lens (when object is left of the lens)

• xi = distance from image to lens center

• xi is a positive number for (real) image on other side of lens from object.

• xi is a negative number for (virtual) image on same side as object.

• Given two, find the third.

• Can use the lens equation to find image position if know object position or vice versa, without any rays

ff

xo

xi

(will be positivenumber for this case)

Here is one example of how to use the lens eqn with a converging lens

• Given:• f = 10 cm

• Object is 15 cm in front of lens: x0 = 15

• Find:– Where is image and is it real

or virtual?

• Solve equation for xi:– Substitute numbers for letters

– Subtract 1/15 from both sides

– Arithmetic on calculator

– Multiply by xi/0.033

— = — + —1 1 1

f xo xi10 15

1510— - — = —1 1 1

xi

0.033 = —1xi

xi = — = 30 cm1.033

Image is 30 cm to right of center of lensand is real because xi is positive

Here is a sketch to show the previous result

• We can verify our result by ray-tracing f = 10

xo = 15

xi = 30

Here is an example of how to use the lens eqn with a diverging lens (see Fig. 3.28)

• Given:• f = —5 cm

• NOTE, THE FOCAL LENGTH OF A DIVERGING LENS IS NEGATIVE

• Object is 12 cm in front of lens: x0 = 12

• Find:– Where is image and is it real or

virtual?

• Solve equation for xi:

– Substitute numbers for letters

– Subtract 1/12 from both sides

– Arithmetic on calculator

– Multiply by xi/(-0.283)

— = — + —1 1 1

f xo xi-5 12

— - — = —12-5

1 1 1xi

-0.283 = —1xi

xi = — = -3.53 cm1-0.283

Image is 3.53 cm to left of center of lensand is virtual because xi is negative

Summary of the meaning of negative number in the lens and magnification equations

• Negative focal length, f, means the lens is diverging. – Otherwise it is converging.

• Negative image distance, xi, from image to lens means the image is to left of the lens, on same side as object and virtual (rays coming from it never really meet)– Otherwise image is real

• Negative object distance, xo, means the object seen by the lens is on the wrong side of the lens (to right of lens)

• Negative magnification, M, means the image is upside down (inverted) relative to the object.

1/f = 1/xi + 1/xo

M = -xi/xo

We can find the magnification of the image relative to the object by using another formula

– The magnification of the image is defined by the following eqn.

• M = Si/So

• So = size of object

• Si = size of image• M is negative for image inverted

relative to object.

• M is positive for image not inverted relative to obect.

• The magnification can found in terms of the distances of the image and object from the lens

• M = -xi/xo– For the situation at the right,

M = -30/15 = -2– Hence, the image is twice as big as

the object and is upside down

f = 10

xo = 15

xi = 30

Now find the magnification using the magnification equation

• M = -xi/xo = -(-10)/5 = 2

– Hence, the image is twice as big as the object and is rightside up

• Compare with our earlier ray-tracing– The image is the same by

both methods

F F'

The "power," P, of a lens is equivalent to the focal length, f. One can be found from the other.

• Definition of P in terms of f is

• Meaning of P• P is a measure of the ray

bending power of the lens

• Large P means the lens bends rays more than if P were small

• Your eyeglass or contact lens prescription is usually given in diopters (P)

P (in diopters) 1

f (meters)

• The power of a converging lens is always positive because f is a positive number for a converging lens– The converging lens always bends

rays towards the axis behind the lens• The power of a diverging (concave)

lens is always negative because f is negative for a diverging lens– The diverging lens always bends rays

away from the axis behind the lens

Demonstration

Demonstration

Question:

The infrared camera in demonstration allows one to see people in the classroom because:

• (A) This is a magic camera;• (B) It detects infrared rays

emitted by people;• (C) Camera is based on total

internal reflection;• (D) Because of dispersion;• (E) This is the camera obscura

Question for class (doesn't count)

• What is your opinion of the class textbook, Seeing the Light?a) It is clear and useful

b) I sometimes have trouble understanding it

c) I can't understand anything in it

d) I don't read (or have) it

e) I “sort of” read it sometimes but don’t spend that much time trying to understand it.

How do you feel about using the clickers (doesn't count)?

a) Scares me to death

b) Ruins the class

c) Not fair

d) Keeps me on my toes

e) Good method to remind me what I need to know for exams but I wish there was some other way

Aside from whether or not you like them, do you find the clicker questions

a) Helpful in learning

b) Especially helpful because I go to the library to review them afterwards

c) Neither helpful nor unhelpful

d) Not too helpful in learning the concepts

e) They confuse me more than they help me

Which phrase best describes your opinion of the class lectures?

a) Can't understand them

b) Not worthwhile

c) Somewhat helpful

d) Very helpful

e) Boring, since I know this stuff!

Exam: September 29, regular class time

• No homework assignment this week;

• Exam: book chapters 1-3;• Will need a calculator;

• The exam will be worth 100 points.  There are 10 multiple choice questions worth 4 points each and 3 problems worth a total of 60 pts. 

• Example of Fall 2007 assignment & solutions for exam #1:

http://www.colorado.edu/physics/phys1230/phys1230_fa08/X1WhiteSoln.htm

To practice, use homeworks;

Use web-based tutorials:http://www.colorado.edu/physics/phys1230/phys1230_fa08/WebTutorials.htm

Especially to learn properties of mirrors, eclipses, and lenses. Examples:

http://micro.magnet.fsu.edu/primer/java/lenses/simplethinlens/index.html

http://antwrp.gsfc.nasa.gov/apod/image/0311/112003lunareclipse_koehn.gif

http://micro.magnet.fsu.edu/primer/java/mirrors/convexmirrors/index.html

http://micro.magnet.fsu.edu/primer/java/mirrors/concave.html

http://micro.magnet.fsu.edu/primer/java/components/perfectlens/index.html

http://micro.magnet.fsu.edu/primer/java/lenses/simplethinlens/index.html

Can verify your calculations and ray diagrams.

Exam information

• Test of understanding;• One page of info allowed;• No need to memorize equations (all

needed equations will be provided, see an example on the left).

Use web-based tutorials:http://www.colorado.edu/physics/phys1230/phys1230_fa08/WebTutorials.htm

Especially to learn properties of mirrors, eclipses, and lenses. Examples:

http://micro.magnet.fsu.edu/primer/java/lenses/simplethinlens/index.html

http://antwrp.gsfc.nasa.gov/apod/image/0311/112003lunareclipse_koehn.gif

http://micro.magnet.fsu.edu/primer/java/mirrors/convexmirrors/index.html

http://micro.magnet.fsu.edu/primer/java/mirrors/concave.html

http://micro.magnet.fsu.edu/primer/java/components/perfectlens/index.html

http://micro.magnet.fsu.edu/primer/java/lenses/simplethinlens/index.html

Can verify your calculations and ray diagrams.