Notes for ray-optics, J. Hedberg © 2019 · 2019. 11. 27. · PHY 21900 - ray-optics - J. Hedberg -...

Optics1. Intro: Models of Light2. The Ray Model

1. Reflection2. Refraction3. Total Internal Reflection

3. Images1. The plane mirror2. Spherical Mirrors (concave)

4. Lenses: an introduction1. The parameters of a lens2. Real Images from lenses3. Virtual Images4. Refraction at a Spherical Boundary5. Two lens systems

5. Optical Instruments

Intro: Models of LightTheWaveModelWave optics treats light as a wave and uses a similar analytical framework as sound waves and other mechanical waves.

TheRayModelRay optics uses the fact that light seems to travel in a straight line. This makes understanding some phenomenon easier.

TheParticle/PhotonModelModern Physics has shown that treating light at a particle, a photon, can be a very successful approach as well. (Requires quantum theory)

Some facts about light1. Light can be considered a wave phenomenon. It will be shown to be created by oscillations of electric and magnetic fields.2. The speed of light in vacuum is constant: m/s3. In vacuum, our same relationship between speed, frequency, and wavelength holds:

4. The visible light we see is just a small section of the whole spectrum of electromagnetic oscillation frequencie.

Quick Question 1

In 1667, Galileo attempted to measure the speed of light by having twopeople hold covered lanterns on hills that were about 1.5 km apart. Oneperson would measure time. One of the people with a lantern woulduncover it. The other person would then uncover his lantern when he sawthe light from the first lantern. Repeated attempts failed. To see why,determine the approximate time it takes light to travel the 1.5 km distance.

The Ray Model

c = 299, 792, 458

c = fλ

a) 5 sμ

b) 50 sμ

c) 5 msd) 50 nse) 5 ns

PHY 21900 - ray-optics - J. Hedberg - 2019

updated on 2019-11-27

Euclid treated lights as a line, called a light ray.

This is a pretty amazing conceptual accomplishment considering they didn’t have lasers or flashlights or other ‘line’ like sources of light.

Like we did before, we have to choose a lot of things to ignore when trying to describe the natural world.

For the ray model, we will ignore the fact that light is a wave phenomena. That means we’ll forget all about diffraction and interference and other waverelated issues.

This can be justified if we restrict our discussions to the interaction between light and objects that are large compared to the wavelength of the light, likemost everyday things.

1. Light Rays travel in straight linesLight rays travel through a vacuum or a transparent material in straight lines called light rays. The speed of the light in the material is given by v = c/n,where n is the index of refraction.

Light Rays can cross2. Light rays do not interact with each other. Two rays can cross without being affected in any way.

3. A light ray travels forever, unless it interacts with matter



vacuum

vacuum

reflection

material

A light ray continues on forever unless it has an interaction with matter. This interaction may cause the ray to be redirected, or absorbed. We’ll discussthese interactions later.

4. An object is a source of light raysRays originate from every point on an object, and each point send rays in all directions.

Sources of raysObjects in the ray model can either be (a) sources of light (self-luminous objects): these are object like the sun, or lightbulbs.

1. Ray Source: like a laser pointer. These emit a singleray in one direction.

2. Point Source: little light sources that project in all directions (christmas tree lights)

3. Extended Source: These also project rays in all directions, but have a non-pointlike extension. (Large light bulbs)

4. Parallel Ray source: These sources create a beam of parallel rays. (flashlights, far away stars)

Reflection... or how do you see anything.

The light from self luminous objects will reflect off other objects creating essentially extended sources of light.

In order to see something, a ray must travel from an object to our eye.



In the case of the laser, the only way you can tell that it is on, is if the light from the laser bounces of something and strikes your eye.

Reflection (Diffusive)When light rays meet rough surfaces, such as a piece of paper, they still obey the lay of reflection microscopically, however, since the surface normals arechanging, the reflected light rays scatter at many angles.

Reflection (Specular)normal

incident ray reflected ray

angle of incidence

angle of reflection

reflective surfaceTheLawofReflection: The incident ray and the reflected ray are both in the same plane, which is perpendicular to the surface The angle of incidence equals the angle of reflection:

(For optics measurements, we will use as defined between the ray and the normal, not the surface.)

Quick Question 2

light ray

Here is a light ray incident on a reflective surface as shown. Which, if any, ofthe following are true?

=θi θr

θ

a) − α = θ90∘

b) α = θ

c) α+ θ = 180∘

d) α ≠ θ

e) Lies! These are all lies I say.



At what angle should the laser beam be aimed at the mirrored ceiling so that it hits the wall half way up?

3.0 m

5.0 m

wall

mirror

RefractionA light ray (the incident ray) encounters a boundary between two media. At that boundary, some of the light is reflected, but most is transmitted.Observation clearly shows that the light inside new medium (the glass) has changed direction.

incidentray

refractedray (glass-air)

refractedray (air-glass)

reflected ray

Index of RefractionThe index of refraction (or refractive index) of a material is a dimensionless parameter, , used to describe how light moves in a particular medium.

The index of refraction of vacuum is and is exactly 1.

Medium IndexVacuum 1 (exactly)Air (0ºC, 1 atm) 1.00029Water 1.33Glass 1.52Saphire 1.77Diamond 2.42

[*Note for labs: The index of refraction can change based on the wavelength of the light]

Quick Question 3

In which one of the following substances does light have the largest speed?

Medium IndexVacuum 1 (exactly)Air (0ºC, 1atm)

1.00029

Water 1.33Glass 1.52Saphire 1.77Diamond 2.42

Example Problem#1:

n

n = =c

v

speedoflightinvacuumspeedoflightinmaterial

n = 1.000

a) Diamondb) Saphirec) Glassd) Watere) None of the above. The speed of light has the same valueeverywhere in the Universe.



air

glass

Many parallel rays enter a new medium. They all change direction in the same manner. In case, we see rays traveling through air, with n = 1, and enteringinto glass with n = 1.5.

medium 1medium 2

angle of refraction

angle ofincidence

normalreflected ray

n2> n1

We can look closely at one ray, and label the interesting parts of the refraction phenomenon. We’ll call the angle between the incident ray and thenormal of the surface. is the angle between the refracted ray and the normal inside the 2nd medium.

Snell's Law

medium 1medium 2

angle of refraction

angle ofincidence

normalreflected ray

n2> n1

Snell’s law describes the geometry of this situation.

is the index of refraction of medium 1 is the index of refraction of medium 2

If the direction of the ray is reversed (i.e. now it is traveling from medium 2 into medium 1) we change the labels of refraction and incidence, but keep theangular definitions and indices of refraction the same.

Snell's Law still holds.

medium 1medium 2

angle of refraction

angle ofincidence

normal

reflected rayn2> n1

A beam of light of wavelength 550 nm in air strikes a slab of transparent material at an angle of 40º to the normal, and the refracted beam makes an

θ1θ2

sin = sinn2 θ2 n1 θ1

n1n2

sin = sinn2 θ2 n1 θ1

Example Problem#2:



angle of 26º to the normal. Find the index of refraction of the material.

... a limit is reachedLet's calculate for this case

n = 1

n = 2

=?

45º

Total Internal ReflectionWhen angle of incidence becomes too large, none of the light will be able to be transmitted through the boundary. When this point is reached, we havetotal internal reflection.

We can find the critical angle when total internal reflection begins by putting into Snell’s Law:

45º45º

n2

n1

n2

n1n2> n1

Total internal reflection occurs only when light attempts to move from a medium of higher index of refraction ( ) to a medium of lower index of refraction( ). The critical angle is determined by Snell’s Law, using 90 degrees for for the refracted angle:

Total internal reflection underwater

image source

θ1

=θ1 90∘

n2n1 θc

= ( )θc sin−1n1

n2

Example Problem



http://en.wikipedia.org/wiki/File:Total_internal_reflection_of_Chelonia_mydas_.jpg

A lightblub is set in the bottom of a 3.0 m deep swimming pool. What is the diameter of the circle of light seen on the water's surface from above?

Images1. Real Images: A realimages exists at a position in space independent of if there is an observer. Ex: images on a movie screen, focused images ofsun using a magnifying glass.2. Virtual Images: These are images which do not really exist at a place in space -- we only perceive them to exist. Ex: images in plane mirrors, whatyou see when you look through binoculars.

The plane mirrorLight strikes a mirror. Each ray coming from the object obeys the law of reflection.

object mirror

The plane mirrorobject

O I

p i

If we didn’t know there was a mirror there, we might be tempted to say the object was actually located at point P.

Images in plane mirrors

#3:



O Ip i

object virtualimage

Plane Mirror - Virtual Image with extended objectIf the object is not a point, but has a spatial extent, then we have to consider each point of the object. Each point on the object will have a correspondingimage point on the opposite side of the mirror.

objectvirtualimage

p i

Quick Question 4

A ball is held 1.5 m in front of a plane mirror. How far is the image of the ballfrom the ball?

Conventions

O Ip i

object virtualimage

We have to establish several signconventions when dealing with ray optics.

The first will be to always call the object distance positive and the image distance for virtual images negative.

a) 0 mb) .75 mc) 1.5 md) 3.0 me) 6.0 m

p = −i



Quick Question 5

An image in a plane mirror is...

Spherical Mirrors (concave)

mirr

or p

lane

optical axis

object

p

C

radius of cu

rvature

focal point ( f )

f

Note: The optical axis (central axis) extends through the center of curvature and the center of the mirror.

The focal point is where the reflected rays from two parallel incident rays meet and is also on the optical axis. It's positive in the case of a concave mirror.

Spherical Mirrors (convex)

mirr

or p

lane

optical axis

object

p

C

radius

of cu

rvatur

e

focal point ( f )

f

Here the focal point is virtual and it gets a negative sign.(f<0)

The mirror plane sets the origin for all distance measurements.

For both convex and concave mirrors, the focal length is half the radius of curvature.

a) flipped left to rightb) flipped top to bottomc) flipped front to backd) not flipped at all.



object virtualimage

p i

C

f

F

object at a distance inside the focal point

We start with an object in between the focal point and the plane of the mirror.

This situation creates a virtual image which appears to be on the other side of the mirror.

object

p

C

f

F

object at the focal point

parallel rays

Now we bring the object exactly to the focal point.

When the rays originate from that distance, then their reflected paths are parallel. Thus, they do not converge until . No image is produced.

object

realimage

pi

C

f

F

object at a distance outside the focal point

Outside the focal distance, and object in front of a concave mirror will create a real image.

The real image is inverted.

To calculate the positions of the images, we can use the following:

∞

+ =1p

1i

1f



Here: is the object distance (always positive) is the image distance ( for real images, for virtual images),

and is the focal length, ( for concave mirrors).

object virtualimage

p i

C

hh’

Define a magnification value:

Which can also be found by using:

Convex mirrormirror plane

optical axis

object

pi

f

virtual image

focal point

An object is 30 cm in front of a convex mirror with a focal length of -20cm. Locate the image. Is it upright or inverted?

Lenses: an introduction

parallel rays

optical axis

focal point

double converging lens

p

i + −f +

|m| =h′

h

m = −i

p

Example Problem#4:



optical axis

focal point

double diverging lens

Lenses can be analyzed much the same as mirrors.

Chart of sign conventions

SphericalMirrorsThinLenses

Focal Length for concave mirror for convex mirror

ObjectDistance

if object is in front mirror if object is behind the mirror

Image Distance if image is in front mirror (real image) if image is behind the mirror (virtual)

Magnification for an upright image with respect to the object for an inverted image with respect to the

object

Imagine an incident ray hitting a piece of glass like this:

Snell's Law tells us how to deal with this situation.

A closer view shows the normals and angles of interest.

incident ray

n > 1n = 1 n = 1

tangent tangent

normal normal

An approximation is needed.

If our lens is shaped like so, and we trace all the incoming rays (assuming they are parallel) we end up noticing that they all converge at a point.

The point is located along the optical axis and is given the name focal point.

This lens is called a converginglens.

= +1f

1p

1i

+−+−+−+−



parallel rays

optical axis

focal point

double converging lens

Another type of lens we have looks like this:

Again, Snell's Law tells us how to deal with this situation.

A closer view shows the normals and angles of interest.

incident ray

normalnormal

n > 1n = 1 n = 1

tangent tangent

optical axis

focal point

double diverging lens

Special Rays

Quick Question 6

We would like to create a beam of light that consists of parallel rays. Whichone of the following arrangements would make that happen?

The parameters of a lens

a) A light bulb is placed at the focal point of a convex mirror.b) A light bulb is placed at the focal point of a diverging lens.c) A light bulb is placed at the focal point of a converging lens.d) A light bulb is located at twice the focal length from a concavemirror.e) A light bulb is located at twice the focal length from a converginglens.



The lens has a shape and an index of refraction.

focal point focal point

focal pointfocal point

same n, different r same r, different n

Quick Question 7

Parallel rays of violet light that are directed at a converging lens are focusedat a point P on the central axis to the right of the lens when the lens issurrounded by air as shown. If the lens is surrounded by water instead ofair, where will the red parallel rays be focused relative to point P?

P

These two lenses are both converging, yet one has a shorter focal length.

This could be due to its geometry or its material makeup.

The focal length is a measure of the strength of the lens:

A shorter focal length implies a stronger optical system, since the light is bent more.

a) above point Pb) below point Pc) to the left of point Pd) to the right of point Pe) at point P



parallel rays

focal point

parallel rays

focal point

f

f

Real Images from lenses

Regarding the imagesWe can see that the real image of the object is smaller than the object.

Additionally, it is inverted.

The magnification is the same for lenses as it was for mirrors:

objectheight = h

optical axis

object plane image plane

p i

image height = h’

Quick Question 8

|m| = = and m = −i

p

h′

h

i

p



What will happen to the image on the screen when half the lens is covered?

converging lens

mask

screen

object

converging lenswall

Virtual ImagesIn order to form a real image, we must have converging rays.

If the object is positioned between the near focal point and the lens, the rays through the lens will not converge. Thus, no real image is formed. Thissituation creates a virtual image.

optical axis

f

virtual image

diverging rays

Converging Lens summaryreal image virtual image

a) It will invertb) The image will get dimmer overall but remain the same otherwisec) The upper half will be darkenedd) The lower half will be darkenede) It will vanish entirely.



Diverging Lens

optical axis

p

i

f f

No matter where the object is, the image is virtual.

The thin lens equationFor a thin lens with focal length , image distance , and object distance , we can use the thin lens equation:

(This was the same for mirrors)

lens is converging, f= 30, p = +50

lens is converging, f= 30, p = +20

lens is diverging, f= -30, p = +50

The Lensmaker's equationIf we have a lens which has two radii of curvatures, then in order to find the focal length, we must use the lensmaker’s equation. This assumes the lens isthin, and inair.

r2

r1

n

objectimagefrom

1st boundary

t

final image

Sign Convention

f i p

= +1f

1p

1i

Example Problem#5:

Example Problem#6:

Example Problem#7:

= (n− 1)( − )1f

1r1

1r2



-

concaveIf the object faces a concave surface, then that surface has a negative radius of curvature.

+

convexIf the object faces a convex surface, then that surface has a positive radius of curvature.

Find the focal Length of this lens

20 cm

30 cm

n =1.33

object

Find the focal length of theses lenses:

20cm

50cm

20cm

30cm

60cm

20cm

object

object

object

Chart of sign conventionsSphericalMirrors ThinLenses

Focal Length for concave mirror for a converging lens for convex mirror for a diverging lens

ObjectDistance

if object is in front mirror for real objects if object is behind the mirror for virutal objects

Image Distance if image is in front mirror (real image) for real image if image is behind the mirror (virtual) for virtual image

Magnification for an upright image with respect to the object for an upright image with respect to the object for an inverted image with respect to the

object for an inverted image with respect to the

object

Example Problem#8:

+ +− −+ +− −+ +− −+ +− −



near focal point

far focal pointobject

p

f f

i

h’

h

Refraction at a Spherical BoundaryHere we have a spherical boundary between two different materials with different index of refraction. Rays from point O will be refracted at the boundaryand converge at point I.

C

p i

n1 n2

spherical surface

object point image point

O I

R

Snell’s law can be used to determine the angles of refraction.

object C image

object C image

n1 n2

n1 n2

There are however, many possible permutations for the objects and images.

We have to account for both indices of refraction for the two media, the location of the object, and the location of the center of curvature.

Here are two situations which produce real images.

In both of theses cases, the real images forms on the opposite side of the boundary from the image.

And these situations will produce virtual images.



object Cimage

n1 n2

OC I

n1 n2

O CI

n1 n2

O CI

n1 n2

Note that for theses virtual images, the image is on the same side of the boundary as the object.

Fortunately, one equation rules them all!

Conventions to use this equation:

is positive. It is the object distance

will be positive for a real image, and negative for a virtual image

will be positive if the object is facing a convex boundary and negative if the object is facing a concave boundary.

This fly is in a ball of amber with a radius of curvature of 12 cm. The amber has an index of refraction of n = 1.7. To an observer outside the amber, thefly appears to be located 2 cm from the surface. Where is the fly actually located?

r = 12cm

n = 1 n = 1.7

Two lens systemsThe image from the first lens acts as the object for the second lens.

Example: The ol' fashioned telescope

objectlens 1

(objective)

real image

Example: The ol' fashioned telescope, cont'd

+ =n1

p

n2

i

−n2 n1

r

p

i

r

Example Problem#9:



objectlens 1 lens 2

(objective) (eyepiece)

virtual image

Optical InstrumentsWhat do we need to make an good image of something?

We need all the rays coming from a particular point on an object to meet again at another place. Certain lenses can accomplish this.

Before lenses were around...Or, we could just block all the other, diverging rays with a piece of material.

Now, we can imagine only one ray leaving each point, and eventually reaching the image plane.

HistoricalThe camera obscura - a very old imaging tool



To improve the focus, the pin-hole needs to be smaller. But, then less light gets through. (Also, if the pin-hole gets too small, then we get diffraction). So,we can do better.

The Modern CameraThe essential features of a modern camera are the lens, which focuses the light, and the image plane, where either the film, or a digital sensor is located.

The CCD DetectorCharge-Coupled Device



The eyeNature has made many different eyeballs. Most operate on the principles of lenses we’ve just looked at.

retina

lens

cornea

iris

aqueoushumor

optic nerve

focusing the eyeFocusing on objects: We cannot adjust the position of the lens with respect to the retina. So, the muscles around the eye change the shape of the lens,which then changes its focal length.

Why some of use put lenses on our face all day.



1. Normal vision. Here everything is set up just right

2. Hyperopia (far-sightedness) “My eyeball is too short”

3. Myopia (near-sightedness) “My eyeball is too long”

These two object are the same size, but since they are located in different positions, the occupy different amounts of our field of vision.

Their corresponding images on our retinal surface are therefore different sizes.

This give the appearance of different sizes. We can say the angle subtended by each object is its angular size.



We don't want to get too close though, because the eye can't focus past its near point. On average, that's 25 cm.

Simple Magnifying LensPlacing on object within the focal length of a converging lens results in a virtual image. The apparent size of this virtual image is larger, and thus appearsto be magnified.

optical axis

f

virtual image

diverging rays

object

pnearpoint

hi

Angular magnification

Since and , we can write:

Compound Microscope

Virtual Image

objective lens eyepiece

diverging rays

f’f

Magnification

The telescopeThe refracting telescope:

=mθθ′

θ

θ ≈ h/25cm ≈ h/fθ′

=mθ25cm

f

= M = m = −mθs

fob

25cm

fey



eyepiece

objective eyepiece

virtual image

distance light source

parallel rays

The reflecting telescope

eyepiece

parabolic mirror

parallel rays

flat mirror

AberrationsLenses aren't perfect.

Since different colors will refract at different angles, the focal point will be slightly different for different wavelengths. This leads to ChromaticAberration.

Spherical AberrationEarlier, our thin lens approximation ignored the fact that the thickness of the lens changed as a function of distance away from the central axis.

blurry focal point

This leads to rays having slightly different focal points depending on where they are incident on the lens. The further the rays are away from the centralaxis, the worse the SphericalAberration effect is.

Fixing aberrationsFortunately, by using a multi lens setup, we can correct these aberrations. For example, to correct the chromatic aberration caused by a converging lenswe can insert a diverging lens after the converging lens to refocus the different colors back to the same point.



Quick Question 9

As I walk away from a plane mirror what will happen to the image of my facein the mirror.

a) It will take up more and more of the mirror surface, i.e. get biggerb) Its size won't changec) It will take up less and less of the mirror, i.e. get smaller



Notes for ray-optics, J. Hedberg © 2019 · 2019. 11. 27. · PHY 21900 - ray-optics - J. Hedberg -...

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Transcript of Notes for ray-optics, J. Hedberg © 2019 · 2019. 11. 27. · PHY 21900 - ray-optics - J. Hedberg -...