Chapter 23- Light: Geometric Optics - University of Reginauregina.ca/~barbi/academic/phys109/2008/2008/notes/... · 2008-11-27 · The Ray Model of Light How do we see objects: •

Chapter 23- Light: Geometric Optics

Changes in OfficeChanges in Office--hours hours

The following changes will take place until the end of the semester

Office-hours:

- Monday, 12:00-13:00h

- Wednesday, 14:00-15:00h

- Friday, 13:00-14:00h

Old assignments and midterm exams

(solutions have been posted on

the web)

can be picked up in my office

(LB-212)

All marks, including assignments, have

been posted on the web.

http://ilc2.phys.uregina.ca/~barbi/academic/phys109/marks.pdfhttp://ilc2.phys.uregina.ca/~barbi/academic/phys109/marks.pdf

Please, verify that all your marks have

been entered in the list.

Chapter 23

• The Ray Model of Light

• Reflection; Image Formed by a Plane Mirror

• Formation of Images by Spherical Mirrors

• Index of Refraction

• Refraction: Snell’s Law

• Total Internal Reflection; Fiber Optics

• Thin Lenses; Ray Tracing

• The Thin Lens Equation; Magnification

• Lensmaker’s Equation

Recalling Recalling LastLast LecturesLecturesRecalling Recalling LastLast LecturesLectures

The Conditions for EquilibriumThe Conditions for Equilibrium

First condition of equilibrium

� No change in translational motion

The first condition for equilibrium is that the forces

along each coordinate axis add to zero.

(9-1)


Second condition of equilibrium

� No change in rotational motion

The second condition of equilibrium is that there be

no torque around any axis; the choice of axis is

arbitrary and we will assume that its direction is

perpendicular to the xy plane of rotation.

(9-2)l

l/2


Recalling an important note:

We consider that the mass of an object with

uniform mass distribution (or uniform object for

short) is such that it can be assumed to be

located at the center of the object.

This point is know as either center of mass or

center of gravity of the object.center of gravity of the object.

l

l/2

For an uniform sphere, its center of mass is located in its center.

Solving Statics ProblemsSolving Statics Problems1. Choose one object at a time, and make a free-body diagram showing all the

forces on it and where they act.

2. Choose a coordinate system and resolve forces into components.

3. Write equilibrium equations for the forces.

4. Choose any axis perpendicular to the plane of the forces and write the torque 4. Choose any axis perpendicular to the plane of the forces and write the torque

equilibrium equation. A clever choice here can simplify the problem

enormously.

5. Make sure you correctly identify the components of the forces that can

contribute to rotational motion. Forces that have zero lever arm do not

contribute to rotational motion.

6. Solve.

TodayTodayTodayToday

Problem 9-1 (textbook): Three forces are applied to a tree sapling, as shown in

Fig. 9–41, to stabilize it. If and , find in magnitude and

direction.

Problem 9-1 :

If the tree is not accelerating, then the net force in all

directions is 0.

( )

A B C

C A B

cos110 0

cos110 310 N 425 N cos110 164.6 N

sin 0

x x

x

F F F F

F F F

F F F

= + + = →

= − − = − − = −

= + = →

∑

∑

And so is 430 N, at an angle of 112o clockwise from .

( )

B C

C B

sin 0

sin110 425 N sin110 399.4 N

y y

y

F F F

F F

= + = →

= − = − = −

∑

( ) ( )2 22 2 2

C C C

C1 1 o o o o o

C

164.6 N 399.4 N 432.0 N 4.3 10 N

399.4 Ntan tan 67.6 , 180 67.6 112.4 112

164.6 N

x y

y

x

F F F

F

Fθ φ− −

= + = − + − = ≈ ×

−= = = = − = ≈

−

Problem 9-27 (textbook): Consider a ladder with a painter climbing up it (Fig. 9–

61). If the mass of the ladder is 12.0 kg, the mass of the painter is 55.0 kg, and the

ladder begins to slip at its base when her feet are 70% of the way up the length of the

ladder, what is the coefficient of static friction between the ladder and the floor?

Assume the wall is frictionless.

Problem 9-27:

The ladder is in equilibrium, so the net torque and net force must be zero.

By stating that the ladder is on the verge of slipping, the static frictional force at the

ground, , is at its maximum value and so:C xF

C Cx s yF Fµ=

Since the person is standing 70% of the way up Since the person is standing 70% of the way up

the ladder, the height of the ladder is:

The width of the ladder is:

0.7 2.8 m 0.7 4.0 my yL d= = =

0.7 2.1 m 0.7 3.0 mx xL d= = =

Problem 9-27

Torques are taken about the point of contact of the ladder

with the ground, and counterclockwise torques are taken

as positive. The three conditions of equilibrium are as follows:

C W C W0

x x xF F F F F= − = → =∑

G0

y yF F Mg mg= − − = →∑( ) ( )( )

( )

G

2

G

1

W 2

67.0 kg 9.80 m s 656.6 N

0

y y

y

y x x

F M m g

F L mg L Mgdτ

= + = =

= − − =

∑

∑

Solving the torque equation gives:

( )( ) ( )( )( )

11222

W

12.0 kg 3.0 m 55.0 kg 2.1 m9.80m s 327.1 N

4.0 m

x x

y

mL MdF g

L

++= = =

The coefficient of friction then is found to beG

G

327.1 N0.50

656.6 N

x

s

y

F

Fµ = = =

The Ray Model of LightThe Ray Model of Light

Light is a form of electromagnetic wave.

• It has electric and magnetic components and propagates according to the

properties of wave

Visible light is a form of electromagnetic wave with wavelength in the so called

visible region of the light spectrum

In fact, light sometimes also behaves as particle � come to my Phys-242 classes

and you will understand some of the particle-like properties of light and how it

changed our perception of the Universe.

The Ray Model of LightThe Ray Model of Light

How do we see objects:

• An object can be a source o light such as stars, candles, etc.

• An object might reflect the light produced by some light source.

In either cases, the light will travel from the object to your eyes.

In this chapter we will assume that light travels in straight paths.

� We represent light using rays which are straight lines emanating from an object.

� This is an idealization, but is very useful for

our purposes.

It is therefore important to understand how light

reflects on objects in order to understand how

we see the world

Since we will be dealing with straight-line rays at various

angles, we call this area geometric optics.

Reflection; Image Formation by a Plane MirrorReflection; Image Formation by a Plane Mirror

Given that one way of observing objects is by light reflection, let’s spend the next

few slides discussing on it.

As light strikes on an object, the following processes can take place:

• Some or all of the light can be reflected

• Some or all of the light can be absorbed (for example, transformed into

thermal energy)thermal energy)

• If the object is transparent (for example, glass or water), some or all of the

light can be transmitted through the object


If a narrow beam of light strikes a flat surface at an angle θi relative to the normal

(perpendicular) to the surface, it can reflect through an angle θr as depicted in the

figure below.

θi is called the angle of incidence;

θr is called the angle of reflection.


It is found that θi and θr both lie in the same plane with the normal to the surface,

and that:

Eq. 23-1 is known as Law of Reflection.

(23-1)


When parallel rays of light reflects from a rough surface, the law of reflection

still holds, but the angles of incidence vary. This is called diffuse reflection.

normal(1)

normal(2)

normal(3)

What happens is that the normal is defined relative to the surface. Depending

where the ray of light strikes the surface, the normal will be pointing to different

directions.

On the other hand, when parallel rays of light reflects on a flat smooth surface, the

angle of incidences do not vary. This is called specular reflection.

normal(1)


The effect of diffuse reflection is that an object can be see at different angles by the

light reflect from it.

With specular reflection (from a mirror), your eye must be in the correct position,

otherwise you do not see the object.


What you see when you look into a plane (flat) mirror is an image, which appears

to be behind the mirror.

You have the perception that the object (or

you) is represented by an image that

reflects the correct dimensions of the

object and appears to be at a distance

“behind” the mirror that corresponds to that

real distance between the object and the

surface of the mirror.

But how does it work?


Let consider the object in the figure (a bottle of something).

Light will be reflected at any point

on the bottle surface that is

illuminated by some beam of light.

Let’s consider two points: the top and

bottom of the bottle.

Light rays will leave each of these

points (and any other point to this

matter) at different angles.

Let’s consider two of such rays leaving each point as shown in the figure such

that they represent the threshold angles between which any other ray will arrive

at an observer’s eye after reflecting on the mirror.

Each set of rays leaving a point on the bottle and reflecting on the mirror will be

seem by the observer as coming from a single point as shown in the figure.


What happen is the following:

Consider point A on the object.

Each of the two rays will leave

this point and strike the mirror at

points B and B’ as shown in the points B and B’ as shown in the

figure.

They will reflect following the law

of reflection.

For instance, at point B the ray

will have angle of incidence θi and

reflection θr such that:


We can now use simple geometry to find some of the properties of the image

formed by the object in the mirror:

It is clear from the figure that both ADB and CDB are right triangles.

The angle ABD can be obtained observing that:

The angle CDB is:The angle CDB is:

Using θABD found above and :

Then:

θABD θCBD


The equality above tell me that the image distance di is equal to the object

distance do.

It is not difficult to show that

the height of the object

You think you are

seeing light

travelling in a

straight path from the height of the object

doesn’t change.

θABD θCBD

This basic explains why we see the

image the way it is!

The image formed “behind” the mirror

is called virtual image opposed to

real image in which the image is

formed on the same side of the

mirror where the real object is located

(we will come back to this in few slides)

straight path from

the image


Example 23-3 (textbook): Two mirrors meet at a 135° angle, Fig. 23–47. If light

rays strike one mirror at 40° as shown, at what angle φ do they leave the second

mirror?


Example 23-3 (textbook):

Using the law of reflection, we will have from the triangle formed by the mirrors and

the first reflected ray:

40 135 180 ,φ° + ° + = °

5 .φ = °


Example 23-4 (textbook): A person whose eyes are 1.68 m above the floor stands

2.20 m in front of a vertical plane mirror whose bottom edge is 43 cm above the

floor, Fig. 23–48. What is the horizontal distance x to the base of the wall

supporting the mirror of the nearest point on the floor that can be seen reflected in

the mirror?


Example 23-4 (textbook):

The angle of incidence is the angle of reflection. Thus we have

Then,

Mirror

H θ

( )tan ;

H h h

L xθ

−= =

H

h

x

L

θ

θ

( )( )

1.68m 0.43m (0.43m),

2.20m x

−=

0.76m 76cm.x = =

Formation of Images by Spherical MirrorsFormation of Images by Spherical Mirrors

Mirrors do not need to be flat. Actually, several mirrors are in fact curved (for

example, rearview mirrors on some cars are curved).

The most common type of curved mirror is the spherical.

A spherical mirror forms a section of a sphere.

There are two types of spherical mirrors:

Concave: The reflecting surface is on

the inner surface of the sphere.

Convex: The reflecting surface is on the

outer surface of the sphere.

Note: the law of reflection still applies for

curved mirrors.


In order to study images formed by spherical mirrors, we will first consider objects

that are located very far from a concave mirror.

Rays coming from a faraway object are effectively parallel.

For example, it applies well to our sun and other stars located very distant from the

Earth.


However, we consider the law of reflection, the rays coming from a point on a

distant object will not all converge at exactly the same place after reflecting on the

mirror. Unlike in flat mirrors, the image of the object formed by a spherical mirror

will be distorted.

This is what we call spherical aberration.This is what we call spherical aberration.


Spherical aberrations can be significantly minimized if we use mirrors which are

smaller than its radius of curvature (r in the figure).

Or, equivalently, a mirror with a small curvature.

In this case, the angle between the incident and reflected rays are small (2θ in the

figure below) � the rays will then cross each other at very nearly a single point

called focus.

In the figure:In the figure:

C is the center of curvature of the mirror

F is the focal point (focus)

A is the center of the mirror

f = FA is called focal length

CA is called the principal axis.

The focal point is the image point of an

object located infinitely far away along the

principal axis.

Documents

Chapter 23- Light: Geometric Optics - University of Reginauregina.ca/~barbi/academic/phys109/2008/2008/notes/... · 2008-11-27 · The Ray Model of Light How do we see objects: •