Chapter 23- Light: Geometric Optics - University of Reginauregina.ca/~barbi/academic/phys109/2010/notes/lecture-23.pdf · 1 W 2 67.0 kg 9.80 m s 656 ... Visible light is a form of

Chapter 23- Light: Geometric Optics

Old assignments and midterm exams (solutions have been posted on

the web)can be picked up in my office

(LB-212)

AssignmentAssignment 1010Textbook (Giancoli, 6 th edition):

Due on Thursday, December 3.

1. On page 222 of Giancoli, problem 48.



4. A girl of mass 50.6 kg stands on the edge of a frictionless merry-go-round of mass 827 kg and radius 3.72 m that is not moving. She throws a 1.13 kg rock in a horizontal direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is 7.82 m/s. Assume that the merry-go-round is a uniform disk. Calculate:

(a) the angular speed (magnitude of angular velocity) of the merry-go-round (Hint: assume conservation of angular momentum) and

(b) the linear speed (magnitude of the tangential velocity) of the girl after the rock is thrown.

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3. If a student does not receive the e-mail by December 1, 2009, he/she is encouraged to contact the Science Office ([email protected]) informing them the e-mail has not been received.them the e-mail has not been received.

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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 EquilibriumFirst 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)

The Conditions for EquilibriumThe Conditions for EquilibriumSecond 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

The Conditions for EquilibriumThe Conditions for Equilibrium

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.

ll/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 9-27 (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= − = → =∑

G 0 y yF F Mg mg= − − = →∑

( ) ( )( )( )

G

2

G

1W 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:

( )( ) ( )( ) ( )11

222W

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

4.0 mx 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 a 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 about 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 at 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 the 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 seen 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

You think you are seeing light travelling in a straight path from that the height of the

object doesn’t change either.

θABD θCBD

This basic explains why we see theimage the way it is!

The image formed “behind ” the mirroris called virtual image opposed toreal 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 ligh t 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 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, if 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.

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Chapter 23- Light: Geometric Optics - University of Reginauregina.ca/~barbi/academic/phys109/2010/notes/lecture-23.pdf · 1 W 2 67.0 kg 9.80 m s 656 ... Visible light is a form of