Chapter 23- Light: Geometric Optics - University of...

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 , 12:00-13:15h- Monday , 12:00-13:15h

- Wednesday , 14:00-15:00h

Old assignments and midterm exams (solutions have been posted on

the web)can be picked up in my office

(LB-212)

Supplemental Instruction

Physics 109 Final Exam Review

Starting Wednesday, December 2nd (in SI Classroom, LB-205) at 11:30am.205) at 11:30am.

Review will continue on: Dec. 4th, Dec. 7th, and Dec. 9th (LB-205, 11:30am)

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

• The Thin Lens Equation; Magnification

• Lensmaker’s Equation

Recalling Recalling LastLast LecturesLecturesRecalling Recalling LastLast LecturesLectures

The Ray Model of LightThe Ray Model of Light

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.

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

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)

• If the object is transparent (for example, glass or water), some or all of the • 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 .(23-1)

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


normal(1)

normal(2)normal(3)

Diffuse reflection � reflection on a rough surface.

Specular reflection � reflection on a flat smooth surface.


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

You think you are

What you see when you look into a plane (flat) mirror is an image, which appears to be behind the mirror.

It is not difficult to show that the height of the object doesn’t change.

θABD θCBD

Virtual image � the image formed “behind ” the mirror

Real image � the image is formedon the same side of the mirror.

You think you are seeing light travelling in a straight path from the image

TodayTodayTodayToday


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.


Using geometry and the fact that 2θθθθ is small, we have:

(23-2)

Eq. 23-2 tells us that the focal length is half the radius of curvature.


Eq. 23-2 applies to any ray that makes a small angle with the principal axis.

(23-2)

Eq. 23-2 applies to any ray that makes a small angle with the principal axis.

� These rays are called paraxial rays .

Spherical aberration can be avoided by using a parabolic reflector; these are more difficult and expensive to make, and so are used only when necessary, such as in research telescopes.


We have defined some of the spherical mirrors characteristics such as radius of curvature, center of curvature, focal length and principal axis.

I have also observed that an image of an object at infinity is located at the focal point of the spherical mirror.

But, what about images of objects not at infinity?

Let’s now see a general method to construct images of objects formed by spherical mirrors.


Consider the object as defined in the figure above:

The object is located at point O between points F (focus) and C (center of curvature). Point O’ is the top of the object.

Where will the image of point O’ be located?

Several rays will leave point O’ and reflect on the mirror. However, for the effect of reconstructing the image, we actually need only three of such rays:

1) a ray passing through the focus (F); 2) a ray passing through O’ and parallel to the principal axis; and3) a ray along the line connecting point O’ and the center of curvature C.

In fact, only two of these rays are needed , but a third one can be used to check whether you have found the right image.

Let’s use the law of reflection and see how this works.


1) A ray parallel to the axis ; after reflection it passes through the focal point;

2) A ray through the focal point ; after reflection it is parallel to the axis

3) A ray perpendicular to the mirror ( radial direction ); it reflects back on itself (the ray is in the direction of the

⇒ i = r = 0 (the ray is in the direction of the normal ⇒ θi = θr = 0 ).

The point where these three rayscross is the image I’ . All other rays from O’ will also cross I’.

To get a full image, we can do the same with other points in the object, though two points suffice for many purposes.

What about the object’s image dimensions and its distance from the mirror?

Again, using geometry:


Let:

ho = height of the real object (OO’)hi = height of the image (I‘I)do = distance of the object from the center of the mirror (OA)di = distance of the image from the center of the mirror (IA)

Let’s now consider two rays from O’ : O’FBI’ and O’AI’(next slide)


Consider the law of reflection and look at ray O’AI’ :

The two right triangles O’AO and I’AI are similar since both have angles θθθθ and 90-θθθθ subtended by their sides and respective hypotenuses as can be seen in the figure:


As we have discussed few slides ago, assuming a mirror much smaller than its radius of curvature � the reflected ray BI’ can be considered as parallel to the principal axis. Take this into account, let’s look at ray O’FBI’ :

The two right triangles O’FO and AFB are similar since both have similar angles O’FO and AFB.

Noting that I’I = BA, like on the previous slide we have:


Let’s summarize what we have gotten so far:

We can combine these two equations:

Dividing the above expression by dodif

(23-3)


(23-3)

Eq. 23-3 is called the mirror equation and relates the object and image distances to the focal length ( f = r/2 )


We can also find the magnification , m, of a mirror as the ratio of image height to the object height.

Note that we have introduced a sign in the above equation.

(23-4)

Note that we have introduced a sign in the above equation.

Do NOT confuse it with the equation obtained few slides ago.

The negative sign here is a convention to indicate that the image is inverted .


In our example, the object is between the center of curvature and the focal point

� its image is larger , inverted , and real (in front of the mirror).


If an object is outside the center of curvature of a concave mirror

� its image will be inverted , smaller , and real (in front of the mirror).


If an object is inside the focal point

� its image will be upright , larger , and virtual (behind the mirror).


Notes:

In the figure below, your eye is located outside the distance AI. What you see are rays diverging from the image. This allows you to see the image well focused because if you trace the rays back to the image, they will converge at the location of the image.

If you were between points A and I, the rays would be converging to form the image at I. This way, you would see the rays diverging (remember, you are looking at the mirror) and the image will look blurry to you.looking at the mirror) and the image will look blurry to you.

You are here

You are now somewherein here


Sign conventions;

Assuming the object height ho to be positive , the sign convention we use are:

• The image height hi is positive if the image is upright, and negative if inverted, relative to the object;

• d ( d ) is positive if image ( object ) is in front of the mirror;• di ( do ) is positive if image ( object ) is in front of the mirror;

• di ( do ) is negative if image ( object ) is behind the mirror.

Note that the magnitude is positive for an upright image and

negative for an inverted image.


Definitions, discussions and equations have been based on concave mirrors so far.

However, we can easily extend them to convex mirrors as well.

Convex mirrors much smaller than their radius ofcurvature will have paraxial rays passing throughits focal point F.

The method to reconstruct an image is the same. Reflected rays are extrapolated back behind the mirror (see the dashed lines in the figure).

For a convex mirror, the image is always virtual , upright , and smaller .

The equations 23-2 to 23-4 also applies. However,both the focal length and radius of curvatureshould be considered negatives for convex mirrors.


Problem Solving: Spherical Mirrors

1. Draw a ray diagram; the image is where the rays intersect.

2. Apply the mirror and magnification equations.

3. Sign conventions: if the object, image, or focal point is on the reflective side of the mirror, its distance is positive, and negative otherwise. Magnification is positive if image is upright, negative otherwise.

4. Check that your solution agrees with the ray diagram.


Problem 23-10 (textbook): A mirror at an amusement park shows an upright image of any person who stands 1.4 m in front of it. If the image is three times the person’s height, what is the radius of curvature?


Problem 23-10:

We can use the mirror equation:

We know d , but we do not know d . However, using the formula for

o i

1 1 1;

d d f

+ =

We know do , but we do not know di . However, using the formula for magnification:

⇒ ⇒

Using di in the mirror equation:

i i

o o

;h d

mh d

−= = ( )i3 ,

1.4m

d−+ = i 4.2m.d = −

( ) ( )1 1 1

,1.4m 4.2m f

+ = −

2.1m.f = The radius of the concave mirror is:

( )2 2 2.1m 4.2m.r f= = =


Problem 23-13 (textbook): A luminous object 3.0 mm high is placed 20.0 cm from a convex mirror of radius of curvature 20.0 cm.

(a) Show by ray tracing that the image is virtual, and estimate the image distance.

(b) Show that the (negative) image distance can be computed from Eq. 23–2 (mirror equation, eq. 23-3 in this note) using a focal length of -10.0 cm.

(c) Compute the image size, using Eq. 23–3 (magnification equation, eq. 23-4 in this note).


Problem 23-13:

(a) We see from the ray diagram that the image is behind the mirror, so it is virtual. We estimate the image distance as ~7 cm.

(b) If we use a focal length of -10 cm we can locate the image from

(c) We find the image size from the magnification

o i

1 1 1;

d d f

+ =

( ) ( )i

1 1 1,

20cm 10cmd

+ = −

i 6.7cm.d = −

i i

o o

;h d

mh d

−= = ( )( )( )

i6.7 cm

,3.00 mm 20 cm

h −= −

i 1.0 mm.h =


Problem 23-14 (textbook): You are standing 3.0 m from a convex security mirror in a store. You estimate the height of your image to be half of your actual height. Estimate the radius of curvature of the mirror.


Problem 23-14:

We find the image distance from the magnification, noting that the image is always upright in convex mirrors.

Using the mirror equation:

( )i i i

o o

0.5 ,3.0m

h d dm

h d

− −= + = = = ( )( )i

3.0m1.5m.

0.5d

−= = −

+

Using the mirror equation:

The radius of curvature (radius of the mirror) is:

Note that both f and r are negative, as expected for convex mirrors.

( ) ( )o i

1 1 1 1 1, so 3.0 m.

3.0 m 1.5 mf

f d d

= + = + = − −

( )2 2 3.0m 6.0m.r f= = − = −

Index of RefractionIndex of Refraction

Light (and any other form of electromagnetic wave) travels at a speed of c = 300,000 Km/s in vacuum.

However, in general it slows somewhat when traveling through a medium.

It is useful to define a quantity called index of refraction n of the medium in which light propagates as the ratio of the speed of light in vacuum to the speed of light in this medium:in vacuum to the speed of light in this medium:

Since v is always smaller than c � n > 1

(23-5)


When light travelling in a transparent medium strikes a boundary with another medium, part of the light is reflected and other part can be transmitted into the other medium. The direction of the ray of light can change direction if the new medium has different index of refraction .

This is called refraction .

The angle the outgoing ray makes with the normal to the surface is called the angle of refraction the surface is called the angle of refraction ( θ2 in the figure) .


Refraction is responsible for some optical illusions. For example, the observer in the figure thinks the foot of the person standing in the water is located at a higher position than it really is.

This happens because the ray is refracted and change direction. However, the observer still thinks that the ray is travelling in a straight path (the ray model) from the foot of the person.

Refraction Refraction –– Snell’s LawSnell’s Law

The angle of refraction depends on the indices of refraction, and was experimentally proven to be related to the angle θ1 of incidence by the formula:

Where

θ = angle of incidence

(23-6)

θ1 = angle of incidenceθ2 = angle of refractionn1 = index of refraction of medium 1n2 = index of refraction of medium 2

Eq. 23-6 is known as Snell’s law or basic law of refraction.

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