chapter 23- light: geometric optics - university of...

Chapter 23- Light: Geometric Optics

Final ExamFinal Exam

Tuesday, Dec. Tuesday, Dec. 1515thth

Time: 14:00Time: 14:00--17:00h17:00hRoom: Room: CK, GYMCK, GYM--22

Equation sheet available atEquation sheet available athttp://ilc2.phys.uregina.ca/~barbi/academic/phys109/2009/phys109.html

There will be 8 problems divided into two sets:

Set 1: 5 problemsSet 2: You have to solve 3 out of 5 problems.

Remember, this is a comprehensive exam .

Tutorial next week will revise several subjects covered over the semester.

Location: TBATime: Wednesday, 4:30pm - 5:30pm

However, feel free to come to my office on the foll owing However, feel free to come to my office on the foll owing days:

o 11th (11am-2:20pm);o 14th (all day) and o 15th (morning).

Also, you can still come during my office hours on Mondays (12:00h-13:15h) and Wednesdays (13:00h-14:15h).

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)

I would STRONGLY recommend that you attend these sessions.

AssignmentAssignment 1111Textbook (Giancoli, 6 th edition), Chapter 23:

page 659: problems 7, 15, 22 and 28

page 660: problems 39 and 47

page 661: problem 52page 661: problem 52

page 662: problem 73

You do not need to hand in this assignment.Solutions will be posted on Dec. 14th.

Comprehensive List of ProblemsComprehensive List of ProblemsIn addition to assignment 11, the following problems are recommended in preparation for the final exam:

Chapter 3: 21, 31, 33Chapter 4: 29, 34, 47, 64Chapter 5: 18, 23, 24Chapter 6: 31, 41, 44, 85Chapter 7: 24, 32, 43, 63Chapter 8: 25, 49, 83Chapter 9: 4, 18, 21, 27Chapter 9: 4, 18, 21, 27

These problems will help but are not enough to prepare you for the final � Try to solve as many other problems as you can.

I also strongly encourage you to solve the following problems:

• On page 222 of Giancoli, problem 60.• On page 221 of Giancoli, problem 32.• On page 190 of Giancoli, problem 45.• On page 131 of Giancoli, problem 19.

Old assignments and midterm exams can be found in the Physics Office

(LB-226)(LB-226)

Solutions have been posted on the web:

All marks, including assignments, will be posted on the web.

http://ilc2.phys.uregina.ca/~barbi/academic/phys109 /2009/grades-web.pdf

You will have between Dec. 9-11 to verify that all You will have between Dec. 9-11 to verify that all your marks have been entered and are correct in

the list.

The link will be removed after 1pm on Dec. 11 th.

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

Formation of Images by Spherical MirrorsFormation of Images by Spherical Mirrors

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 .


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’.


(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).


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.


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.

TodayTodayTodayToday


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, 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 changes 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 θθθθ1 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.

Total Internal Reflection; Fiber OpticsTotal Internal Reflection; Fiber Optics

We can use eq. 23-6 and write:

This equation tells us that for the same angle of incidence θθθθ1 and index of refraction n1 , the angle of refraction θθθθ2 is larger for smaller index of refraction n2 .

Also, for given indices of refraction n and n , there is a critical angle of Also, for given indices of refraction n1 and n2 , there is a critical angle of incidence θθθθc for which the angle of refraction is 900.

(23-7)


If the angle of incidence is larger than the critical angle, NO transmission occurs.

This is called total internal reflection .


Total internal reflection is the principle behind fiber optics .

Light will be transmitted along the fiber even if it is not straight.

An image can be formed using multiple small fibers.

Thin Lenses; Ray TracingThin Lenses; Ray Tracing

Lenses are very important in optics: they are used in eyeglasses, telescopes, cameras, medical instruments, etc.

The most common lenses are circular with two faces, each being portion of a sphere. Like in mirrors, the lenses surfaces can be convex, concave or plane.

Thin lenses are those whose thickness is small compared to their radius of curvature. They may compared to their radius of curvature. They may be either converging (a) or diverging (b).

Lenses work based on the Snell’s law.

A lens is usually made of material such as glassor transparent plastic such that its indexof refraction is greater than that of the air.


Some properties of lenses are similar to those defined for mirrors:

• The axis of a lens is the straight line passing through the center of the length and perpendicular to its two surfaces.

• Focal point F is the point where parallel rays converge after passing through a lens.

• Focal length is the distance from F to the lens.• Focal length is the distance from F to the lens.The focal length is the same on both sides ofthe lens, even if they have different radius of curvature. If parallel rays falls on a lens at an angle, they will converge at a point Fa as shownin figure (b).

In addition, we can define:

• Focal plane : this is the plane that contains the points F and Fa.


Converging lenses are those thicker in the center than at the edges.Snell’s law apply to these rays

n2

n1n1

Parallel rays are brought to a focus by a converging lens.


Diverging lenses are those thinner in the center than at the edges.

Parallel light diverge; the focal point is that point where the diverging rays that point where the diverging rays would converge if projected back.


Optometrists and ophthalmologists specify the strength of an eyeglass by the reciprocal of the focal length. This quantity is called power P of the lens:

Lens power is measured in diopters, D.

(23-8)

1 D = 1 m-1


Ray tracing for thin lenses is similar to that for mirrors. We have three key rays:

1- This ray comes in parallel to the axis and exits through the focal point.

2- This ray comes in through the focal point and exits parallel to the axis.

3- This ray goes through the center of the lens and is undeflected.


For a diverging lens , we can use the same three rays; the image is uprigh t and virtual (the rays do not actually pass through the image).

Converging lenses form real images as the rays do really pass through the image.


The thin lens equation is the same as the mirror equation:

The sign conventions are slightly different:

(23-9)

1. The focal length is positive for converging lenses and negative fordiverging .

2. The object distance is positive when the object is on the same side as the light entering the lens otherwise it is negative.

3. The image distance is positive if the image is on the opposite side from the light entering the lens (the image is said real) ; otherwise it is negative (the image is said virtua l).

4. The height of the image is positive if the image is upright and negative otherwise.


The magnification formula is also the same as that for a mirror:

The magnification is positive if the image is upright and negative otherwise.

(23-10)

The Lensmaker’s equation relates the lens’s index of refraction and the radii of curvatures of its two surfaces to its focal length:

Where:

Lensmaker’sLensmaker’s EquationEquation

(23-11)

Where:

R1 and R2 = radii of curvature of each of the lens’s surfaces.

n = index of refraction of the length.

Sign convention:

The radius of curvature is positive for convex surfaces (center of curvature behind the lens) and negative for concave surfaces (center of curvature in front of the lens).

OpticsOptics

Problem 23-24 (textbook): The speed of light in ice is 2.29 × 108 m/s. What is the index of refraction of ice?

OpticsOptics

Problem 23-24 (textbook):

We find the index of refraction from

;c

vn

=

( )83.00 10 m s×( )8

83.00 10 m s

2.29 10 m s ,n

×× =

1.31.n =

OpticsOptics

Problem 23-27 (textbook): A diver shines a flashlight upward from beneath the water at a 42.5°angle to the vertical. At what angle does the light leave the water?

OpticsOptics


We find the angle of refraction in the water from

1 1 2 2sin sin ;n nθ θ=

( ) ( )1.33 sin 42.5 sin ,θ° = 1.00( ) ( ) 21.33 sin 42.5 sin ,θ° = 1.00

2 64.0 .θ = °

OpticsOptics

Problem 23-45 (textbook): A certain lens focuses light from an object 2.75 m away as an image 48.3 cm on the other side of the lens. What type of lens is it and what is its focal length? Is the image real or virtual?

OpticsOptics


The image is formed on the opposite side of the lens, therefore it is a converging lens.

We find the focal length of the lens from

1 1 1;

+ =

According to the sign convention for lens, images formed on the opposite side of a lens are such that d i > 0 � the image is real.

o i

;d d f

+ =

1 1 1,

275cm 48.3cm f

+ =

41.1cm.f = +

OpticsOptics

Problem 23-11 (textbook): A dentist wants a small mirror that, when 2.20 cm from a tooth, will produce a 4.5 х upright image. What kind of mirror must be used and what must its radius of curvature be?

OpticsOptics


We find the image distance from the magnification:

i i

o o

;h d

mh d

−= = ( )i4.5 ,

2.20cm

d−+ = i 9.90cm.d = −

We find the focal length from

Because the focal length is positive , the mirror is concave with a radius of

o i

1 1 1;

d d f

+ = ( ) ( )

1 1 1,

2.20cm 9.90cm f

+ = −

2.83cm.f =

( )2 2 2.83cm 5.7cm.r f= = =

OpticsOptics

Problem 23-30 (textbook): An aquarium filled with water has flat glass sides whose index of refraction is 1.52. A beam of light from outside the aquarium strikes the glass at a 43.5°angle to the perpendicular (F ig. 23–49). What is the angle of this light ray when it enters

(a) the glass, and then

(b) the water? (b) the water?

(c) What would be the refracted angle if the ray entered the water directly?

OpticsOpticsProblem 23-30 (textbook):

(a) We find the angle in the glass from the refraction at the air–glass surface:

(b) Because the surfaces are parallel, the refraction angle from the first surface is the incident angle at the second surface. We find the angle in the water from the refraction at the glass–water surface:

1 1 2 2sin sin ;n nθ θ= ( ) ( ) 21.00 sin 43.5 sin ,θ° = 1.52 2 26.9 .θ = °

(c) If there were no glass, we would have

Note that, because the sides are parallel, is independent of the presence of the glass.

2 2 3 3sin sin ;n nθ θ= ( ) ( ) 31.52 sin 26.9 sin ,θ° = 1.33 3 31.2 .θ = °

1 1 3 3sin sin ;n nθ θ ′= ( ) ( ) 31.00 sin 43.5 sin ,θ ′° = 1.33

3 31.2 .θ ′ = °

OpticsOptics

Problem 23-47 (textbook): A stamp collector uses a converging lens with focal length 24 cm to view a stamp 18 cm in front of the lens.

(a) Where is the image located?

(b) What is the magnification?

OpticsOptics


(a) We locate the image from

o i

1 1 1;

d d f

+ = i

1 1 1,

18cm 24cmd

+ =

i 72cm.d = −

The negative sign means the image is 72 cm behind the lens (or on the same

side as the light entering the lens) � the image is virtual .

(b) We find the magnification from

( )( )

i

o

72cm4.0.

18cm

dm

d

−−= = − = +

chapter 23- light: geometric optics - university of...

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