Chapter 23- Light: Geometric Optics
Final Exam
Tuesday, Dec. 15th
Time: 14:0014:00-17:00h
Room: CK, GYMGYM-2
Equation sheet available at
http://ilc2.phys.uregina.ca/~barbi/academic/phys109/2009/phys109.html
There will be 8 problems divided into two sets:
Set 1: 5 problems
Set 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: TBA
Time: Wednesday, 4:30pm - 5:30pm
However, feel free to come to my office on the following
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, LB205) 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.
Assignment 11
Textbook (Giancoli, 6th edition), Chapter 23:
page 659: problems 7, 15, 22 and 28
page 660: problems 39 and 47
page 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 Problems
In addition to assignment 11, the following problems are
recommended in preparation for the final exam:
Chapter 3:
Chapter 4:
Chapter 5:
Chapter 6:
Chapter 7:
Chapter 8:
Chapter 9:
21, 31, 33
29, 34, 47, 64
18, 23, 24
31, 41, 44, 85
24, 32, 43, 63
25, 49, 83
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)
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
your marks have been entered and are correct in
the list.
The link will be removed after 1pm on Dec. 11th.
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 Last Lectures
Formation of Images by Spherical Mirrors
(23-2)
Eq. 23-2 applies to any ray that makes a small angle with the principal axis.
These rays are called paraxial rays.
Formation of Images by Spherical Mirrors
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
normal ⇒ θi = θr = 0 ).
The point where these three rays
cross is the image I’ . All other rays
from O’ will also cross I’.
Formation of Images by Spherical Mirrors
(23-3)
Eq. 23-3 is called the mirror equation and relates the object and image
distances to the focal length ( f = r/2 )
Formation of Images by Spherical Mirrors
We can also find the magnification, m, of a mirror as the ratio of image height
to the object height.
(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.
Formation of Images by Spherical Mirrors
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).
Formation of Images by Spherical Mirrors
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).
Formation of Images by Spherical Mirrors
If an object is inside the focal point
its image will be upright, larger, and virtual (behind the mirror).
Formation of Images by Spherical Mirrors
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;
• 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
negative for an inverted image.
is positive for an upright image and
Formation of Images by Spherical Mirrors
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 curvature
should be considered negatives for convex mirrors.
Today
Formation of Images by Spherical Mirrors
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?
Formation of Images by Spherical Mirrors
Problem 23-10:
We can use the mirror equation:
 1  1
 +
 d o   di
 1
= ;
 f
We know do , but we do not know di . However, using the formula for
magnification:
hi − di
m= =
;
ho
do
⇒
+3 =
− di
,
(1.4 m )
⇒
d i = − 4.2 m.
Using di in the mirror equation:
 1  
 1
1

+
= ,
 (1.4 m )   ( − 4.2 m )  f
f = 2.1m.
The radius of the concave mirror
is:
r = 2 f = 2 ( 2.1m ) = 4.2m.
Formation of Images by Spherical Mirrors
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).
Formation of Images by Spherical Mirrors
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
 1  1
 +
 d o   di
 1
= ;
 f
 1  1

+
 ( 20cm )   d i

1
=
,

 ( −10cm )
d i = − 6.7 cm.
(c) We find the image size from the magnification
h
− di
m= i =
;
ho
do
− 6.7 cm )
(
hi
=−
,
( 3.00 m m )
( 20 cm )
hi = 1.0 m m .
Formation of Images by Spherical Mirrors
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.
Formation of Images by Spherical Mirrors
Problem 23-14:
We find the image distance from the magnification, noting that the image is
always upright in convex mirrors.
hi − di
− di
m = + 0.5 = =
=
,
ho
do
( 3.0 m )
− 3.0 m )
(
di =
= −1.5m.
( + 0.5)
Using the mirror equation:
1  1   1 
=
+  =
f  do   di 

 

1
1
+

 
 , so f = − 3.0 m.
 ( 3.0 m )   ( − 1.5 m ) 
The radius of curvature (radius of the mirror) is:
r = 2 f = 2 ( −3.0 m ) = − 6.0m.
Note that both f and r are negative, as expected for convex mirrors.
Index 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:
(23-5)
Since v is always smaller than c n > 1
Index of Refraction
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
( θ2 in the figure) .
Index of Refraction
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 – Snell’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:
(23-6)
Where
θ1 = angle of incidence
θ2 = angle of refraction
n1 = index of refraction of medium 1
n2 = index of refraction of medium 2
Eq. 23-6 is known as Snell’s law or basic law of refraction.
Total Internal Reflection; Fiber Optics
We can use eq. 23-6 and write:
This equation tells us that for the same angle of incidence θ1 and index of
refraction n1 , the angle of refraction θ2 is larger for smaller index of refraction n2 .
Also, for given indices of refraction n1 and n2 , there is a critical angle of
incidence θc for which the angle of refraction is 900.
(23-7)
Total Internal Reflection; Fiber Optics
If the angle of incidence is larger than the critical angle, NO transmission
occurs.
This is called total internal reflection.
Total Internal Reflection; Fiber Optics
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 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
be either converging (a) or diverging (b).
Lenses work based on the Snell’s law.
A lens is usually made of material such as glass
or transparent plastic such that its index
of refraction is greater than that of the air.
Thin Lenses; Ray Tracing
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.
The focal length is the same on both sides of
the 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 shown
in figure (b).
In addition, we can define:
• Focal plane: this is the plane that contains the
points F and Fa.
Thin Lenses; Ray Tracing
Converging lenses are those thicker in the center than at the edges.
Snell’s law
apply to
these rays
n1
Parallel rays are brought to a focus by
a converging lens.
n2
n1
Thin Lenses; Ray Tracing
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
would converge if projected back.
Thin Lenses; Ray Tracing
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:
(23-8)
Lens power is measured in diopters, D.
1 D = 1 m-1
Thin Lenses; Ray Tracing
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.
Thin Lenses; Ray Tracing
For a diverging lens, we can use the same three rays; the image is upright
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.
Thin Lenses; Ray Tracing
The thin lens equation is the same as the mirror equation:
(23-9)
The sign conventions are slightly different:
1. The focal length is positive for converging lenses and negative for
diverging.
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 virtual).
4. The height of the image is positive if the image is upright and negative
otherwise.
Thin Lenses; Ray Tracing
The magnification formula is also the same as that for a mirror:
(23-10)
The magnification is positive if the image is upright and negative otherwise.
Lensmaker’s Equation
The Lensmaker’s equation relates the lens’s index of refraction and the radii of
curvatures of its two surfaces to its focal length:
(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).
Optics
Problem 23-24 (textbook): The speed of light in ice is 2.29 × 108 m/s. What is
the index of refraction of ice?
Optics
Problem 23-24 (textbook):
We find the index of refraction from
c
v= ;
n
3.00 × 10
(
m s=
8
2.29 × 10
8
n = 1.31.
n
m s)
,
Optics
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?
Optics
Problem 23-27 (textbook):
We find the angle of refraction in the water from
n1 sin θ1 = n2 sin θ 2 ;
(1.33) sin 42.5° = (1.00 ) sin θ 2 ,
θ 2 = 64.0°.
Optics
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?
Optics
Problem 23-45 (textbook):
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

 do
 1
+
  di
 1
= ;
 f
 1   1  1

+
= ,
 275cm   48.3cm  f
f = + 41.1cm.
According to the sign convention for lens, images formed on the opposite side of
a lens are such that di > 0 the image is real.
Optics
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?
Optics
Problem 23-11 (textbook):
We find the image distance from the magnification:
hi − di
m= =
;
ho
do
− di
+ 4.5 =
,
( 2.20cm )
d i = − 9.90cm.
We find the focal length from
 1

 do
 1
+
  di
 1
= ;
 f

 
 1
1
1

+
= ,
 ( 2.20cm )   ( − 9.90cm )  f
f = 2.83cm.
Because the focal length is positive, the mirror is concave with a radius of
r = 2 f = 2 ( 2.83cm ) = 5.7cm.
Optics
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?
(c) What would be the refracted angle if the ray entered the water directly?
Optics
Problem 23-30 (textbook):
(a) We find the angle in the glass from the refraction at the air–glass surface:
n1 sin θ1 = n2 sin θ 2 ;
(1.00 ) sin 43.5° = (1.52 ) sin θ 2 ,
θ 2 = 26.9°.
(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:
n2 sin θ 2 = n3 sin θ 3 ;
(1.52 ) sin 26.9° = (1.33) sin θ 3 , θ 3 =
31.2°.
(c) If there were no glass, we would have
n1 sin θ1 = n3 sin θ 3′ ;
(1.00 ) sin 43.5° = (1.33) sin θ 3′,
θ 3′ = 31.2°.
Note that, because the sides are parallel, is independent of the presence of the
glass.
Optics
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?
Optics
Problem 23-47 (textbook):
(a) We locate the image from
 1

 do
 1
+
  di
 1
= ;
 f
 1  1

+
 18cm   di

1
=
,

 24cm
d i = − 72cm.
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
m=
( − 72cm ) = + 4.0.
−di
=−
do
(18cm )