Chapter # 36 Image Formation Forming Images in a plane mirror

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Chapter # 36 Image Formation
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36.1: Images Formed by Flat Mirrors (6)
36.2: Images Formed by Spherical Mirrors
36.3: Images Formed by Refraction (10)
36.4: Images formed by Thin Lenses (14)
36.5: Lens Aberrations (2)
36.6: The Camera (1)
36.7: The Eye (2)
36.8: The Simple Magnifier (1)
36.9: The Compound Microscope (1)
36.10: The Telescope
Forming Images in a plane mirror
•  The image is upright.
Geometrical Construction of an
Image
•  In order to
form image in
plane mirror
me must follow
two different
rays from the
same point
(e.g. top).
• 
d i = d0
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Image Properties
•  Upright
•  Reversed right to
left
•  At the same
distance behind
the mirror as the
object is
•  The same size
•  Virtual
One example more
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Virtual Image versus Real Image
•  Real Image - light rays path through the
image. It is always possible to project
such image on the screen.
•  Virtual Image - no light rays path through
the image. It is impossible to project such
image on the screen. It is exist in our
eye only.
•  Does it mean that virtual image is bad or
wrong?
•  NO! We use virtual image in our mirror
every day and a lot of times!
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Conceptual Question 26-2
•  Two flat
mirrors meet
at right angle.
How many
images do we
have?
How Many? Question 26-2
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Questions
•  Two flat mirrors meet at angle 720. How
many images do we have?
•  Two flat mirrors meet at angle 600. How
many images do we have?
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Mirrors meet at angle 720
•  It will be 5
images if
mirrors meet
at angle 600
Conceptual Question 26-4
•  If you watch
a clock in a
mirror, do the
hands rotate
clockwise or
counterclock
wise?
Test questions 5, 6, 7
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5) The image formed in a plane mirror is
A) always inverted.
B) always upright.
C) sometimes inverted.
D) sometimes upright.
6) The image formed in a plane mirror is
A) always behind the mirror.
B) always in front of the mirror.
C) sometimes behind the mirror.
D) sometimes in front of the mirror.
7) The image formed in a plane mirror is
A) at the same distance in front of the mirror as the object is in front of the mirror.
B) at a shorter distance in front of the mirror than the distance that the object is in
front of the mirror.
•  C) at the same distance behind the mirror as the object is in front of the mirror.
•  D) at a shorter distance behind the mirror than the distance the object is in front of
the mirror.
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Test questions 14, 15
•  14) How tall must a plane mirror be in order for you to be
able to see your full image in it?
•  A) the same height as you
•  B) half of your height
•  C) 3/4 of your height
•  D) twice your height
•  15) Which one of the following is the correct number for
the magnification of a plane mirror?
•  A) 1.0
•  B) 0.5
•  C) 1.5
•  D) 2.0
Test questions 8, 18, 19
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8) A person approaches a plane mirror at a speed of 6 m/s. How fast does he approach
his image?
A) 9 m/s
B) 10 m/s
C) 11 m/s
D) 12 m/s
18) A ray of light strikes a plane mirror perpendicular to the mirror. What is the angle of
incidence?
A) 360°
B) 0°
C) 180°
D) 90°
19) A ray of light strikes a plane mirror perpendicular to the mirror. What is the angle of
reflection?
A) 0°
B) 90°
C) 180°
D) 360°
Spherical Surface
•  A spherical mirror
is a section of a
spherical surface.
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Concave and Convex
Spherical Mirrors
•  Inside reflecting surface - Concave Mirror
•  Outside reflecting surface - Convex Mirror
Question
•  What type of spherical mirror is this?
Center of Curvature and
Principle Axis
•  The center of curvature
is the center of the
sphere with radius R of
which the mirror is a
section.
•  The principle axis is a
straight line drawn
through the center of
curvature and the
midpoint of the mirror.
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Focus
•  Focus is a point to which a
parallel light rays converge
(or from which they appear
to diverge). f = R/2
Sign convention
See also p. 1165-1166
•  For concave
mirror:
f = R/2 > 0
focal point is in
the front of the
mirror
•  For convex
mirror:
f = -R/2 < 0
focal point is
behind the mirror
Forming Images in spherical mirror
•  In order to find orientation, size and
location of an image in a spherical
mirror geometrical optics uses two
techniques:
•  Ray tracing
•  Mirror equation
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Ray Tracing
•  Ray Tracing is a graphical method of
finding orientation, size and location of
an image.
•  The are three rays of light with wellknown path: P-ray, F-ray, and C-ray
Ray Tracing for Concave and
Convex Mirrors
•  P-ray is drawn parallel to the principal axis and then
passes through the focus
•  F-ray is drawn through the focus and after reflection
goes parallel to the principal axis
•  C-ray is drawn towards the center of curvature and
reflected back in the same direction
Convex
•  Image properties:
•  Virtual
•  Smaller than
object
•  Upright
•  Located between
the mirror and
focal point
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Ray Tracing and real photo
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Question
•  What type of mirror
is this?
•  Convex mirror
always form image
which is smaller
than object. So,
this is concave
mirror.
Concave: take a look at the FIG
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Mirror Equation
•  Mirror equation is an exact mathematical
expression between the object distance and
the image distance for a given mirror.
h0
h
=− i
d0
di
h0 d0
=
−hi di
€
m=
hi
d
=− i
h0
d0
€
€
h0 d0 − R
=
−hi R− di
Mirror Equation
d0 d 0 − R
=
di R− di
€
R
1 1
1−
−
1
d0 R d0
=
=
di R− di 1− di
R
€
1 1 1 1
− = −
di R R d 0
⇒
1 1 2 1
+€ = =
di d0 R
f
€
1 1
−
R d0
1=
1 1
−
di R
€
Concave Mirror: final table
1 1
1
+
=
di d 0
f
€
⇒ di = d0
f
d
f
, h i = − h 0 i = − h0
d0 − f
d0
d0 − f
d0
di
|hi|
m
2f < d0<∞
f < di < 2f
|hi| < h0
m<1
real, inverted,
reduced
f < d0 < 2f
2 f < di
|hi| > h0
m>1
real, inverted,
enlarge
d0 < f
di < 0
|hi| > h0
m>1
virtual, upright,
enlarge
di = d 0
f
d0 − f
⇒
xi =
x0
f
, h i = − h0
x 0 −1
d0 − f
⇒
hi
−1
=
h0 x 0 −1
€
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Convex Mirror: final table
1 1
1
+
=−
di d 0
f
€
⇒ di = − d0
f
d
f
, h i = − h 0 i = h0
d0 + f
d0
d0 + f
d0
di
hi
m
∞ > d0 >0
|di| < f
hi < h 0
m<1
virtual, upright,
reduced
What type of mirror is this?
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Can we use spherical mirror
equation for plane mirror?
f →∞ ⇒
1 1
+
= 0 ⇒ di = − d0
di d 0
• 
€
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Problems 21, 22, 23, 24
•  An object with a height of 42 cm is placed 2.0 m in front of a
concave mirror with a focal length of 0.50 m. (a) Determine the
approximate location and size of the image using a ray
diagram. (b) Is the image upright or inverted?
•  Find the location and magnification of the image produced by
the mirror in Problem 21 using the mirror and magnification
equations.
•  An object with a height of 42 cm is placed 2.0 m in front of a
convex mirror with a focal length of -0.50 m. (a) Determine the
approximate location and size of the image using a ray
diagram. (b) Is the image upright or inverted?
•  Find the location and magnification of the image produced by
the mirror in Problem 23 using the mirror and magnification
equations.
Problems 26, 30, 35, 39
•  A concave mirror produces a virtual image that is
three times as tall as the object. (a) If the object is 22
cm in front of the mirror, what is the image
distance? (b) What is the focal length of this mirror?
•  An object is placed between C and F in front of a
concave mirror with a focal length of 10 cm. What is
the position of its image in relation to the mirror and
what are its characteristics?
•  An object is placed 70 cm in front of a concave mirror
with a focal length of 35 cm. What is the magnification?
Test Problem 88
•  An object is placed in front of a convex mirror at a
distance larger than twice the focal length of the
mirror. The image will appear
•  A) in front of the mirror.
•  B) inverted and reduced.
•  C) upright and reduced.
•  D) upright and enlarged.
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Test problems 26-43, 44
•  John's face is 20 cm in front of a concave shaving
mirror. If he observes his image to be twice as big and
erect, what is the focal length of the mirror?
•  John's face is 20 cm in front of a concave shaving
mirror of focal length 30 cm. How large an image
does he observe?
Nice picture. Isn’t it?
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Conceptual question 26-4
•  After passing through the prism, is the
ray (a) deflecting upward, (b) still
horizontal, (c) deflected downward ?
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Lenses: devices used to focus
light and form image
•  Converging lenses
take parallel rays of
light and bring them
together at a focus.
•  Diverging lenses cause
parallel rays of light to
spread out as if
diverging from a point
source.
Convex (converging) Lens
•  We can treat convex lens as two prisms
placed back-to-back. After refraction a
parallel rays of light are brought together
at a focal point.
Convex Lens
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Concave (diverging) Lens
•  We can treat concave lens as two prisms
placed point-to-point. When diverging rays
from a lens are extended back, they appear
to originate at a focal point.
Concave Lens
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Images formed by Lenses
•  In order to find orientation, size and
location of an image formed by lens
geometrical optics uses two techniques:
•  Ray tracing
•  Thin lens equation
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Ray tracing
•  Ray Tracing is a graphical method of
finding orientation, size and location of an
image.
•  The are three rays of light with well-known
path: P-ray, F-ray, and M-ray
Ray Tracing for Diverging and
Converging Lenses
•  P-ray is drawn parallel to the principal axis. After refraction it
passes through the focus.
•  F-ray is drawn through focus and then it passes parallel to
the principal axis
•  M-ray goes through the midpoint of the lens and then
continues in its original direction.
Examples of
ray tracing
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Thin lens equation
•  For convex lenses
•  For concave lenses
1 1
1
+
=
di d 0
f
f
di = d 0
d0 − f
hi
d
f
m=
=− i =−
h0
d0
d0 − f
€
1 1
1
+
=−
di d 0
f
f
di = − d 0
d0 + f
hi
d
f
m=
=− i =
h0
d0 d0 + f
•  f>0 for convex lenses; f<0 for concave lenses
•  m>0 for upright image; m<0 for inverted image
€
•  d>0 when image is on opposite side; d<0 otherwise
Converging Lens: final table
1 1
1
+
=
di d 0
f
€
⇒ di = d0
f
d
f
, h i = − h 0 i = − h0
d0 − f
d0
d0 − f
d0
di
|hi|
m
2f < d0<∞
f < di < 2f
|hi| < h0
|m| < 1
real, inverted,
reduced
f < d0 < 2f
2 f < di
|hi| > h0
|m| > 1
real, inverted,
enlarged
d0 < f
di < 0
hi > h 0
m>1
virtual, upright,
enlarged
Diverging Lens: final table
1 1
1
+
=−
di d 0
f
€
⇒ di = − d0
f
d
f
, h i = − h 0 i = h0
d0 + f
d0
d0 + f
d0
di
hi
m
0 < d0<∞
0 < di < f
hi < h 0
m<1
virtual, upright,
reduced
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Test problems 64, 70
•  The index of refraction of a certain medium is 1.5. What is
the speed of light in that medium?
•  If a lens is submerged in water, what happens to its focal
length? (a) decreases, (b) stays the same, (c) becomes
negative, (d) increases
•  75) An object is placed 10 cm from a convex lens of focal
length 20 cm. What is the magnification?
•  76) An object is placed 60 cm from a convex lens of focal
length 10 cm. What is the magnification?
•  78) An object is placed 16 cm from a concave lens of focal
length 20 cm. What is the magnification?
Test problem
•  Which of the following statements are true?
•  A) For total internal reflection to take place, light
must go from a medium of lower index of
refraction to a medium of higher index of
refraction.
•  B) A convex lens always produces a virtual image.
•  C) A concave lens always produces a virtual image.
•  D) The index of refraction for a given material is
directly proportional to the frequency of light.
See also Example 34.7 page 1173
The Camera
•  The main parts are a lens equipped with shutter,
light-tight enclosure, and light-sensitive film
(CCD array) that records an image.
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Projector - “camera in reverse”
•  A carbon-arc lamp
shines through the
film (or Liquid
Crystal Display),
and the projection
lens forms a real,
enlarged, inverted
image of the film.
Problem
•  An ordinary 35 mm color slide has a picture
area of 24x36 mm. What focal length
would a projection lens need in order to
project a 1.2 m x 1.8 m image of this picture
on a screen 5.0 m from the lens?
The Human Eye
•  Air 1.00
•  Cornea 1.38
•  Aqueous humor
1.33
•  Lens 1.40
•  Vitreous humor
1.34
•  Retina consists of
rods and cones
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Human Eye picture
•  The amount of
light is
controlled by
iris. In bright
light it can be
1mm. At night
7mm.
Retinal rods and cones
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Sensitivity of rods and cones
• 
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Image in the eye
•  Image is
real,
reduced,
inverted.
•  Our eye can not focus an image if the object
is closer then near point distance, N. For
young people with normal eye N is 25 cm.
Accommodation in the human eye
•  Ciliary's muscles
change the shape
of the lens in
order to see
object closer.
This process is
named
accommodation.
Nearsightedness
•  Refractive power
is the ability of the
lens to refract light.
•  Refractive power:
• 
1
refractive power =
f
−1
[m ] = [diopter]
€
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Farsightedness
•  For diverging lens
R.P. < 0
•  For converging
lens R.P. > 0
The Maximum Angular size of
the image
•  If object is
closer to the eye
the image at the
retina becomes
greater.
•  For (a)
h
θ≈
•  For (b)
Maximum possible
θ≈
h0
N
0
d0
h
θ$ ≈ 0
d0$
€
€
The Magnifying Glass
•  We put object at focus. So,
angular size is θ " ≈ h0 f
•  Angular
magnification:
M≈
€
θ # h0 f N
=
=
θ h0 N f
€
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Greater Magnification
• 
M≈
θ# N N
= =
θ p d0
€
Problem 27-4
•  What is the angular magnification when
image produced (a) at infinity, (b) at the near
point? N = 26 cm, f = 4.3 cm.
•  At infinity:
M=
N 26cm
=
=6
f 4.3cm
•  At near point:
€
M=
N 26cm
=
=7
d0 3.7cm
1 1
1
+
=
di d 0
f
1
1
1
= −
d0
f (−di )
€
€
Test 27 - 19
•  The power of a lens is 4.0 diopters. What is the
focal length of this lens?
•  What is the power of a lens that has a focal
length of -40 cm?
•  The focal length of a magnifying glass is 15 cm.
If the near point of a person is 25 cm, what is the
magnification of this glass for this person when
his eyes are focused at infinity (at near point)?
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The compound microscope
• 
Compound Microscope Magnification
•  Objective is a lens
with a short focal
distance. Image:
real, inverted,
enlarged.
•  Eyepiece is a
magnifier which that
views the image of
the objective.
mobj = −
M =−
di
d
≈− i
d0
f obj
M eye = −
N
f eye
di N
f obj f eye
€
Test Problems 50, 68
•  The focal lengths of the objective and the eyepiece in a microscope are 0.29
cm and 2.50 cm, respectively. An object is placed at 0.30 cm from the
objective lens and the image of this object is viewed with the eyepiece
adjusted for minimum eyestrain. What is the final magnification of the
microscope?
•  A) -190
•  B) -300
•  C) -310
•  D) -470
•  The distance between the object and the eyepiece of a compound microscope
is 18.0 cm. The focal length of its objective lens is 0.80 cm and the eyepiece
has a focal length of 2.3 cm. What is the total magnification of the
microscope?
•  A) -120
•  B) -200
•  C) -360
•  D) -480
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Why in this way?
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