Download Optics Refraction Dispersion

Optics
Refraction
Dispersion
Image formation from Mirrors
Lana Sheridan
De Anza College
June 12, 2017
Last time
• speed of light
• ray optics
• reflection
Overview
• refraction
• dispersion
• image formation from mirrors
Refraction
When light rays pass from one medium into another, they are often
observed to bend.
1
Image from Wikipedia, by Zátonyi Sándor.
zon due to the curvature of the Earth. Explain this
phenomenon.
Refraction
© Cengage Learning/Charles D. Winters
13.
Figure
CQ35.13
shows
pencil partially
immersed
a
When
light rays
pass from
onea medium
into another,
they areinoften
cup
of
water.
Why
does
the
pencil
appear
to
be
bent?
observed to bend.
7
ss
us
re
to
Figure CQ35.13
Courtesy of Henry Leap and Jim Lehman
Refraction
a
"
!
$
%
#
Refraction
action
s on strings
cident wave
s 16.15 and
th another
nto the secransparent
nt medium
t enters the
mitted wave
nal nature
es its direcd. The inci-
All rays and the normal lie in the
same plane, and the refracted
ray is bent toward the normal
because v2 ! v1.
Incident
ray
Normal
Reflected
ray
A
u1
Air
Glass
u2
u"1
B
v1
v2
Refracted
ray
Refraction
Why does this happen?
Refraction
Why does this happen?
Recall that Arago proposed to Fizeau and Foucault that they
might measure the speed of light in water as well as air.
That is because the answer to the puzzle was already suspected.
Refraction
Why does this happen?
Recall that Arago proposed to Fizeau and Foucault that they
might measure the speed of light in water as well as air.
That is because the answer to the puzzle was already suspected.
Foucault did the experiment in water (1850), and Fizeau (1851)
went further investigating light moving water.
Both found visible light had a slower speed in water than in air.
This agreed with the wave model of light.
Refraction
Why does this happen?
Recall that Arago proposed to Fizeau and Foucault that they
might measure the speed of light in water as well as air.
That is because the answer to the puzzle was already suspected.
Foucault did the experiment in water (1850), and Fizeau (1851)
went further investigating light moving water.
Both found visible light had a slower speed in water than in air.
This agreed with the wave model of light.
However, Fizeau could not explain the magnitude of the speed
change that occurred for light in moving water...
Refraction
If a wave enters a medium where it moves more slowly, what
happens?
1
the frequency cannot change – the source still “updates” the
medium about a new wave front every T seconds.
Refraction
If a wave enters a medium where it moves more slowly, what
happens?
1
the frequency cannot change – the source still “updates” the
medium about a new wave front every T seconds.
2
the wavelength changes (v = f λ)
When the wavefronts slow, they bend.
Refraction
1
c 2002 by The
McGraw-Hill Concise Encyclopedia of Physics. McGraw-Hill Companies, Inc.
Refraction
electrons, and the vertical arrows
represent their oscillations.
Concrete
v1
v2
Grass
v2 ! v1
This end slows first; as a
result, the barrel turns.
1
Figure
Overhead view
Serway & Jewett,
9th ed,35.13
page 1066.
may encounter a
light is absorbed
sented by the do
an antenna and
again absorbed.
in terms of quan
light passing fro
from one atom t
place cause the
2 3 108 m/s. On
and the light tra
A mechanica
of the rolling ba
on the concrete
the barrel to piv
Index of Refr
In general, the
Refraction
The same thing happens when waves move into shallower water on
a beach.
1
http://astarmathsandphysics.com/o-level-physics-notes/297-refraction-ofwaves.html
Refractive Index
Light at a particular frequency moves at different speeds in
different media.
Light interacts with the charges that constitute the medium, and
the net effect is a wave that moves more slowly.
Refractive index of a medium, n
n=
c
v
where v = ω
k is the phase velocity of light with angular frequency
ω in that medium.
Materials with a higher n are said to be more optically dense.
Refractive Index
The index of refraction also relates to the ratio of the wavelength
of light at a particular frequency in the medium, λn , to that same
light’s wavelength in a vacuum, λ.
n=
fλ
c
=
v
f λn
so,
n=
λ
λn
Refractive Index
The index of refraction also relates to the ratio of the wavelength
of light at a particular frequency in the medium, λn , to that same
light’s wavelength in a vacuum, λ.
n=
fλ
c
=
v
f λn
so,
n=
λ
λn
If the light passes from one medium with refractive index n1 to
another with index n2 :
n1
λ2
=
n2
λ1
Snell’s Law: Angle relationship in Refraction
How do the refractive indices relate to the angles of incidence and
refraction?
Let h be the hypotenuse of the two triangles.
λ1
λ2
h=
=
sin θ1
sin θ2
1
Figure by Richard Fitzpatrick, http://farside.ph.utexas.edu/teaching/
Snell’s Law
λ1
λ2
=
sin θ1
sin θ2
Rearranging,
λ2
n1
sin θ2
=
=
sin θ1
λ1
n2
And rearranging again gives Snell’s Law:
n1 sin θ1 = n2 sin θ2
1
Willebrord Snell discovered this law experimentally.
Snell’s Law and Refraction
If n1 < n2 the
bends
towards the normal, if n1 > n2 the ray
the Principles
ofray
Ray
Optics
bends away from the normal.
Normal
m
g the
bent
Normal
u1
u1
u2 ! u1 v1
v2
Air
Glass
u2
v2 ! v1
Glass
Air
u2 " u1 v1
v2
v2 " v1
u2
Whe
mov
into
spee
ente
and
ben
the
ln 5
n
5
1.52
5 388 nm
Example 35.4: Light Passing Through a Slab
er speed of the light found in part (B). In part (C), we see
the air.A light beam passes from medium 1 (refractive index n1 ) to
AM
medium 2, with the latter medium being a thick slab of material
whose index of refraction is n2 . Show that the beam emerging into
medium 1 from the other side is parallel to the incident beam.
latter medium being
35.15). Show that the
to the incident beam.
35.15
(Example 35.4) The
ine drawn parallel to the ray
out the bottom of the slab
ts the path the light would
e the slab not there.
u1
n1
n2
g
t u2
a
u3
d
m in which
apply
underonrefraction
Also,wefind
howthe
d wave
depends
t (and model.
θ1 , θ2 ).
Measuring Refractive Index
way to measure refractive indices of solids is
and theAnother
Principlesstandard
of Ray Optics
using a prism:
Figure 35.16
The apex angle ! is the angle
between the sides of the prism
through which the light enters
and leaves.
!
d
A prism
efracts a single-wavelength
ight ray through an angle
of deviation d.
is called
the angle
of deviation.
e 35.4,δthe
light passes
through
a slab of material with parallel sides.
1
s when light
strikes
a
prism
with
sides as shown in Figure
See example 35.5, on page nonparallel
1070.
Fermat’s Principle
Also called the Principle of Least Time.
Fermat’s Principle
The path a light ray follows between at starting point S and an
end point T is the path between S and T that is travelled in the
least time.
This principle correctly predicts the reflection and refraction
equations.
This also relates to the calculus of variations, another approach to
solving mechanics problems.
1
Do questions 84 and 85 in the textbook.
Dispersion
We have already said that the speed of light for a given frequency
of light is different in different media.
However, the speed of light is also different for different
frequencies of light in the same medium.
This means the refractive index is a function of frequency, n(ω).
refraction.
sin u 1
Dispersion
A plot of refractive index vs. wavelength for some Dividing
kinds of the first equat
transparent solids:
From Equation 35.4, ho
n
1.54
Crown glass
1.52
1.50
and
Acrylic
which is Snell’s law of r
1.48
Fused quartz
1.46
35.7 Dispersio
An important propert
the index varies with t
Figure 35.21 Variation of index Figure 35.21 shows. Th
400
500
600
l (nm)
700
Dispersion
All materials exhibit dispersion, to varying degrees, except the
vacuum.
This has important effects: rainbow creation!
Dispersion can also have detrimental effects. In camera lenses,
dispersion causes a blurring of the image.
1
Image from http://imaging.nikon.com
Prism Dispersion
of light both reflects and
at the surface between
glass as shown in Figure
If the refractive index
lass is ng , find the angle
ence u1 in the air that
esult in the reflected ray
refracted ray being perlar to each other.
θ1
Andy Ryan/Stone/Get
am hits a target at the top of the building.
al is to find the height of the target above sea
) Draw a diagram of the situation, identifying
triangles that are important in finding the
. (b) Find the angle of incidence of the beam
the water–air interface. (c) Find the angle of
on. (d) What angle does the refracted beam
ith the horizontal? (e) Find the height of the
bove sea level.
Figure P35.38
39. The index of refraction for violet light in silica flint
M glass is 1.66, and that for red light is 1.62. What is the
angular spread of visible light passing through a prism
of apex angle 60.0° if the angle of incidence is 50.0°?
See Figure P35.39.
Deviation of
red light
ng
Figure P35.35
Huygens’s Principle
Dispersion
ex of refraction for red light in water is 1.331
t for blue light is 1.340. If a ray of white light
he water at an angle of incidence of 83.0°, what
underwater angles of refraction for the (a) red
blue components of the light?
beam containing red and violet wavelengths is
on a slab of quartz at an angle of incidence
The index of refraction of quartz is 1.455 at
1 its index of refraction is 1.468
(red light), and
Left, Wikipedia by Spigget;
Visible
light
Angular
spread
Figure P35.39
R
O
Y
G
B
V
Screen
Problems 39 and 40.
40. The index of refraction for violet light in silica flint glass
S is n V , and that for red light is n R . What is the angular
spread of visible light passing through a prism of apex
angle F if the angle of incidence is u? See Figure P35.39.
Section 35.8 Total Internal Reflection
right,
Serway & Jewett, page 1083.
ow a rainbow is formed, consider FigDispersion
in Rainbows
wave
under reflection
and wave under
white light) passing overhead strikes a
cted and reflected as follows. It is first
ith the violet light deviating the most
violet
light refracts
e of the drop, theThe
light
is reflected
and
through larger angles
undergoes refraction
as
it
moves
from
than the red light.
h that the angle between the incident
iolet ray is 40° and
the angle between
Sunlight
e returning red ray is 42°. This small
ys causes us to see a colored bow.
bow as shown in Figure 35.24. If a rainmost intense red light returning from
eviated the least; the most intense vioecause it is deviated
40! 42! the most. Hence,
R
drop. Similarly, a drop lower in theVsky
he observer and appears violet to the
is drop passes below the observer’s eye
V
m other colors of the spectrum reaches
R
hese two extreme
positions.
nbow. The secondary rainbow is fainter
Figure 35.23 Path of sunlight
35.7 Dispersion
1073
The highest-intensity light
traveling from higher raindrops
towardPitfall
the eyes
of the observer
Prevention
35.5
is red, whereas the most intense
A Rainbow of Many Light Rays
light from lower drops is violet.
Pictorial representations such as
Figure 35.23 are subject to misinterpretation. The figure shows one
White
ray of light entering the raindrop
42!
and undergoing
reflection and
40!
refraction, exiting the raindrop
White
in a range of 40° to 42° from the
42! illustration
entering ray. This
40!
R be interpreted incorrectly as
might
V meaning that all light entering the
raindrop
exits
V
R in this small range
of angles. In reality, light exits the
raindrop over a much larger range
of angles, from 0° to 42°. A careful analysis of the reflection and
refraction from the spherical rainFigure 35.24 The formation of a
Dispersion Question
Figure 35.20
Huygens’s construction for proving Snell’s law of
lenses
in a camera use
refraction.
is the w
v1 Dt, wh
From
Quick Quiz 35.41 In photography,
refraction to form an image on a light-sensitive surface. Ideally, you
want all the colors in the light from the object being photographed
Dividing
to be refracted by the same amount. Of the materials shown,
which would you choose for a single-element camera lens?
From Eq
n
1.54
(A) crown glass
1.52
(B) acrylic
1.50
Crown glass
Acrylic
(C) fused quartz
(D) impossible to determine
Fused quartz
400
Serway & Jewett, page1074.
which is
1.48
1.46
1
and
500
Figure 35.21
600
l (nm)
700
Variation of index
35.7
An imp
the inde
Figure 3
Dispersion Question
Figure 35.20
Huygens’s construction for proving Snell’s law of
lenses
in a camera use
refraction.
is the w
v1 Dt, wh
From
Quick Quiz 35.41 In photography,
refraction to form an image on a light-sensitive surface. Ideally, you
want all the colors in the light from the object being photographed
Dividing
to be refracted by the same amount. Of the materials shown,
which would you choose for a single-element camera lens?
From Eq
n
1.54
(B) acrylic
(C) fused quartz
Crown glass
1.52
(A) crown glass
1.50
←
(D) impossible to determine
Acrylic
Fused quartz
400
Serway & Jewett, page1074.
which is
1.48
1.46
1
and
500
Figure 35.21
600
l (nm)
700
Variation of index
35.7
An imp
the inde
Figure 3
Question
A water resistant flashlight is switched on under water in a pool.
The ray from the flashlight strikes the top surface of the water,
and makes an angle of 30◦ with the water’s surface. What is the
angle of refraction?
The refractive index of water is n = 1.333.
Question
A water resistant flashlight is switched on under water in a pool.
The ray from the flashlight strikes the top surface of the water,
and makes an angle of 30◦ with the water’s surface. What is the
angle of refraction?
The refractive index of water is n = 1.333.
Snell’s law: n1 sin θ1 = n2 sin θ2 .
Question
A water resistant flashlight is switched on under water in a pool.
The ray from the flashlight strikes the top surface of the water,
and makes an angle of 30◦ with the water’s surface. What is the
angle of refraction?
The refractive index of water is n = 1.333.
Snell’s law: n1 sin θ1 = n2 sin θ2 .
What’s happening?
Total Internal Reflection
When a light ray travels from a medium with a higher refractive
index to one with a lower refractive index, and it strikes the
interface at a sufficiently large incident angle, there is no valid
solution for the refracted ray.
In fact, in this case we don’t see one! All of the light is reflected at
the boundary.
This is called total internal reflection.
of incidence u1 increases,
The angle of incidence producing
refraction u2 increases
an angle of refraction equal to 90!
! (ray 4). The dashed line
is the critical angle uc . For angles
The critical angle,greater
θc , isthan
the umaximum
angle of incidence such
t no energy actually
c , all the energy of the
that
there
could
be
a
refracted
ray.
The
n this direction.
incident light is reflected. ray would just skim along
Total Internal Reflection
the surface between the media.
mal
Normal
n1 " n2
n1 " n2
2
u2
3
4
5
For even larger angles of
In total
this internal
case, the
incidence,
reflection occurs (ray 5).
n2
n1
uc
angle of refraction θ2 = 90◦ .
(D) four
7, five light rays enter a glass prism from the left.
dergo total internal reflection at the slanted surwo (c) three (d) four (e) five (ii) Suppose the
otated in the plane of the paper. For all five rays
lection from the slanted surface, should the
or (b) counterclockwise?
2
Serway & Jewett, page1075.
Courtesy of Henry Leap and Jim Lehman
. 1, which is a meaningless result because the sine
Total
han
unity. Internal Reflection (TIR) Question
ernal reflection is small when n 1 is considerably
e critical angle for a diamond in air is 24°. Any ray
hes the surface at an angle greater
than 24° is comQuick Quiz 35.52 In the
picture, five light rays enter a glass
stal. This property, combined with proper faceting,
fromare
thecutleft.
(i) light
Howismany
of these rays undergo total
angles ofprism
the facets
so that
“caught”
internal
reflection
atmultiple
the slanted
surface of the prism?
le internal
reflections.
These
reflections
h the medium, and substantial dispersion of colors
s through the top surface of the crystal, the rays
ave been fairly widely separated from one another.
ndex(A)
of refraction
one and can be made to sparkle very
jewel is immersed in corn syrup, the difference in
(B)
two
or the
corn
syrup is small and the critical angle is
escape
(C)sooner;
three as a result, the sparkle completely
not lose all its sparkle when placed in corn syrup.
Figure 35.27 (Quick Quiz 35.5)
Five nonparallel light rays enter a
glass prism from the left.
(D) four
7, five light rays enter a glass prism from the left.
dergo total internal reflection at the slanted surwo (c) three (d) four (e) five (ii) Suppose the
otated in the plane of the paper. For all five rays
lection from the slanted surface, should the
or (b) counterclockwise?
2
Serway & Jewett, page1075.
Courtesy of Henry Leap and Jim Lehman
. 1, which is a meaningless result because the sine
Total
han
unity. Internal Reflection (TIR) Question
ernal reflection is small when n 1 is considerably
e critical angle for a diamond in air is 24°. Any ray
hes the surface at an angle greater
than 24° is comQuick Quiz 35.52 In the
picture, five light rays enter a glass
stal. This property, combined with proper faceting,
fromare
thecutleft.
(i) light
Howismany
of these rays undergo total
angles ofprism
the facets
so that
“caught”
internal
reflection
atmultiple
the slanted
surface of the prism?
le internal
reflections.
These
reflections
h the medium, and substantial dispersion of colors
s through the top surface of the crystal, the rays
ave been fairly widely separated from one another.
ndex(A)
of refraction
one and can be made to sparkle very
jewel is immersed in corn syrup, the difference in
(B)
two
←is small and the critical angle is
or the
corn
syrup
escape
(C)sooner;
three as a result, the sparkle completely
not lose all its sparkle when placed in corn syrup.
Figure 35.27 (Quick Quiz 35.5)
Five nonparallel light rays enter a
glass prism from the left.
7, five light rays enter a glass prism from the left.
dergo total internal reflection at the slanted surwo (c) three (d) four (e) five (ii) Suppose the
otated in the plane of the paper. For all five rays
lection from the slanted surface, should the
or (b) counterclockwise?
3
Serway & Jewett, page1075.
Courtesy of Henry Leap and Jim Lehman
(TIR)
Question
. 1,Total
which is Internal
a meaninglessReflection
result because the
sine
han unity.
ernal reflection is small when n 1 is considerably
e criticalQuick
angle for
a diamond
Any ray
Quiz
35.53 inInair
theis 24°.
picture,
five light rays enter a glass
hes the surface at an angle greater than 24° is comprism from the left. (ii) Suppose the prism can be rotated in the
stal. This property, combined with proper faceting,
of the
For light
all five
rays to experience total internal
angles ofplane
the facets
arepaper.
cut so that
is “caught”
reflection
fromThese
the slanted
the prism should be rotated
le internal
reflections.
multiplesurface,
reflections
h the medium, and substantial dispersion of colors
s through the top surface of the crystal, the rays
ave been fairly widely separated from one another.
ndex of refraction and can be made to sparkle very
jewel is immersed in corn syrup, the difference in
(A)
clockwise
or the
corn
syrup is small and the critical angle is
escape
as a result, the sparkle completely
(B)sooner;
counterclockwise
not lose all its sparkle when placed in corn syrup.
Figure 35.27 (Quick Quiz 35.5)
Five nonparallel light rays enter a
glass prism from the left.
7, five light rays enter a glass prism from the left.
dergo total internal reflection at the slanted surwo (c) three (d) four (e) five (ii) Suppose the
otated in the plane of the paper. For all five rays
lection from the slanted surface, should the
or (b) counterclockwise?
3
Serway & Jewett, page1075.
Courtesy of Henry Leap and Jim Lehman
(TIR)
Question
. 1,Total
which is Internal
a meaninglessReflection
result because the
sine
han unity.
ernal reflection is small when n 1 is considerably
e criticalQuick
angle for
a diamond
Any ray
Quiz
35.53 inInair
theis 24°.
picture,
five light rays enter a glass
hes the surface at an angle greater than 24° is comprism from the left. (ii) Suppose the prism can be rotated in the
stal. This property, combined with proper faceting,
of the
For light
all five
rays to experience total internal
angles ofplane
the facets
arepaper.
cut so that
is “caught”
reflection
fromThese
the slanted
the prism should be rotated
le internal
reflections.
multiplesurface,
reflections
h the medium, and substantial dispersion of colors
s through the top surface of the crystal, the rays
ave been fairly widely separated from one another.
ndex of refraction and can be made to sparkle very
jewel is immersed in corn syrup, the difference in
(A)
clockwise
or the
corn
syrup is small and the critical angle is
escape
as a result, the←
sparkle completely
(B)sooner;
counterclockwise
not lose all its sparkle when placed in corn syrup.
Figure 35.27 (Quick Quiz 35.5)
Five nonparallel light rays enter a
glass prism from the left.
a ma
be su
Application of TIR: Optical Fibers
a cut
is les
1076
Chapter 35 The
of Light and the Principle
GlassNature
or
reflec
plastic core
of inc
core
if thin fibers are used rathe
An
cal fiber. If a bundle oftwo
pare
line, images can be transfer
usefu
use su
Prize in Physics was awarded
out m
transmit light signals over
l
Jacket
coaxi
has led to the development
Cladding
volum
A practical optical fiberFig
Figure 35.30 The construction
Figure 35.29 Light travels in
a material thattravels
hasmore
i
can b
ina lower
Optical
fibers
are
mainly
used
for
telecommunications:
a curved transparent rod by mul- of an optical fiber. Lightmuch
the core,
which
is surrounded
ing th
be
surrounded
bybyaaone
plastic
j
information
be carried by an optical
fiber
than
an electrical
tiple
internalcan
reflections.
cladding and a protective jacket.
other
in a given amount of time.
a cutaway view of this constr
trans
a curved transparent rod by multiple internal reflections.
is less than that of the core
Optical
Glass
or fibers are also used in medicine.reflection if it arrives at the
plastic core
Ray Optics and Image Formation
Simple geometric ray optics can be used to understand how images
are formed by simple optical devices: mirrors and lenses.
1
Wikipedia user Heptagon.
Images Formed by Flat Mirrors
When we see an object “in the mirror”, we are not actually seeing
something that is behind the mirror.
We are seeing an image of an object that is placed in front of the
mirror.
The image seems to be the same distance behind the mirror as the
object is in front of it.
The image seems to be the same size as the object.
This is not true for all optical devices, but we can work out things
about how the image will form for many different optical devices.
Image Formation Terminology
object distance, p
The perpendicular (shortest) distance from the object to the
device.
image distance, q
The perpendicular (shortest) distance from where the image
appears to be to the device.
(lateral) magnification, M
The factor by which the image size exceeds the object’s size
M=
image height
object height
For mirrors and lenses
M=−
q
p
Image Formation Terminology
real image
An image that can be displayed on a screen formed when the light
rays pass through and diverge from the image point.
virtual image
An image that cannot be displayed on a screen, but can be seen
“in the device” because the light rays appear to diverge from the
image point.
The image in a flat mirror is virtual.
Image Formation Terminology
upright image
An image that appears to be right-side-up with respect to the
object.
inverted image
An image that appears to be upside-down with respect to the
object.
The image in a flat mirror is upright.
focal length, f
The distance from the optical device to where a set parallel rays
striking the device head on (perpendicularly) will be focused.
For a flat mirror, the f is infinite.
path perpen| p | ! | q | and h ! h".
Ray Diagram for a Flat Mirror
ay follows the
law of reflected rays back
p Q
q
P
P'
9 behind the
object would
h'
R
h
u
object behind
Image
Object u
, so |p | 5 |q |.
irror is as far
h equals
the have the optical device sketched vertically in cross
Ray diagrams
Figure 36.2 A geometric confollows:
section, the object on the left represented by an arrow pointing up.
struction that is used to locate the
image of an object placed in front
The image is also
of arepresented
flat mirror.by an arrow, but it may be to the
left(36.1)
or right, pointing up or down depending on how the image is
formed.
We sketch rays with paths we know to find the image.
a spherical
d a concave
e R, and its
ction, and a
6.6a shows a
olid, curved
port for the
flecting suro a point as
Henry Leap and Jim Lehman
Images formed by Spherical Mirrors: Concave
Mirrors
Figure
36.7
focus
light. Reflection of paral6.6b,Concave
wheremirrorslelcanrays
from a concave mirror.
ght rays that
6
Images formed by Spherical Mirrors: Concave
If the rays diverge from
Mirrors
O at small angles, they
all reflect through the
same image point I.
Center of
curvature
Mirror
Mirror
R
C
C
V
O
I
Principal
axis
a
b
(a) A concave mirror of radius R. The center of curvature C is located on th
oint object placed at O in front of a concave spherical mirror of radius R, wher
principal axis farther than R from the mirror surface, forms a real image at I.
Assumption: paraxial rays
In studying curved mirrors and thin lenses we assume that all rays
are paraxial rays.
Paraxial rays are rays close to the principle optical axis of our
optical device. (Rays that strike close to the middle.)
For a spherical lens, rays that strike further from the axis are not
focused to the same point (spherical aberration).
Mirror Equation and Thin Lens Equation
The same equation can help us to find the location and
magnification of the image that will be formed by curved mirrors
and thin lenses.
1
1 1
= +
f
p q
Spherical Concave Mirrors
The focal point is where the detectors are place on satellite dishes
and radio telescopes.
1
Parkes telescope in Australia.
Spherical Concave Mirrors Focal Length
For a spherical concave mirror of radius R
f =
R
2
Concave Mirrors and the Mirror Equation
Simple geometry shows why the mirror equation is true for a
concave mirror.
ge Formation
The real image lies at the
location at which the
reflected rays cross.
h
a
O
C
V
I
u
a h' u
Principal
axis
q
R
p
Figure 36.9 The image formed by a spherical concave mirror when the object O lies outside the center of curvature
This this
geometric
construction is used to derive Equation 36.4.
We willC.use
ray diagram.
Summary
• refraction
• dispersion
• total internal reflection
• image terminology
• images formed by spherical mirrors
4th Collected Homework!
due Monday June 19.
Homework Serway & Jewett:
• Ch 35, onward from page 1077. OQs: 9, 15; CQs: 7, 9, 13;
Probs: 7, 13, 27, 31 37, 39, 41, 47, 59, 69, 75, 84, 85, 64 (in
this problem the description means that one side of the mirror
reflects 10% of the light and transmits the other 90%, the
other side reflects 90%)
• Carefully read all of Chapter 36.