phys4 chap2 light
TRANSCRIPT
PROGRAM OF PROGRAM OF “PHYSICS”“PHYSICS”Lecturer: Dr. DO Xuan Hoi
Room 413 (New Building)E-mail : [email protected] , Tel : 0918 220 217
PHYSICS 4 (Wave and Modern Physics)
02 credits (30 periods)
Chapter 1 Mechanical Wave
Chapter 2 Properties of Light
Chapter 3 Introduction to Quantum Physics
Chapter 4 Atomic Physics
Chapter 5 Relativity and Nuclear Physics
References :
Halliday D., Resnick R. and Merrill, J. (1988), Fundamentals of Physics, Extended third edition. John Willey and Sons, Inc.
Alonso M. and Finn E.J. (1992). Physics, Addison-Wesley Publishing Company
Hecht, E. (2000). Physics. Calculus, Second Edition. Brooks/Cole.
Faughn/Serway (2006), Serway’s College Physics, Brooks/Cole.
Roger Muncaster (1994), A-Level Physics, Stanley Thornes.
http://ocw.mit.edu/OcwWeb/Physics/index.htmhttp://www.opensourcephysics.org/index.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/HFrame.htmlhttp://www.practicalphysics.org/go/Default.htmlhttp://www.msm.cam.ac.uk/http://www.iop.org/index.html...
PHYSICS 4 PHYSICS 4 Chapter 2Chapter 2 Properties of Light
B. GEOMETRIC OPTICS
A. WAVE OPTICS
Interference of Light Waves
Diffraction Patterns Polarization
The Nature of Light
The Laws of Reflection and Refraction MirrorsThin Lenses
Light Rays
A. WAVE OPTICS
Interference of Light Waves Diffraction Patterns
Polarization
The Nature of Light
1 The Nature of Light1.1 Dual nature of
light• In some cases light behaves like a wave
(classical E & M – light propagation)• In some cases light behaves like a particle
(photoelectric effect)• Einstein formulated theory of light:
346 63 10
E hf
h . J s
Planck’s constant
1.2 Huygens’s Principle
Light travels at 3.00 x 108 m/s in vacuum(travels slower in liquids and solids) In order to describe propagation:Huygens method : All points on given wave front taken aspoint sources for propagation of sphericalwaves Assume wave moves throughmedium in straight line in directionof rays
2 Interference of Light Waves
2.1 Conditions for interference
light sources must be coherent (must maintain a constant phase with each other)
sources must have identical wavelength superposition principle must apply
2.2 Young’s Double-Slit Experiment
• Setup: light shines at the plane with two slits
• Result: a series of parallel dark and bright bands called fringes
In phases Opposite phases
Interference of Sinusoidal Waves
1 1sin( ) ;y A t Kd 2 2sin( ) ;y A t Kd
1 2y y y 1 2sin( ) sin( )A t Kd A t Kd
1 2sin( ) sin( )A t Kd t Kd
2 1 1 22 cos sin( )2 2
d d d dA K t K
2
K
S1 S2
M
d2
d1
Amplitude of y depends on the path difference : = d2 - d1
Amplitude of y maximum :
2 1cos 1;2
d dK
2 1 ;
2d d
K k
2 1
2;
2 /k
d d
2 1d d k
Amplitude of y equal to 0 :
2 1
1( )
2d d k
The path difference in Young’s Double-Slit
Experiment
2 1 sind d d tand
y
dL
d1
d2
Positions of fringes on the screen
sind
;BRIGHT
dy k
L
1
( ) ;2DARK
dy k
L
Bright fringes (constructive interference) :
BRIGHT
Ly k ki
d
Li
d
Dark fringes (destructive interference) :
12DARKy k i
Angular Positions of Fringes
y
dL
sind k
1
sin ( )2
d k
Bright regions (constructive interference) :
Dark regions (destructive interference) :
A viewing screen is separated from a double-
lit source by 1.2 m. The distance between the two slits is
0.030 mm. The second-order bright fringe is 4.5 cm from the
center line. (a) Determine the wavelength of the light.
SOLUTION
The position of second-order bright fringe : (a)
PROBLEM 1
2
Ly k
d
2
2dy
L
2Ld
5 2(3.0 10 )(4.5 10 )2(1.2 )m m
m
75.6 10 560m nm
A viewing screen is separated from a double-
lit source by 1.2 m. The distance between the two slits is
0.030 mm. The second-order bright fringe is 4.5 cm from the
center line. (b) Calculate the distance between adjacent bright
fringes.
SOLUTION
(b)
PROBLEM 1
1 ( 1)k k
L L Ly y k k i
d d d
Li
d
7
5
(5.6 10 )(1.2 )3.0 10
m mm
2.2cm
A light source emits visible light of twowavelengths: 430 nm and 510 nm. The source is used
in adouble-slit interference experiment in which L = 1.5 m
and d = 0.025 mm. Find the separation distance between
the third order bright fringes.
SOLUTION
PROBLEM 2
3 3L
yd
9
3
(1.5 )(430 10 )3
0.025 10m m
m
27.74 10 m
3' 3 'L
yd
9
3
(1.5 )(510 10 )3
0.025 10m m
m
29.18 10 m
3 3'y y y 2 29.18 10 7.74 10m m 21.40 10 1.40m cm
PROBLEM 3
SOLUTION
2.3 Intensity distribution of the double-slit.Interference pattern
light : electromagnetic wave• suppose that the two slits represent coherent sources of
sinusoidal waves : the two waves from the slits have the same angular frequency
• assume that the two waves have the same amplitude E0 of the electric field at point P
d1
d2
d1
d2
1 0 1sin( ) ;E E t Kd 2 0 2sin( ) ;E E t Kd
1 2E E E
2 1 1 2
02 cos sin( )2 2
d d d dE K t K
2
K
At point P :
Amplitude of E :
2 102 cos
2d d
E K
02 cos2
E K
At point P :Amplitude of E :
2 102 cos
2d d
E K
The intensity of a wave is proportional to the square ofthe electric field magnitude at that point :
02 cos2
E K
2 1 sind d d
20 cos
2I I K
20
2 sincos
2d
I I
20
sincos
dI I
sin tan ;yL
20 cos
dyI I
L
d1
d2
The intensity maximum :
sin
;d
n
dyn
L
sin ;nd
Ly n
d
PROBLEM 4
SOLUTION
I
PROBLEM 4
SOLUTION
PROBLEM 4
SOLUTION
I
PROBLEM 4
SOLUTION
2.4 INTERFERENCE IN THIN FILMS
• Interference effects are commonly observed in thin films, such as thin layers of oil on water or the thin surface of a soap bubble
• The varied colors observed when white light is incident on such films result from the interference of waves reflected from the two surfaces of the film
Interference pattern with Lloyd’s mirrorP’ is equidistant from points S and S :the path difference is zero
There a 180° phase changeproduced by reflection
We observe a dark fringe at P’
bright fringe at P’ ?
CONCLUSION : an electromagnetic wave undergoes a
phase change of 180° upon reflection from a mediumthat has a higher index of refraction than the one inwhich the wave is traveling.
a. Change of phase due to reflection
180° phase change No phase change
Change of phase due to reflection
If is the wavelength of the light in free space
and if the medium has the refraction index n :
cf
;cnf
n
vf
cn
v
The wavelength of light in this medium :
n n
• Consider a film of uniform thickness t and index of refraction n. Assume that the light rays traveling in air are nearly normal to the two surfaces of the film.
b. Interference in thin films
• Reflected ray 1 : phase change of 180° with respect to the incident wave.
• Reflected ray 2 : no phase change
• Ray 1 is 180° out of phase with ray 2 a path difference of n /2.
• ray 2 travels an extra distance 2t before the waves recombine in the air
• Ray 1 is 180° out of phase with ray 2 a path difference of n /2.
• ray 2 travels an extra distance 2t before the waves recombine in the air
Condition for constructive interference :
12 ( )
2 nt m
1( )
2m
n
12 ( )
2nt m
2 nnt m
Condition for destructive interference :
0;1; 2; 3;...m
Calculate the minimum thickness of a soap-
bubble (n = 1.33) film that results in constructiveinterference in the reflected light if the film is
illuminated withlight whose wavelength in free space is = 600 nm.SOLUTION
The minimum film thickness : m = 0
PROBLEM 5
4t
n
6004(1.33)
nm
12 ( )
2nt m Constructive interference :
22
nt
113nm
A thin, wedge-shaped film of refractive index
n is illuminated with monochromatic light of wavelength , as
illustrated in the figure. Describe the interference pattern
observed for this case.
SOLUTION
PROBLEM 6
2 ;mnt mDark fringes :2m
mt
n
12 ( )
2mnt m Bright fringes :
0 0 ;t 1
1.;
2t
n
2
2.;...
2t
n
0 ;4
tn
1
3.;
4t
n
2
5.;...
4t
n
A thin, wedge-shaped film of refractive index
n is illuminated with monochromatic light of wavelength , as
illustrated in the figure. Describe the interference pattern
observed for this case.
SOLUTION
PROBLEM 6
2 ;mnt m
White light :
2m
mt
n
12 ( )
2mnt m Bright fringes :
Dark fringes :
Suppose the two glass plates in the figure are two
microscope slides 10.0 cm long. At one end they are in contact; at
the other end they are separated by a piece of paper 0.0200 mm
thick. What is the spacing of the interference fringes seen byreflection? Is the fringe at the line of contact bright or dark?Assume monochromatic light with a wavelength in air of =
500nm.
SOLUTION
PROBLEM 7
2 2nt t m
;t hx l
2l
x mh
Dark fringes :
;2m h
x l
9
3
(0.100 )(500 10 )2(0.002 10 )
m mm
m
1.25 ( )m mm 0;1.25 ; 2.50 ;...mm mm
Suppose the two glass plates in the figure are two
microscope slides 10.0 cm long. At one end they are in contact; at
the other end they are separated by a piece of paper 0.0200 mm
thick. What is the spacing of the interference fringes seen byreflection? Is the fringe at the line of contact bright or dark?Assume monochromatic light with a wavelength in air of =
500nm.Suppose the glass plates have n = 1.52 and the space
betweenplates contains water (n = 1.33) instead of air. What happens
now?
SOLUTION
PROBLEM 7
In the film of water (n = 1.33), the wavelength :
500376
1.33nm
nm
1.25 /1.33( )m mm
'n
''
2l
x mh
9
3
(0.100 )(500 10 )2(0.002 10 )(1.33)
m mm
m
2l
mn h
0.0150 ( )m mm
0; 0.015 ; 0.030 ;...mm mm
Suppose the two glass plates in the figure are two
microscope slides 10.0 cm long. At one end they are in contact; at
the other end they are separated by a piece of paper 0.0200 mm
thick. What is the spacing of the interference fringes seen byreflection? Is the fringe at the line of contact bright or dark?Assume monochromatic light with a wavelength in air of =
500nm.Suppose the upper of the two plates is a plastic material with
n = 1.40,the wedge is filled with a silicone grease having n = 1.50, and
thebottom plate is a dense flint glass with n = 1.60. What
happens now?
SOLUTION
PROBLEM 8
SOLUTION
The wavelength in the silicone grease:500
3331.50
nmnm '
n
' 0.833x mm
The two reflected waves from the line of contact are in phase
(they both undergo the same phase shift), so the line of contact
is at a bright fringe.
The convex surface of a lens in contact with a plane glassplate.
c. Newton’s Rings
A thin film of air is formed between the two surfaces.
View the setup with monochromatic light :Circular interference fringes
Ray 1 : a phase changeof 180° upon reflection;ray 2 : no phase change the conditions forconstructive and destructive interference
Principle : splits a light beam into two parts and thenrecombines the parts to form an interference pattern
Use : To measure wavelengths or other lengths withgreat precision
The observer sees aninterference patternthat results from thedifference in pathlengths for rays 1and 2.
2.5 THE MICHELSON INTERFEROMETER
3.1 Introduction
3. Light Diffraction If a wave encounters a barrier that has an opening of dimensions similar to the wavelength, the part of the wave that passes through the opening will spread out (diffract) into the region beyond the barrier
That is another proof of wave nature
Diffraction Phenomenon WAVE
Light produces also the diffraction: a light (with appropriate wavelength) beam going through a hole gives a number of circular fringes
3.2 Fraunhofer diffraction
All the rays passing through a narrow slit are approximatelyparallel to one another.
Diffraction from narrow slit
Huygens’s principle : Each portion of the slit acts as a source of light waves
Rays 1 and 3 : ray 1 travels farther than ray 3 by an amount equal to the path difference (a/2) sin
Divide the slit into two halves
Rays 2 and 4 : ray 2 travels farther than ray 4 by an amount equal to the path difference (a/2) sin
Waves from the upper half of the slit interfere destructively with waves from the lower half when :
sina sin ;
2 2a
Divide the slit into four parts :Dark fringes :
2sin
a
Divide the slit into six parts :Dark
fringes : 3sin
a
Divide the slit into two halves
sina sin ;
2 2a
sin ;4 2a
sin ;6 2a
Divide the slit into four parts :Dark fringes : 2
sina
Divide the slit into six parts :Dark
fringes :3
sina
The general condition for destructive interference :
sin ; 1; 2; 3;...m ma
a sin tan myL
Positions of dark fringes :
; 1; 2; 3;...m
Ly m m
a
Divide the slit into two halves : sin
a
Light of wavelength 580 nm is incident on a slit
having a width of 0.300 mm. The viewing screen is 2.00 m from
the slit. Find the positions of the first dark fringes and the width
of the central bright fringe.
SOLUTION
PROBLEM 9
sin ; 1; 2; 3;...m ma
The two dark fringes that flank thecentral bright fringe : m = 1
The width of the central bright fringe :
12 2(3.87 ) 7.74y mm mm
You pass 633-nm laser light through a narrow
slit and observe the diffraction pattern on a screen 6.0 m away.
You find that the distance on the screen between the centers of
the first minima outside the central bright fringe is 32 mm. How
wide is the slit?
SOLUTION
PROBLEM 10
The first minimum : m = 1
Dark fringes :
; 1; 2; 3;...m
Ly m m
a
1
321
2L mm
ya
9
3
(6 )(633 10 )32 10 / 2m m
m
1
La
y
0.24 mm
Intensity of Single-Slit Diffraction Patterns
The intensity at each point on the screen :
where Imax is the intensity at = 0 (the central
maximum).
2
max
sin( sin / )sin /a
I Ia
We see that minima occur when : sin /a m
sin ma
(The general condition for destructive interference)
a sin/
Width of the Single-Slit Pattern
The single-slit diffraction pattern depends
on the ratio :
2
max
sin( sin / )sin /a
I Ia
a
a =
sin 1a First dark fringes :
a = : sin = 1 = /2
a = 5 a = 8
Consequence :
We can consider practically all the light to be concentrated at the geometrical focus.
a >> : sin << 1 0 :
The central maximum spreads over 180o : the fringe pattern is not seen at all.
a < : /2 :
Example : The sound waves in speech : 1 m. Doorway < 1 m : a < The central intensity maximum extends over 180o.The sounds coming through an open doorway or can bend around the head of an instructor.
Visible light ( 5 10-7 m), doorway (a 1 m) : << a No diffraction of light; you cannot see around corners
Find the ratio of the intensities of the secondary
maxima to the intensity of the central maximum for the single-
slit Fraunhofer diffraction pattern.
SOLUTION
PROBLEM 11
This corresponds to a sin/ values of 3/2, 5/2, 7/2, . . .
To a good approximation, the secondary maxima lie midway between the zero points
a sin/ 2
max
sin( sin / )sin /a
I Ia
2
1
max
sin(3 / 2)(3 / 2)
II
0.045
2
2
max
sin(5 / 2)(5 / 2)
II
0.016
The first secondary maxima (the ones
adjacent to the central maximum) have
an intensity of 4.5% that of the central
maximum, and the next secondarymaxima : 1.6%
Intensity of Two-Slit Diffraction Patterns
Must consider not only diffraction due to the individual slits butalso the interference of the waves coming from different slits. The combined effects of diffraction and
interference
The pattern produced when 650-nm light waves pass
throughtwo 3.0-m slits that are 18 m
apart
3.3 THE DIFFRACTION GRATING
An array of a large number of parallel slits, all with the same widtha and spaced equal distances d between centers, is called adiffraction grating
The condition for maxima in the interference pattern at the angle :
Use this expression to calculate the wavelength if we know the grating spacing d and the angle
sin 0;1;2 ;...d m m
The wavelengths of the visible spectrum areapproximately 400 nm (violet) to 700 nm (red).Find the angular width of the first-order visible spectrum
producedby a plane grating with 600 slits per millimeter when white
light fallsnormally on the grating.
SOLUTION
PROBLEM 12
The grating spacing :
611.67 10
600 /d m
slits mm
9
6
400 10;
1.67 10mm
m = 1, the angular deviation of the violet light :
The angular deviation of the red light :
sin 1. VV d
013.9V
9
6
700 10;
1.67 10mm
sin 1. R
R d 024.8R
The angular width of the first-order visible spectrum : 0 0 024.8 13.9 10.9R V
Monochromatic light from a helium-neon laser( = 632.8 nm) is incident normally on a diffraction gratingcontaining 6 000 lines per centimeter.Find the angles at which the first order, second-order, and
third-ordermaxima are observed.
SOLUTION
PROBLEM 13
The slit separation :
11667
6000d cm nm
632.8;
1667nmnm
For the first-order maximum (m = 1) we obtain :
1sin 1.d 0
1 22.31
632.82 ;
1667nmnm
For the second-order maximum (m = 2) we find:
2sin 2.d 0
2 49.39
For m = 3 we find that sin3 > 1 : only zeroth-, first-, and second-order maxima are observed
3.4 Circular Apertures The diffraction pattern formed by a circular aperture consists of a central bright spot surrounded by a series of bright and dark rings
If the aperture diameter is D and the wavelength is , the angular radius 1 of the first , the second dark ring,… are given by :
1sin 1.22 ;D
Between these are bright rings with angular radii given by :
2sin 2.23D
sin 1.63 ;2.68 ;...D D
(The central bright spot is called the Airy disk)
Monochromatic light with wavelength 620 nm
passes through a circular aperture with diameter 7.4 m. The
resulting diffraction pattern is observed on a screen that is 4.5 m
from the aperture. What is the diameter of the Airy disk on the
screen?
SOLUTION
PROBLEM 14
The angular size of the first dark ring : 9
6
620 101.22 ;
7.4 10mm
1sin 1.22
d
1 0.1022 rad
0.462m
The radius of the Airy disk (central bright spot) :
1(4.5 ) tanr m
The diameter : 2r = 0.92 m = 92 cm.
3.5 DIFFRACTION OF X-RAYS BY CRYSTALSX-rays are electromagnetic waves of very short wavelength (of the order of 0.1 nm)
How to construct a grating having such a small spacing
??
Crystalline structureof sodium chloride (NaCl) a 0.56 nm
Na+Cl
-
The beam reflected from the lower plane travels farther thanthe one reflected from the upper plane by a distance 2d sin The diffracted beams are very intense in certain directions constructive interference : Laue pattern
You direct a beam of x rays with wavelength0.154 nm at certain planes of a silicon crystal. As you
increase theangle of incidence from zero, you find the first strong
interferencemaximum from these planes when the beam makes an
angle of34.5o with the planes.(a) How far apart are the planes?
SOLUTION
PROBLEM 15
The Bragg equation :
0
(1)(154 )2.sin34.5
nm
2 sind m
0.136 nm
The distance between adjacent planes :
2sinm
d
(a)
You direct a beam of x rays with wavelength0.154 nm at certain planes of a silicon crystal. As you
increase theangle of incidence from zero, you find the first strong
interferencemaximum from these planes when the beam makes an
angle of34.5o with the planes.(b) Other interference maxima from these planes at larger
angles?
SOLUTION
PROBLEM 15
The Bragg equation :
0
(1)(154 )2.sin34.5
nm
2 sind m
0.136 nm
The distance between adjacent planes :
2sinm
d
(a)
(b)154
2(0.136 )nm
mnm
sin2m
d 0.566m
m = 2 : sin > 1 there are no other angles for interference maxima for this particular set of crystal planes.
4.1 Introduction Light wave is electromagnetic (EM) wave traveling at the speed of light
• The E and B fields are perpendicular to each other
• Both fields are perpendicular to the direction of motion
Therefore, em wavesare transverse waves
Ec
B
cf
The wavelength of light in the vacuum :
4. POLARIZATION OF LIGHT WAVES
The EMSpectrum
• Note the overlap between types of waves
• Visible light is a small portion of the spectrum
• Types are distinguished by frequency or wavelength
Polarized waves on a string.
(a) Transverse wave linearly polarized in the y-direction
(b) Transverse wave linearly polarized in the z-direction
(c) The slot functions as a polarizing filter, passing only components polarized in the y-direction.
We always define the direction of polarization of anelectromagnetic wave to be the direction of the electric-field vector E ; not the magnetic field BThe plane formed by E and the direction of propagation iscalled the plane of polarization of the wave.
4.2 Polarization of an electromagnetic wave
E
A linearly polarized light beam viewed along the direction of propagation (perpendicular to the screen)
Circular polarization may be referred to as right-handed or left-handed, depending on the direction in which the electric field vector rotates
Polarization : 1st meaning : to describe the the shifting of electric charge within a body, such as in response to a nearby charged body 2nd meaning : to describe the direction of E in an electromagnetic wave
CAUTION
These two concepts have the same name,they do not describe the same phenomenon
A symmetricmolecule has no permanent polarization
An external electric field induces a polarization in themolecule.
4.3 Polarizing Filters Waves emitted by a radio transmitter are usually linearlypolarized.
An ordinary beam of light consists of a large number ofwaves emitted by the atoms of the light source. Each atomproduces a particular direction of E
All directions of vibration from a wave source are possible,the resultant electromagnetic wave is a superposition ofwaves vibrating in many different directions.
An unpolarized light beam
Polarizing Filters : To create polarized light from unpolarized natural light
Polarizing Filter
The most common polarizing filter for visible light : called Polaroid Polaroid : material is fabricated in thin sheets of long-chain hydrocarbons which transmits 80% or more of the intensity of a wave that is polarized parallel to a certain axis in the material, called the polarizing axis
What happens when the linearly polarized light emerging from a polarizer passes through a second polarizer (analyser)
?
: angle between the polarizing axes of the polarizer and analizer The intensity of the light transmitted through the analyzer :
2cosMAXI I (Malus’s law)
(Imax is the intensity of the polarized beam incident on the analyzer)
0 cosE
analyser
= 0 ; I = IMAX
2cosMAXI I (Malus’s law)
polarizer analyser
= 450 ; I < IMAX
polarizer analyser
= 900 ; I = 0
polarizer
If the incident unpolarized light has intensity
I0 , find the intensities transmitted by the first and second
polarizers if the angle between the axes of the two filters is
30°.
SOLUTION
PROBLEM 16
This problem involves a polarizer (a polarizing filter on which
unpolarized light shines, producing polarized light) and an analyzer
(a second polarizing filter on which the polarized light shines).
0
38
I I
The angle between the polarizing axes :
0
2I
0I
030
Malus's law : 2cosMAXI I
2 00 cos 302
I
4.4 Polarization by ReflectionUnpolarized light can be polarized, either partially or totally, by reflection.
The angle of incidence at which the reflected beam is completely polarized is called the polarizing angle p .
(the reflected ray and the refracted ray are perpendicular to each other)Snell’s law of refraction :
Brewster’s angle :
tan p n
A. WAVE OPTICS
B. GEOMETRIC OPTICS
The Laws of Reflection and Refraction MirrorsThin Lenses
Light Rays
A wave front is a surface of constant phase;wave fronts move with a speed equal to the propagation speed of the wave.
5. Light Rays
A ray is an imaginary line along the direction of travel of the wave
Spherical wave fronts :the rays are the radii of the sphere
Planar wave fronts : the rays are straight lines perpendicular to the wave fronts.
When a light wave strikes a smooth interface separating two transparent materials (such as air and glass or water and glass), the wave is in general partly reflected and partly refracted (transmitted) into the second material
6. Reflection and Refraction
6.1 Basic notions
If the interface is rough, both the transmitted light and the reflected light are scattered in various directionsReflection at a definite angle from a very smooth surface is called specular reflection (from the Latin word for "mirror"); scattered reflection from a rough surface is called diffuse reflection.
The directions of the incident, reflected, and refracted rays at a interface between two optical materials in terms of the angles they make with the normal (perpendicular) to the surface at the point of incidenceThe index of refraction of an optical material (also called the refractive index), denoted by n.
(the ratio of the speed of light c in vacuum to the speed v in the material)
c v n 1
c
nv
The incident, reflected, and refracted rays and the normal to the surface all lie in the same plane
The angle of reflection is equal to the angle of incidence for all wavelengths and for any pair of materials.
(law of reflection)
a b
6.2 The Laws of Reflection and Refraction
(law of refraction - Snell's law)
For monochromatic light and for a given pair of materials, a and b, on opposite sides of the interface, the ratio of the sins of the angles a and b, is equal to the inverse ratio of the two indexes of refraction: sin sina b b bn n
a
b
a
normalincident ray
refracted ray
reflected ray
na
nb
In the figure, material a is water and material
is a glass with index of refraction 1.52. If the incident ray
makes an angle of 600 with the normal, find the directions of
the reflected and refracted rays. SOLUTION
PROBLEM 17
The wavelength of the red light from ahelium-neon laser is 633 m in air but 474 m in the
aqueoushumor inside your eye-ball. Calculate the index of
refraction ofthe aqueous humor and the speed and frequency of
the lightin this substance.
SOLUTION
PROBLEM 18
Light traveling in water strikes a glass plate at
an angle of incidence of 53.00; part of the beam is reflected
and part is refracted. If the reflected and refracted portions
make an angle of 90.00 with each other, what is the index of
refraction of the glass?
SOLUTION
PROBLEM 19
A ray of light traveling with speed c leaves
point 1 shown in the figure and is reflected to point 2. The ray
strikes the reflecting surface a horizontal distance x from point
1. (a) What is the time t required for the light to travel from 1
to 2 ? (b) When does this time reaches its minimum value ?
SOLUTION
PROBLEM 20
A ray of light traveling with speed c leaves
point 1 shown in the figure and is reflected to point 2. The ray
strikes the reflecting surface a horizontal distance x from point
1. (a) What is the time t required for the light to travel from 1
to 2 ? (b) When does this time reaches its minimum value ?
SOLUTION
PROBLEM 20
A ray of light traveling with speed c leaves
point 1 shown in the figure and is reflected to point 2. The ray
strikes the reflecting surface a horizontal distance x from point
1. (a) What is the time t required for the light to travel from 1
to 2 ? (b) When does this time reaches its minimum value ?
SOLUTION
PROBLEM 20
Fermat's Principle of Least Time : among all possible paths between two points, the one actually taken by a ray of light is that for which the time of travel is a minimum.
( The law of reflection corresponding to the actual path taken by the light )
7. Mirrors
The distance p is called the object distance.
The distance q is called the image distance.
7.1 Flat mirrors
Real object Virtual image
7. Mirrors
A spherical mirror has the shape of a section of a sphere.
Light reflected from the inner (concave surface): concave mirror.
7.2 Spherical mirrors
Light reflected from the outer (convex surface) : convex mirror.
VC
R
O
C
R
F : focal point ; f : focal length
Mirror equation :
Real object p > 0
Virtual object p < 0
Real image q > 0
Virtual image q < 0
Concave mirror :
OC
F
A
B
A’
B’
OF
A
B
A’
B’
Convex mirror :
O
F
A
B
A’
B’
Assume that a certain spherical mirror has a
focal length of 10.0 cm. Locate and describe the image for
object distances of (a) 25.0 cm, (b) 10.0 cm, and (c) 5.00 cm.
SOLUTION
PROBLEM 21
(a)
Assume that a certain spherical mirror has a
focal length of 10.0 cm. Locate and describe the image for
object distances of (a) 25.0 cm, (b) 10.0 cm, and (c) 5.00 cm.
SOLUTION
PROBLEM 21
(b)
(c)
A woman who is 1.5 m tall is located 3.0 m
from an antishoplifting mirror. The focal length of the mirror is
0.25 m. Find the position of her image and the magnification.
SOLUTION
PROBLEM 22
8. Thin Lenses
8.1 NotionsA lens is an optical system with two refracting surfaces. If two spherical surfaces close enough together that we can neglect the distance between them (the thickness of the lens) : thin lens.
The focal length f :
(lens makers’ equation)
R1 R2R1R2
(lens makers’ equation)
p q
Thin-lens equation :
Lateral magnification :
p q
Real object p > 0 Virtual object p < 0
Real image q > 0 Virtual image q < 0
A diverging lens has a focal length of 20.0 cm.
An object 2.00 cm tall is placed 30.0 cm in front of the lens.
Locate the image. Determine both the magnification and the
height of the image.SOLUTION
PROBLEM 23
A converging lens of focal length 10.0 cm forms
an image of each of three objects placed(a) 30.0 cm, (b) 10.0 cm, and (c) 5.00 cm in front of the
lens.In each case, find the image distance and describe the
image.
SOLUTION
PROBLEM 24
(a)
A converging lens of focal length 10.0 cm forms
an image of each of three objects placed(a) 30.0 cm, (b) 10.0 cm, and (c) 5.00 cm in front of the
lens.In each case, find the image distance and describe the
image.
SOLUTION
PROBLEM 24
(b)
When the object is placed at the focal point, the image is formed at infinity.
(c)
8. Thin Lenses
8.2 Simple magnifier
The simple magnifier consists of a single converging lens : this device increases the apparent size of an object.
O F’
A
B
A’
B’
F
OM
p q
25 cmCCOM
B
0A
Angular magnification:
8. Thin Lenses
8.3 The compound microscope
The objective has a very short focal length
The eyepiece has a focal length of a few centimeters.
B2
O1F’1
A
B
A1
B1
F1
O2F2
A2
L1
L2
Objective Eyepiece
F’2
8. Thin Lenses
8.4 The Telescope