PROGRAM OF PROGRAM OF “PHYSICS”“PHYSICS”Lecturer: Dr. DO Xuan Hoi

Room 413 (New Building)E-mail : dxhoi@hcmiu.edu.vn , 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