EP 225 Waves, Optics, and FieldsWebsite: http://physics.usask.ca/~hirose/ep225/contains

• Course outline• Laboratory instruction• Notes• Past exams• Animation

Instructor: Akira Hirose Office Physics 66akira.hirose@usask.caphone: 966 6414

Marks

• Assignments (9 or10) 15%• Laboratory 15%• Midterm 20%• Final 50%

EP 225 Laboratory

• Instructor: Brian Zulkoskey, Room 115 Physics, ph. 6439brian.zulkoskey@usask.ca

• Lab orientation sessions

MON 5 JAN 1:30 p.m. L2A & B RM 130 PHYSICSWED 7 JAN 1:30 p.m. L4A & B RM 129 PHYSICS

• Students should bring a copy of "A Laboratory Manual for Engineering Physics 225.3, REVISED 1997". These are available at the Book Store.

• If a student misses the lab period for which he/she is scheduled, he/she should see Brian Zulkoskey in Room 115 Physics as soon as possible.

Lab Schedule

L11 L13 L16 W2 L12 Make-up Labs

L2A & B(Mon)

Jan 12

Jan 26

Feb 9

Mar 2

Mar 16

by appt.

L4Α & Β(Wed)

Jan 14

Jan 28

Feb 11

Mar 4

Mar 18

by appt.

Lab Introductions: L2A & B, 1:30 p.m., Monday, 5 January, Rm 130 PhysicsL4A & B, 1:30 p.m., Wednesday, 7 January, Rm 129 Physics

Lab Titles:L11 Prism SpectrometerL13 Geometric OpticsL16 Optical InstrumentsW2 Microwave OpticsL12 Interference and Diffraction Patterns

Subjects to be covered

• Geometrical optics: reflection, refraction, mirrors, lenses, aberration, optical instruments

• Oscillations: mechanical (mass-spring, pendulums, energy tossing)

• Oscillations: E&M (LC circuits, energy tossing, damped oscillations, forced oscillations)

• Mechanical waves: waves in spring and string, sound waves, water waves, wave reflection, standing waves

• E&M waves: LC transmission line, characteristic impedance, wave reflection due to impedance mismatch, radiation of EM waves

• Wave optics: interference, diffraction, resolving power of optical devices

Common questions• Water waves increase amplitude as they approach a beach. Why?• What determine the velocity of light? Sound wave? • How does the police radar measure the speed of a car? • The frequency of electromagnetic waves in telecommunication has been

steadily increasing. Why is high frequency wave more beneficial? • How much power is the earth receiving from the sun and in what form? • Lenses of high quality optical instruments are coated with dielectric films.

Why? • What is the fraction of light power reflected at a glass surface? • Soap films, CD, DVD appear colored. Why? Is it due to the same

mechanism as prism and rainbow? • Why are telescopes for astronomical observation so large?• Electron microscopes can see better than optical microscopes. Why? • How can CAT (computer assisted tomography) and NMRI (nuclear

magnetic resonance imaging) image internal organs? • How does the antenna work? What is the basic mechanism of radiation of

electromagnetic waves? • How does the laser work?

Light Wave

• Light velocity in vacuum• In medium with permittivity ε

• Harmonic wave c = fλ, f (oscillations/sec = Hz) is the frequency and λ (m) is the wavelength. http://physics.usask.ca/~hirose/ep225/

• Index of refraction

• When light wave enters from air to water (n = 1.33), the wavelength is shortened by the factor n. The frequency remains unchanged because it is determined by wave sources, e.g., microwave generator, laser, etc.

8

0 0

1 3.0 10 m/scε µ

= = ×

0

1'c cεµ

= <

0

1'

cnc

εε

= = >

Past Efforts to find c

• Roemer (Danish astronomer) observed change in the rotation period of the moon Io revolving around Jupiter. Based on Roemer’s data, Huygens deduced

• Bradley (British astronomer) found a fairly accurate estimate of

by aberration effect, shift in the angular location of distant star by

• Fizeau made the first laboratory measurement of the speed of light using rotating toothed wheel.

• Currently accepted value

82.3 10 m/sc ×

82.8 10 m/sc ×

4/ where is the earth velocity 3 10 m/s.v c vθ∆ ×

8

0 0

1 2.99792458 10 m/scε µ

= = ×

12 70 08.85 10 F/m, 4 10 H/mε µ π− −= × = ×

Light reflection and refraction at dielectric boundary

• Light path obeys the law of minimum transit time.

• Reflection angle = incident angle• Snell’s law for the refraction angle

In the figure, 1 2 1 2

1 2 1 2

1 1 2

2 2 1sin sin sin ssinsinin

c nc

c cn

θθ

λ λθ θ θ θ

= == → = →

Prism

( )

( )

min

min

1cos '2sin sin 12 2 cos '2

when ', '

sin sin2 2

r rn

i i

r r i i

n

δ α α

δδ α α

−+ = −

= =+

=

min

min

min

min

Example: A prism has =60 and 1.5.What is the minimum deviation angle ?

From sin sin2 2

1.5 0.5 sin 302

30 / 2

n

n

α δ

α δα

δ

δ

=

+=

× = +

+

min

min

arcsin 0.75 37.2

If 1.51, 38.05

n

δ

δ

= → =

=

=

Prism and Rainbow

• Prism and rainbow: splitting of white light into color spectrum. n (blue) > n (red) (blue light is refracted more than red)

• Colors on CD and DVD are due to something else: diffraction.

min

min

sin2 , minimum deviation

sin2

n

δ α

δα

+

= =

Total reflection• When light emerges from glass to air,

Snell’s law becomes

• Maximum of is 90 degrees. For incident angles > arcsin(1/1.5) = 41.8 deg, light is totally reflected.

• Optical fiber waveguide

2θ

2

1

sin 1.5sin

gg

air

nn

nθθ

= = =

2θ

1 1

2

sincnn

θ − =

Total reflection

Example: Find the range of incident angle for total reflection at the vertical surface.

( )

10 2

11

1sin 50.3 , 90 50.3 39.71.3

sin 1.3sin 39.7 56.1

θ θ

θ

−

−

> = = − =

< =

Spherical Mirrors

• Assume small aperture h << R• Reflection is symmetric about the

radial line • Place an object at axial position o

(not zero) and find the image location i.

Example: Concave mirror with f = 10 cm. Object at 15 cm, real image at 30 cm. Magnification is m=-i/o=-2 (inverted image).

1 1 2 1 ,

Magn

focal

ific

leng

ation

, 2 ,

th

2

im

fo

h h ho i R

R f

o

i

α θ γ θ γ β α β γ+ = + = → + =

+

+ = =

= −

=

=

Mirrors (cont)

• Example: Convex mirror (R, f< 0)

1 1 120 15

8.6 cm = 8.6 cm behind the mirror8.6 0.4320

ii

m

+ = −

= −−

= − = +

f = -15

R = -30

O20 8.6

Spherical and Parabolic Mirror

• Spherical mirrors are subject to spherical aberration for large apertures (top).

• For a large object distance (e.g., stars), parabolic mirror is used for better focusing (bottom).

• Schmidt camera(telescope) corrects spherical aberration by placing a lens corrector.

Sign Convention, Magnification

• Mirrori, o > 0 real object and real image distance in front of the mirroro, i < 0 virtual object and image distance behind the mirrorR > 0, f = R/2 > 0 concave mirrorR < 0, f = R/2 < 0 convex mirror

• Lateral (or angular) magnificationm = -i/o, m > 0 erect image, m < 0 inverted

• Axial magnification 2

2 2 from 0adi i do dimdo o o i

= = − − − =

Refraction at a Spherical Boundary

Snell’s law

Then

Example: Find the image of a fish at the center of spherical bowl.

n (water) = 4/3.From

sin sini rnθ θ=

,i rθ α γ θ β γ= + + =

sin sin For small , i r i rn nθ θ θ θ θ= =

( )1n nα β γ+ = −

( )1 11n no i R

+ = −

( )1 11

4 / 3 1 1 13

(at the bowl center)

Magnification is = = +4/3. Ray tracing confirms the result.3/4

n no i R

R i Ri R

i Ro n R

+ = −

+ =

= −−

− −×

R3Rn=4/3

Thin Lens

First refraction

Second refraction

If t is negligible (thin lens)

If in water,

( )2

1 11 , ' 0 in the case shown'n n i

i i t R+ = − <

− +

( )1

1 11'

n no i R

+ = −

( )2 2

1 1 1 1 11 . Lens maker's formulano i R R f

+ = − − =

1 2

1 1 1g w

w

n nf n R R

− = −

( ) ( )2 2

1 1 11 1 ( ' 0 in the case shown)'n n n i

i t i R R+ = − = − − <

− +

Sign Convention for Lenses

• For lenses, f > 0 converging lens, f < 0 diverging lensR > 0 convex surface, R < 0 concaveo > 0 real object distance in front of the lensi > 0 real image distance behind the lens

• Magnification/ /

In the case of the fish in a bowl, , , 4 / 3, 1and 4 / 3.

i

o

o i

i nmo n

o R i R n nm

= −

= = − = == +

Thin Lens

R1>0R2<0f > 0

R1>0R2>0R1<|R2|f > 0

R1>0R2>0R1>R2f < 0

R1<0R2>0f < 0

( )1 2

1 1 11nf R R

= − −

ExamplesExample: When a converging lens

of f = +30 cm is immersed in a liquid, it becomes a diverging lens of f = - 130 cm. If n (lens glass) = 1.5, what is n of the liquid?

( )1 2 1 2

1 2

1 1 1 1 1 1 11 ,30 15

In liquid,

1 1 1 1130

Then 1.696.

gair

g liq

liq liq

liq

nf R R R R

n nf n R R

n

= − − = − =

− = − = −

=

Example: Design lenses with f = +25 cm and – 25 cm given n(glass) = 1.5.

( )1 2 1 2

1 2

1 2

1 2

1 2

1 2

1 2

1 1 1 1 1 11 0.525

12.5 cm, (plano convex)25 cm (symmetric)

15 cm, 75 cm (meniscus)

1 1 1 10.525

12.5 cm, (plano-concave)25 cm, 8.33 c

nf R R R RR RR RR R

f R RR RR R

= − − = − =

= = ∞= − == =

= − = −

= − = ∞= = m, etc. (meniscus)

Physical Meaning of Focusing

Focusing requires the transit time along OHI be independent of the height h

or angle φ.

Proof:

φ

( )( )

( ) ( )

( )

( )

2 22 2

2

2 2 2 22 22 2

2

=

1where . 2

1 1,2 2 2 2

1 11 (independent of )2

1 1where 1 0 is used.

o hOH HI nn i hc c c c

hR

h h h ho h o i h iR o R i

o i h n o in n n hc c o i R c c

n no i R

δτ δ

δ

δ δ

τ

+ ++ = + − +

=

+ + = + + − + = − +

= + + + − − = +

+ − − =

Example: Two lenses

1 1 1First lens: , ' 75 cm50 ' 30

1 1 1Second lens: , 14 cm35 10

ii

ii

+ = =

− + = − = −

30 cm -10 cm

50 cm 40 cm

O

1 275 14Magnification: 0.6 (erect, virtual)50 35

m m m − = = − − = −

( ) ( )

1 1 1Lens , 60.030 20

1 1 1Mirror , ' 6.67 (-20 means a virtual object)20 ' 10

1 1 1Lens again , '' 50.0 (final image)33.33 '' 20

50Magnification 2 0.333 1.0 (erect, real)33.33

ii

ii

ii

m

+ = =

+ = =−

+ = =

= − + − = +

Lens-Mirror

Compound lens

• Two lenses touching: Effective focal length

• Two lenses separated by d (to be shown later)

1 2 1 2

1 1 1 1 1 1, efff i f f f f

− + = = +

1 2 1 2

1 1 1 df f f f f

= + −f1 f2

d

n = 1.34

R = 5.7 mm

22.5 mm

n = 1.0

Human Eye

1 1 1.34 0.34 22.5 mm.5.7

The curvature radius R is adjustable.

n n i fi R i

−+ = → = → = =

∞

Myopia

Myopia correction

d

f = -d

Myopia is caused by too short a focal length of the eye lens.Placing a diverging lens can correct it. If the farpoint of the eye is , the focal length of the correcting lens is since the effectiv

df d= − e focal length becomes longer

1 1 1

eff ef f d= −

Hyperopia

Hyperopia correction

Hyperopia is caused by too long a focal length of the eye lens.Placing a converging lens can correct it. If the nearpoint of the eye is - , the focal length of the correcting lens should be .

df d=

d

Image at infinity

f

25 cm

25 cm

o

Magnifying Glass

' 25mf

θθ

= =

251mf

= +

Microscope

image at infinity

Fo Fo Fe

eye

L

25 if image is at

251 if image is at 25 cm

o e

o e

Lmf f

Lmf f

= − × ∞

= − +

Telescope

image at infinity

eyeθ

θ'

f f1 2

θ'

The eyepiece can be a divergent lens < 0.

0 (erect image, Galileo type)

o o

e e

e

o

e

f fmf f

ffmf

∞= − × = −

∞

= − >

Matrix Method

( ) ( )

'

' ' ' 1' '

In matrix form1 0

'1'

' '

h hn h nn nn R n

h h hn h nn R n

φ γ φ γ φ φ

φ φ φ

=

+ = + = − +

= = −

R

hh'=h+t

t

φφ

R

φ

φ'

n n'

h γ

• Refraction

• Transmission over distance t' 1' 0 1

h t h hφ φ φ

= =

T

Matrix of Thin Lens

1 1 1 1'

1' and '

1 0'

1 1'

ho i f f

h h hf

h h

f

φ φ

φ φ

φ φ

+ = → − =

= − + =

= −

φ φ' < 0o

if

γh

Single Thin Lens

If an object is at from the first lens and image at from the second lens, total matrix is

1 01 1

1 10 1 0 1

1 0

11 1

0 de

oi

i o

fi oi io i

A Bf f ooo C D

f if fB

−

− + − − = = = − −− − = termines the image distance because focusing is independent of

1 1 1 , lens formula

Magnification = 1

i f oi iAf o

φ

= −

= − = −

f

o i

d

f1 f2

f'

2 1

1

1 2 1 2 1

1 2 1 2 2

Two lenses separated by 1 0 1 0

11 11 10 1

11 1 1 , , ' 1

1 1 1eff

eff

d

d

f f

d df d A dC f f

d d f f f f f C ff f f f f

− −

− = = − = + − = − = − − − + −

Two Thin Lenses

( )2 1

1 2 1 2 1

1 2

1 2 1 2

Calcualtion of '

1 1 1 1' 11 1'

where 1 1 1

' is not a proper focal length because for a ray coming from right,focal point is not at ' to the

eff

eff

f

f f d df ff d f f f f d f

f d f

df f f f f

ff

− − + = → = = = − − + − +

−

= + −

left of the first lens.

Revisit30 cm -10 cm

50 cm 40 cm

O

The system matrix is1 0 1 01 1 40 1 501 10 1 0 1 0 11 1

10 300.33 0.067 23.33 1.67

0.067 1.667From 0 14 cm

0.6 (magnification)Focal length = 1 / 15cm

i

i i

B iA

C

− − − +

= − = → = −

=− =

Principal Planes (thin compound lens)

f’B: back focal positionf’F: front focal positionf: effective focal length

If the object distance is relative to H1 and image distance relative to H2, the formal lens formula still holds

f

f'B

d

H2f1 f2

fd/f1

H1

fd/f2ff'F

1 1 1o i f

+ =

1

2

For the compound lens in the previous example,15 40H2 at 20 cm to the left of lens 2

3015 40H1 at 60 cm to the right of lens 1 = 60 cm

10to the left of lens 1Then the object distance relat

fdffdf

×= =

×= = −

−

ive to H1 is 10 cm.1 1 1From , we find 6 cm to the right of H2 or14 cm to the left of lens 2.10 15

ii

−

+ = =−

Thick Lens

Total lens matrix is

( )

( ) ( ) ( )

( ) ( )

2 1

1

2

1 2

2

1 2 1

1 2 2

1 01 01

1 1 1 11 10 1

11 1

11 11 1 1

The effective foca

1

l length s

1

i

11

1 gg

g gg g

g g

gg g

g g

g

tn n

R n R n

t tR n n

n t tn nR R n

n tn

f R

R R R

R n

n

R

− −

− −

= −

− − − − + −

− = − − +

1 22 1 2 1

1

2 1 2

2

1 1 1cf. (compound lens)

H2 at to the left of second vertex. H1 at

1

to the right of first vert

1

.

,

ex

g

g

g

df f f f f

ft ftf

tR f f

f n

n f f

n

= + − = + −

R R

t

1 2

gnf'f

H2

Example: A lens 3 cm thick has R1=10 cm, R2=-5 cm, n (glass) = 1.5. Determine its focal length and location of the principal planes for the two surfaces.

( )

( )

2 1

-1

The lens matrix is1 01 0

11 1 1 11 10 1

1 01 0 1 3 0.900 2.00cm1 1 11 10 1 0.140cm 0.8001.5 1 1.5

1.5 10 1.551 1The focal length is

0

tn n

R n R n

fC

− −

= = − −− −

= − = −−

7.143 cm..140

Paraxial ray from left is focused at ' 6.429 cm.

The principal plane H2 is at 7.143-6.429=0.714 cm = 7.14 mm from the vertex to the leftSimilarly, H1 is at 1.43 cm to the right of t

AfC

=

= − =

he first vertex.

( )

-1

For light coming from right, the order of matrices is reversed.1 01 0 1 3

1 1 11 10 11.5 1 1.51.5 5 1.510

0.8 2.00cm0.140cm 0.900

1 1The focal length is 7.143 cm0.140

fC

−− −

= −

= − = − =−

(unchanged)

Paraxial ray from right is focused at ' 5.714 cm.

The principal plane H1 is at 7.143-5.714=0.714 cm = 1.429 cm from the vertex to the right

AfC

= − =

Thick Lens Paraxial Ray Diagram

3 cm 6.43

7.14

H2

5.71

H1

7.14

10 18.33

7.147.14

If object at 10 cm from the first vertex, the image is at 18.33 cm from the second vertex.

Example: An object is placed at 20 cm in front of the thick lens of the previous example. Determine the final image location.

In the lens formula, 1 1 1 , is to be measured from H1 and from H2.

20 1.43 21.43 cm1 1 1From 10.71 cm.

21.43 7.143Magnification is 10.71 / 21.43 0.5In matrix method,1 ' 0.900 2.00 1

eff

o io i fo

ii

m

i

+ =

= + =

+ = → =

= − = −

0 1 200.140 0.800 0 1

0.9 0.14 ' 20-2 '0.140 2

From 0, ' 10 measured from the second vertex.Magnification is 0.9 1.4 0.5.

i i

B iA

−

− = − −

= → == − = −

Chromatic Aberration

Prism effect in single lensBlue light is refracted more than red light and the imageis blurred.

Achromatic Compound Lens

( ) ( )

( )

1 2 1 2

2

1 2 3 4 1 2 3 4

1 2 3 4 1 2 3 4

The effective focal length of compound lens is1 1 1=

1 1 1 1 1 1 1 11 - 1

1 1 0 yields

1 1 1 1 1 1 1 12 1

df f f f f

n n dR R R R R R R R

d dn dd f d dn f

n dR R R R R R R R

λ λ

+ −

= − − + − − − −

= =

− + − − − − −

( )1 2

1 2 3 4

0

1 1 11 1 1 12 1 2

f fdn

R R R R

=

+ = + =

− − −

Achromatic DoubletChromatic aberration of lens can be corrected by combining converging and diverging lenses made of different glasses as shown. The focal length is

( )

( )

( ) ( )

1 2 3 4

1 2 3 4

1 2 3 4

1 1 1 1 1Red: ( 1) ' 1

1 1 1 1 1Blue: ( 1) ' 1

From ,

1 1 1 1' '

R RR

B BB

R B

B R R B

n nf R R R R

n nf R R R R

f f

n n n nR R R R

= − − + − −

= − − + − −

=

− − = − −

1 2

Example: Converging lens made of Crown glass ( 1.505,1.510) and diverging lens made of Flint glass ( 1.615, 1.630).

If focal length 50 mm is needed, determine the radii , .Achromatic condit

R

B R B

nn n n

R R

== = =

2 11 2 2

1 2 2

1 2

ion

1 1 1 10.005 0.015 2 (1)

Focal length of blue light (= red light)

1 1 1 10.51 0.63 (2) 50From (1) and (2), 22.5 mm, 45 mm

R RR R R

R R RR R

− = − − → = − ∞

= − +

= =

Telescope with Erector (for Rifle)

To get erect image as needed in rifle telescope, place another lens in between. The total telescope length of ~30 cm and magnification of 8 or 10 is reasonable. An example is shown. In this example, the erector has magnification of -2 and the total magnification is +8. (homework)

120 32 30

120 48 96 30