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CONTENTS

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6

REFRACTION The

of

95

Positions

Image

Change

of

Curvature

95 of

Wavefront

115

120

Reflection Rainbow

128 130

da c

Derivations Partial The

of Method

SURFACES

CURVED

AT Determination

ce


The

Cornu

281

5»iral

Double-Slit

Dif raction

Dif raction

Gratings

Resolving

287

290 302

Power

Half

Period

The

Color

The

Huygen-Fresnel

Elements the

303 304

y

Theory

305

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of

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Plane

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Number Between

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703

705 706

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Luminance nlumination

and

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ce


1

da c

CHAPTER

ce


da c

.

PROBLIM

1-2

ce


y(z,t)

+

The

1)

wave

values

of order

to

/(b/a)

+2

zt)

/ (b / a) is

traveling

z,

since

keep

.t)

2

-a(z+

=

Thus,

]. toward t

as

the

t)2.

/(b/a)

exponent

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left,

increases, the

toward

or z

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get

same.

da c

in

2

t

a

exp[-a(z

=

negative smaller

b

_a(z2+

becomes

ce


(d)

the

speed

(e)

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direction

the

maximum

(f)

of

the

wave;

of transverse

propagation speed

of

the

of

a

wave;

particle

and of

the

da c

wave.

ce


y

with

respect

to

t

gives

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Dif erentiating

ce


da c

or

ce


RESULTANT

OF

TWO

OR

MORE

WAVES

da c

THE

ce


P

2

0N

( 2;t

=

through parallel

to

radius

the

to

Applying

OP

the

I

OP

Drop

.

principle

displacement

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radius

the

l

OP

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2

2

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by

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drawing

PIP

through

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perpendicular

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2

the

to

PC

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2

C

displacement

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superposition,

of

time

at

OP

t

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OA

P

P

2

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=

2 are

+

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2

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parallel),

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PROBLEM

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12

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2

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and

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Figure

3

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1-12

ce


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(34)/2)

where

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in

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first distance

102,

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2

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=

l

again

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every 51

first

situation when

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51

"wavelengths"),

times

=

2

100, for

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the

is

da c

m

m

ce


2

ac

CHAPTER

ce d


OF IN

A

GIVEN

THE

DISTANCE TIME

TRAVELLED

PERIOD

ac

DETERMINATION BY LIGHT

ce d


of

speed are

index

the

given

refraction

of

in

light to

be

a

1.50

of vacuum

=

and

500

a

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10

=

8

m/sec,

1,

=

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n

G

the and

AV

respectively.

ac

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=-

2wQ, 8

ac

c

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(dark), two

divided

were

beams.

mirror

struck

finally

D.

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bright

M M

1

'2

1

M

back

mirror

and

M

2

arrived

troughs

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at

the

chopping

was

detector

from

the

output

the

M, the

later

moment

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to

reached a

,

1

reflected

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beam

flashes

together.

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,

photothe

two

signal

frequency.

ac

strongly

M

and

1

M

beam

equidistant

were

light

the

M

into

1

mirror

other

the

simultaneously,

cell

through

reflected l

M

by

The

D.

struck

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mirror downwards

detector

which

,

of

crests

reflected

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mirrors

the

electric fluctuated

mirror

to

silvered

reflected

was

to

through

back

half

was

,

2

traveled

transmit ed

dark

beam

One M

the

by

ce d


ABERRATION

ac

STELLAR

ce d


(1)

equation

Therefore,

substituting

for

v

e

becomes

and

for

e

see

)

( 60

see

e

-

and

see

)(

1

degree 60

min

)(

1T

180

ac

(20.47

min

or

miles

=

1

gives

18.53 e

=

radians

degrees

)

ce d


Since

8

is

small,

the

approximation

tan

8

=

e

is

valid,

so,

ac

and

ce d


The

nodes

Applying

is Y between

=

a

2TIX

1nteger,

for

x

two

A -

4 successive

3A '

li

Since

mTI

=

-X-

5A '

li nodes

'

and

etc., .

1S

=

A -

4

_ -

lGr

_

where mA

li

m

Thus

distance

the -

t.

result

0,

=

mTIA

so,

3A

li

all

the

1T x

so,

for

a

gives

cos

and

=

(4)

( 2X ).

cos

·

odd

an

y

equation

to

( 2;t )

sin

2r

=

condition

the

by

condition

A

2

·

ac

o

given

are

this

ce d


equation

(3) =

2r

sin

[ 221T (

=

2r

sin

( 2;t )

-

cos

+

+

TI cos[

221T ( -(

( 2X ) -

ac

y

becomes

-

+

) (5)

ce d


3

da

c

CHAPTER

ce


{l

-

: J

da

c

=

ce


light)

or

da

c

of

ce


is

da

c

fect,

ce


Solving

for

Sine

-

Sine

the

using

r

0

0

43

formula,

quadratic

J(

043)2

+

4(3003)

(0.98) (3)

=

r

Sine

2(3003) =

r =

r

30°

or

0

da

c

a

0.50,

ce


We at:

wil

dimensional

use

now

for

the

first

analysis equation momentum

to

see

how

the

units

were

arrived

had

we =

Energy

speed J

m/sec

da

c

=

ce


1.2

X

10

13

pho;ons

1.33

X

10-

27

kg-m/sec

photon

da

c

m

X

ce


PROBLEM

3-8

da

c

.

ce


The c

a

of

flux

watts/cnf

20

A

where

the

the

want

we

energy

the

is

1mm

=

(

20

c

wattS

cnfA

) (l\1

(3

X

10

cuP X

10

-4

uP

1

3.3

m/sec) =

20

beam mm

we

3

have

volume,

e

and

FAt

=

T t

is

lmm. =

-, )

the

the

travel

to

:)

of

X

10-

X

10

4

watts

uP

12

sec.

=

20

X

da

c

F

.001

=

=

of

beam

velocity

F.

(time) 3

mm

surface

one

flux

(area) 1

in

the

with

traveling

beam

this

Ener

=

area

of

arc

for t

Power

=

enclosed

required

te

Call

.

F

If

The

situationo

the

indicates

diagram has

ce


PROBLEM

3 -10

da

c

.

ce


POYNTING

VECTOR

da

c

THE

ce


PROBLEM

3 -12

da

c

.

ce


in

electromagnetic

an

fal ing initial y in the

field 1

on

m

2

of in

all of

space

this 1

second.

Here,

i>

the

6QE2

=

X

1.

X

35

X

of

10

-12

7

l W.l m

·

energy

is

section

of

The

.

the

energy 3 X

and

the

cylinder

near

the

10

8 m

in earth

is

-2

s

space.

3

2 C.N.m

second

density

free

10

X

tu

1

cross

end

E2

=

-2

W.m

-2

-1

X

3

W

C

2 ·

N

-l

.m.

s

-1.

da

c

26.55

10

10

X

ty

1.35 8.85

the 3

3 vi

in

energy

1.35

=

ti

permi

the

2

m

to

Hence

H2 is

1 travels

energy

in

atmosphere

cylinder

a

H2

is

vacuum

earth's

the

contained

length;

in

ce


60

km/hr

to

da

c

Converting

ce


or

the

of

angle

total

the 2 e

As

velocity radiation

developed

we

V

was

in

the

=

c/nj

then

is

cone

(1),

equation for

or

medium.

v/v said the

we

light,

So

e

also

can

we

.

that

phase write

c =

-

·

nv

da

c

Sin

l

Sin-

2

=

ce


2rr

=

x

10-

i+

mm

da

c

5

ce


invariant.

t'

fol owing

the

In

0

=

to

of

light

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that

time

t

0

=

and is

frame

and

frame

S

coincide

S'

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origin.

the

in

both

of

origins

moving

a

with

time

light

velocity

right. in

observer is

emit ed

frame

either

from

the

frame

his

sees

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as

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pulse

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get

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x

of

kind

same

2

ya

+

the

expanding +

2

2

t

2x'yt'

+

ytt]3

yt3

+

+

equation,

above +

y

3

t'3]

we

+

y'3

+

z'3

=

now

have

zt3

c

=

3

c

y2 3y2

[

t'

[t'3

+

system. the

in

substituting

By

.

S'

the

in

x:J

inverse

12

.

+

da

c

y2[x'3

c

=

the

on

form

wave

get

we

y2[x' By

z2

+

transformations

these

using

by spherical

that

show

to

want

ce


4

ce da c

CHAPTER


ce da c

}6IN.


ce da c

.

PROBLEM

4-4


ce da c

VERGENCE


Solution the

the

in

brightness

of

the

shaped,

uniformly its

at

size,

the

and

shadow

the

on

and

the

of

part

is

object The

screen.

penumbra

is

an

umbra annular

part

dark

it

at

becomes

that

the

outside disk

bright

a

disk of

inner

its

equal fal s

is

source

opaque

being called

peripheral

a

screen

the

of

part,

from where

the

figure,

the

In

and

graded edge,

outer

the

to

shadow.

is

instead

size

central

a

dark,

which

penumbra,

bright

to

divided

is is

finite into

of

is

source

which

umbra,

edge

the

shadow

the

called

finite its

If

:

point,

which shadow

of

throws

is

disk-

(ring-shaped).

ce da c

a


of

edges due

not

see

see

a

penumbra

shadow

a

part

any

of

portion the

by

cast

penumbra.

the

to

entire

of the

the source

within

observer

from

while can

be

lamp the

within

one

source, source,

incandescent

bulb

frosted

a

An

the

points

umbra

can-

penumbra outside

can

the

seen.

ce da c

is


the

Thus, it d X.

is

length given

also the

and

The

distance of

is

the ABC

and

between

CDE

disk

opaque is

penumbra

is

AB

segment distance the

are

represented similar,

2r.

addition,

In

the and

two

disks

the

screen

Y

as so

the

in

is is the

fol owing

valid:

ce da c

Triangles

relation

line

the between

width

figure.

the

of

that


AB

the

radius

of

the

sun

=

1

2

(864,000

miles)

=

432,000

da

c

miles;

=

ce


( (

< ( (

( ( ( (

the

represent that clockwise

complete circumference

the

of

plane

a

viewed

and

equator

is

earth

section

rotating Since

direction.

revolution, this

circle

the

its on

wil

axis

own

24 the rotate

to

This

pole.

takes

parallel

earth,

south

point

any

of

of the about it

from

hours

equator

means

in

a

to

make

or

on

counterone

the

through

ce da c

the


da c

.

PROBLEM

4-12

ce


5

da c

CHAPTER

ce


p litude

=

1(1

+

180°

cos

2

)

=

1 2

(1

-

1)

da c

Am

=

0

ce


decreases

velocity light deviated

earth's

the

as

is

atmosphere and

with

decreases

entering

waves as

shown

increases. at

Fig.

earth's 1.

The

line

As are

A-A'

the

of

surface

atmosphere

represents

of

density

The the

elevation.

increasing the

in

density greatest

da c

the earth

a

result,

continuously a

ce


for

its

the

ground

production

which

is

sun's

rays. the

reach

hot

sun, a

in

glancing conditions

body

of

the the

a

rise

in

the

Mirages

angle. arise,

interprets of

road

ahead also

are

which

is

its

sometimes

a

though

as

and

observer.

the

reflecting

the

actual

it,

phenomenon

same

surface

in

below

object

the

This

water.

of

appearance when

image

by may the

in

seen

for

accounts

highway

smooth

it

permits produced the

da c

verse

"wet"

a

is

heated

object shown

paths

two

inverted

an

of

situation

a

of

object

between desert

lay

traveler as

the

the

that its

with surface

weary

by

result

above,

portion

upper

surface

the

nearer

intensely

area

an

the

observer

an

the

than

over

from

of

eye

reflecting

The

at

Light

air that

the

that

dense

less found

together

surface the a

be

sometimes

with

figure, position, a

require shall

under

be

to

when

the

case

over

seen re-

ce


n

da c

w

ce


c

17

-

-

y

1.0

21°

tan

·

+

+

a

=

0.5

+

+

5

-

D

5.0

·

b

+

1

-

then

c

m

2

45°

tan

.

6.5

tan

14°

+

5

·

tan

45°

em

da c

y

tan

since

and

Y

5.

-

ce


the

internal

angles

Hence,

Cf"

Draw

BA,

sin

air

Cf"

=

(n n

sin

e

=

n =

l

e

1)

sin

l and

A

at

equal n

0

180

to

up

0

.

line also

is

Law,

of

e

add

sin

n

=

is

parallel to

that

to

e

', of

.

where

glass.

.

da c

index n

Snell's

by

+

BAC

Angle

But

triangle

a

0

90

-

construction

a

mirror.

of

0

90

=

ce


8

2

an

incidence

of

angle

strikes

and

'

8

angle

critical

8

;

wil to

yield be

must

be

transmit ed

the

3

equal the

into

interface

equal is

1

the

to

be

e

critical

through

reflected wil

3

the

8

1

which

at

the

down

the

decrease; and

angle of

B

than)

total y

value

largest

point

greater

increased,

angle

at at

(or

to

to

ray 8

fiber

the

solving permits

fiber.

da c

l

core-coating

if e

refracted

'

1

the

for

setting e

3

that

Note

thus,

light

3

angle

fiber.

for

8

ce


angle

of

critical flected

if

incidence;

angle, Since

ratio

of

sine

the the

this

indices

is

angle wil

rays

interface,

the

at

mum.

all

thus of

the

of

refraction

equal

set

necessarily reducing

be losses

critical

is

angle of

the

to

the

to

a

total y

two

re-

mini-

given

by

media

in-

the

da c

volved,

ce


=

tan

S

da c

a

ce


object

fraction incidence basic

so

it

is

at

the

the

also the

at

first the

the of

the

at

emergence,

of

angle angle second

slab, incidence of

e

is

sur-

the

by

angle of

angle the

start

apparent the

or

similarly, We

the

first

the

refraction,

and, surface.

be

x

on

surface.

surface second

and

applying

re-

of

angle some

trigonometry.

da c

of

is

face,

side

front

the

to

distance.

ce


account

the

refraction

of

the

light

ray

da c

into face.

at

each

sur-

ce


Dividing

(1

)

( 3)

by sin

in

results

1.65

sin

y

17.4

tan

y

-

tan

cp

da c

x

ce


·

sin

8

=

1.50

·

sin

.

da c

1.33

ce


of

refraction,

we

have:

da c

law

ce


da c

.

PROBLEM

5-15

ce


Oil the

of beam

index

1.2

is

now

spread

on

the

water.

Where

go?

da c

(b) does

ce


-

64

da c

ce


Using

(d)

angle the

air

just

89 above

refraction

of

law

the

is

0

we

,

the

given

and

can

find

hot

pavement

the

index

of

that refraction

the

by

da c

critical of

ce


in rays rise defined

region

then

I

wil

able reflection.

internal

to

by

the

wil

there

be

not

to

angle

go

critical

region

rg

-

-

cg n

=

0,

or

then

region

to

critical

angle the

which

above

angle I

equation

giving

I

then

can

for

be

cg

is

n

0.9999785

da c

problem

a

from

This

cos this

be

ce


example,

of

determined

curvature

for

the

case

by il ustrated

index

the

in

the

gradient.

figure

da c

radius

For

shown,

in

ce


da c

and

ce


da c

.

PROBLEM

5-21

ce


-

-

-

+

d

-

o'A

+

d

-

d

-

o'B

(o'A

+

d

'

)

d'

da c

x

o'A

ce


equation

(3)

can

be

reduced

to

one

dimension:

da c

x-axis,

ce


DISPLACEMENT

da c

TRANSVERSE

ce


Solution cribe

according

the

From

:

wil ,

ray an

of

arc

to

the

radius

Schlieren

of

theory entering

upon

the R

determined

inci-

the

optics, stratified

des-

medium,

by

the

index

gradient

relation

da c

dent

ce


is

R

n(A)

large

so =

1.48,

very and

lit le

error

is

made

if

we

assume

then

da c

since

n(B)

ce


Solution

First

:

(see

figure

construct

1).

a

A

ray

diagram

ray

from

the

of

picture object

A

proceeds

this

probto

G

da

c

lem

ce


Snell's

law

da c

Using

ce


da c

.

PROBLEM

5-27

ce


face transmit ed

has

a

index,

higher wave

does

there not

is

experience

a

a

of

change

phase phase

change

n.

in

The

either

da c

case.

ce


8

da c

CHAPTER

ce


-

1.0 20

S'

Hence,

Thus,

the

image

1.5 S'

+ rom

-

24 appears

_

1.5

-

-

40

1

nun

mm.

24

rom

to

the

right

of

the

spherical

da c

surface.

ce


length length

of of

the a

spherical spherical

of

radius

the

that

Recall

interface.

mirror

da c

focal focal

R

is

R/2.

ce


da

c

",

ce


Solution surface

equation

equation

The

:

wil

holds

in

used

be

only

for

for this small

the

refraction

problem, distances

at

a

from

da c

axis:

spherical

that

noting the

this

optical

ce


1 u

1

1.5

+

front 1

front

-

50

0.102

-

1

rom

0.02

-

0.102

-

-

0.082

da c

1

-

-

25

nun

1 -

u

1.5 -

14.58

nun

-1

ce


u

l

and

n ,

and

ture

three

denote

the

surface, of

the curved

indices the

represent

u

curved

2

of

object

respectively, curved

surfaces,

surface) with

refraction

and

image

and

R

separation

the

distances

is

media, from

radius

the

each

for

successively zero

of

at

da c

(n

each

the

of

curva-

of

the surface.

ce


da c

or

ce


=

3

d

_

15d

-

2d

-

360 63

2d

2 -

63d

-

_

2d

-

15d 63

+

360

da c

u

ce


gives:

da c

mirror

ce


r

n

n '

S

da

c

S

+

ce


Solution

:

(a)

problem

is

solved

surface

is

(VI)

surface

refraction for

the

is

image given

by

at

formed the

at

fol owing

first

right

the surface

refraction a

The

refraction

the the

Considering

+20cm).

1.

figure the

distance

the

at

s

from

equation:

da c

first

(=

the

surface,

first the

-

and distance

object

in

shown

considering

by first

The

next.

is

rod

glass

easily

surface

left

the

at

The

ce


da c

.

PROBLEM

6-8

ce


4

1 -

2

90

+

1 -

15

10 90

da

c

U'

ce


80mrn

first air.

angles

the

air and

the

of

(b)

of

water

focal

of

of

vertex

magnification second

axis

the

to

of

left and and

position (a)

right

at to

the

the

of

1.33. the

is

rod,

located

surface.

Find if

image

index

lengths

the

the

(c)

spherical

da c

high,

the

is the

rod

Find surface

in in

ce


in l85mrn

water

to

=

of the

-185mrn.

when

Hence,

index

1.33, left

of

a

the

virtual surface

the

rod

glass

image of

wil

the

is appear

immersed about

glass.

da c

Sl

Thus,

ce


the

is

radius

the

the

fish, normal),

a.

and

the the

of is

8t

OS'

sphere, angle is

the

of

angle

=

b

incidence of

is

the

(measured

size

apparent from

refraction.

da c

R

of

ce


da c

and

ce


Another a

the surface surface radial be

image

which

first, the

at

fish

and 8

(i.e., rays

converging

radial

the

since

i the

is

therefore

8t of

center -

0.5

the the

spherical

incident =

observer

sees

sphere.

So,

the

at

the

Since

0).

is from

rays

the

at

normally

are

rays

and

0

refracted,

not

are

to

distance

=

all

refracted

of

use

fish

small

Then

sphere. not

are

make

to

the

that

says

the

of

center

radial

are

is

point

image

the

find

to

way

approximation point

them

again

to

the

ft.

da c

an

ce


=

Coss.

da c

Rll Ro

ce


(11)

values

n

=

i

1.5

and

nt

-

1.0

into

gives

da c

equation

given

the

Substituting

ce


da c

FI G.

,

ce


2

Figure acts

with

air,

all

plane

/

tions

c

action

air,

time

distance necessary

index

the

for

of

speed

n,

relative

axis

Dosi

of

part

the

Fermat's

the

time medium

the

wave

the

in

x

for

is

x

move

plane

principle

necessary to

con-

the

refraction,

surface. that

us

m

the

off

the

convex

wil of

part (the

of

determine

to

when the

tel s a

move

of

in

front

wave

with

to

amount

Since

that c

convex

axis

the

while

of

part

the

velocity

a

used

be

can

the

least

wave

xl

m

this

the

When

off

wave c

with

moves

medium).

the ,

of interacts

wave

axis

plane

the

in with

right

encounters

velocity

a

the

to

the

the

same a

distance

or

da c

c

with

the

in

light

part

first

is

wave

air).

in

light

axis

the of

inter-

wave

plane

move

wave

of

speed

move

on

plane

plane

a

the

When

the

on

the to

wave

of

(the wave

surface, tinue

=

of

c

the

surface.

parts

velocity

when

happens

what

shows

convex

a

ce


solving

for

Rl

yields

da

c

Then,

ce


in

d

the

=

l

r

preceding

problem,

2

2R

da c

derived

ce


represents

the

distance

from

window

da c

where

to

window.

ce


is

da c

face

ce


law

of

refraction

states:

da c

Snell's

ce


I

da c

FIG.

ce


the

mirror,

da c

For

ce


the

u

final

l

=

image

u,

initial

the

position,

object

equation

distance, (10)

and

u

' -

3

u',

becomes

da c

Letting

ce


sine

sinS.

a -

1

(u

2

a" -

t

2

(a"

+

ull2)

a" -

(u,2

+

sinS. sinS

a2)

+

=

t

n

a"2)

(a)

sinS.

(b)

(c)

sinS

sinS

a

'

b

-

(e)

-

1

[(a

b)2

_

+

u2]

b

'

(f)

-

t

(b

2

+

r2) a"

' -

t

[(r-u")

2

+a"]

2

(g)

(d)

da c

sinS.

ce


gives

da c

Cross-multiplying

ce


b

2

a"

2

terms

cancel

and

so,

da c

The

ce


13.

A

incidence there

of

partly that

Show at

right

entering passing

light, and

60° reflected

the

and

the

angles

ray

each

at

the

other

the

emergent

an

of

angle side,

is

refracted.

partly

reflected to

sphere to

over

and

ray

are

other.

da c

(a)

of

ray

ce


to

ray

B

striking

the

surface

so

rays

E

and

F

wil

be

as

da

c

shown.

ce


light between

180°

138°

-

is

40°,

=

while

42°.

The

that

angle

corresponding for

other

colors

lies

for intermediate

violet

these.

da c

to

ce


7

da

c

CHAPTER

ce


these

into

values

equation

(1)

gives

da

c

Substituting

ce


The the

power

of

the

first

surface

is

found

by

making

use

relation

da

c

of

ce


da

c

and

ce


use

of

the

fol owing

expression:

da

c

making

ce


PROBLEM

7-5

da

c

.

ce


Thus

the =

(n

-

1)

is

( -.L Rl

by

given -

-

the

fol owing

equation:

R12 )

da

c

D

D

power

ce


in f

with

contact can

be

calculated

the

other,

each as

resulting

focal

fol ows;

da

c

placed length

ce


PROBLEM

7-10

da

c

.

ce


0.6

x

16

da

c

-

ce


The thickness

the

of

problem

considering -s

of

from

the

first

surface

the

at an

shown

in

at

fig.

solved surfaces

two

object

of

terms

be

can

light as

in

length

curvature

Consider

consecutively.

lens

focal of

a

by of

distance

1.

da

c

of

radius

refraction

the

the

the

finding the

and

ce


Hence, 1

1

1

f

-

n

_

f

s

and

sIt

t

-

1

1

.

1

+

-

so

1

(

1

t -

1

1:f .)

5

-

n

(IT 5

1

+

1 f

)

Thus, t

n

n -

1 -

-

1

5"

+

1

(4)

f

da

c

5

-

1 f

ce


+

f

( nf

+

t

2nf

+

t

)

s

da

c

ss"

ce


-

hi

=

2f

l

+

2f(nf

t)

+

2nf

+

t

da

c

h

ce


f'

:t

-

:t

-

2nf

t

n

2n

+.. .t.

f'

f2 f'

0

-

t

+

f

when

2

nf

f

=

-

2

.

,

hence,

n -

2n

+

t -

f2

da

c

f

ce


da

c

or

ce


8

da c

CHAPTER

ce


PI

=

eff

1

-

dP

·

1

da c

P

ce


POINT

OF

EYE

da c

FAR

ce


.1

1

=

Si

f

'

eff

da c

So

+..l:.

ce


da c

=

ce


focus

which

eyes the

focal

see

the

from

is

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1

=

7

10-

x

1

ern

6.71

x

10-

6.71

x

10

6

rad

em

m,

1.22 -

x

550

10-

x

7

em -

100

em

da c

ex

10-

x

-

100

for

550

x

-7

rad

ce


13

ce da c

CHAPTER


ce da c

·

PROBLEM

13-2


5

and

intensity

cm,

10

values

of 25

cm,

cm,

the

light and

after 50

cm

traveling in

the

medium

initial y.

ce da c

tude of their

distances

compared

to


I

o

o

-a.x

C

ce da c

E

e


ce da c

.

PROBLEM

13-4


M

=

i

n

sin

Kd

ce da c

L

Kd

os


=

27T A

ce da c

K


Let

.

be

the

amplitude the

of

surface

incident

the at

A

(not

beam at

an

angle

of

its incidence

.

ce da c

8

a

striking

intensity)


reflected. I

rays 2nd

cosS

2nd

CDS

2

and -

r

(Jr

-

expression

the

interference

be-

minima.

mA

(m

for

is

+

)

A

maxima.

ce da c

and

the

Thus,

tween


IT

-

Intensity

of

the

transmit ed

beam

ce da c

Then,


=

1

-

2

sin

2

ce da c

coso


IT I

-

1

T

-

1

0

4 ( .04)

+

(1

-

.04)2

27f(l.S)

2

sin

5

x

10-

001] 7

1 -

1 x

10

is

a

multiple

sin of

2

360

[108

x 0

or

10 21T

5

] radians.

ce da c

108

5

+

0.16 .9216

Since


so,

ce da c

and


n

Thus, R

( (

-

n

n

n

-

2 +

2

n

(1.62

-

index

refractive

(the

2

(1.

l

glass)

-

1.62

)2 1.333)2 1.333)2

-

62

flint

2

)

l

of

+

(.287)2 (2.953)2

-

8.237

10

x

-2

-

8.720

0.94

-

R

-

10-

2

0.94%.

ce da c

or

x

.

PROBLEM

13-8


film.

ce da c

the


the

tively. (one

of

(n

The index

overall

the

tiplying each

high

layer.

H n

L

L

one

of

transfer M

materials,

index

) matrix

transfer and

individual

Then,

(n

low

and

)

index

for n

matrices

H

)

respec-

such

two

is

calculated

corresponding

layers by

multo

-

ce da c

for


-nH I1r,

-n

()N

2

( ) \ jN ( H )N (-nn: J (

_

H

)

-

+

substitution

N

=

4

into

of

the

preceding

the

values

equation,

n

( -n:) ( :H )

-

+

1

=

1.4,

L we

n

see

=

H

2.8,

2N -

2N

+

1

1

and

that

ce da c

Upon

1

-

(:L )N ( -n:H)N ( :L )N

L

n

2


fi

=

n

jK

-

into

ft*

and the

=

preceding

n

+

jK,

and

equation

substituting

these

gives

ce da c

(2),

expressions


change

Ce

form

jbi

then

,

y

write is

To

it

simply

;

b/a.

real

his

is

part

Using

the

done, ratio

amplitude

the

get a

as

t

once

fi

+

an

=

n

-

ce da c

+

iy

i y

Ce

tan-Iy.

just

is

the a

form

standard

the

into

imaginary jK,

phase into

part,


ce da c

Thus,


Taking

the

exponential

of

both

sides

of

this

equation

ce da c

gives


1.5141 2.303

-

0.6575

-

(2)

When

T

.

0.52

=

,

1

O.D.

-

2.303

R.n

.1 )

R.n

(O.8)

52

0.6539 -

2.303

0.2839

-

When

T

98%

-

0.98,

=

1

O.D.

-

ce da c

2.303

.0202

-

2.303

.0088

-

When

T

=

.

2%

=

02

NON-ACTIVATED VERSION 2.03 C2 ) www.avs4you.com O.D.

.

R.n

-

-

,

3.9120 2.303

ce da c

FIG.


T

2

T

(1

T

R)2

-

2

-

4R

+

2

1

2

T

-

-

R

2

-

-

2R

+

-

0.0007

(R

1

+

1)2

2R

(0.05)2

(1.9)2

+

2

R

+

4R

25 -

x

10-

4

3.61

.

ce da c

-

+

-


ce da c

.

PROBLEM

13-15


14

da c

CHAPTER

ce


For

this

problem

we

can

use

the

relation

da c

Solution:

ce


from number

the

equal

right

be

left

a

dif erence

in

circularly

polarized

the

light

to

L This

of

waves

wil

there and

crystal,

quartz of

-

path

).:

=

d

(). ).lR)

dif erence,

expressed =

2).1Td

(n

-

L

as

n

R

)

n

-

a

R

phase

]

=

path angle

dif erence. is

,

da c

8

Ad [nL

=

-

ce


centimeter

by

a

10

active

cm

of

column

liquid

with

a

1

gram/cubic

substance.)

da c

duced

ce


Solution netic

traverses/£,

:

flux

The

rotation

faraday B,

density and

=

B

constant

V

of

related

V,

the

to

medium

the

by

the the

mag-

light

relation

£

da c

8

length

Verdet

the

is

8,

the

ce


12.57

x

-

6.25

60

1.3

0

da c

-

ce


Solution:

To

first

The

amplitude

lar

(1..)

investigate

look

at

and

parallel

polarization the

Fresnel

on

equations

for

coefficients

reflection

(

I

for

)

are

found

reflection reflection. E

to

waves

let

perpendicu-

be

da c

us

ce


Since

8.

sin

and

1

8

8t

are

.

1

sin

8.

cos

8.

1 =

-

(90

-

8

.

1

)

tan

8.

1

=

1

( 3)

,

nt -

n.

1

da c

sin

equation

through

connected

ce


parallel

Eo where

is

flection

the

in

vector

of

is

this

the

vectors.

Hr

and

are

co

incidence,

s

r

r

of

E'

=

is

co

the

s

(2)

<1>' of

angle

refraction.

The

re-

magnetic

satisfy H

=

r

the

(3)

H'

reflected

and

transmi

da c

where

+

0..L H'

Er

-

angle must

case

H

netic

cos

I

angle 4>'

and

incidence

of

plane

the

to

t ed

mag-

ce


-

n

cos

Eo

+

n

cos

Er

-

-

n'

cos

da c

or

'

E'

ce


Two

added

circularly to

form

a

polarized linearly

waves

polarized

(a

right wave:

and

left)

a

At

da c

2) be

x

=

can

0,

ce


(2'ITvt

cas

cas

=

where

of

one

+

(21T\)t we

a)

a

2

)

made

formulas,

angle (2'ITvt

cas

=

+

have

double

the

+

a

cas

2 use

2

sin

-

-

a

of

)

cas

21\\)t the

find

we a -

-

2 cas

sin ().

2

(2'ITvt +

cas

+

21T\)t

a

2

a

sin

) sin

2"

1)

sin

a.

2

formula

da c

Using

ce


everything

by

-1)

.

da c

multiplies

ce


da c

y

ce


da c

Hence

ce


the

axes.

is It

has

linear,

amplitude

Eo/l2

along

a

line

at

45

degrees

.

da c

to

polarization

the

Thus

ce


1/4

da c

1/8

ce


vertical,

and

light

linearly

(c)

transmission

polarized

in

axis the

at x

-

45° direction.

to

da c

on

the

horizontal,

ce


linearly

polarized

in

the

horizontal

direction.

da c

light

ce


da c

.

PROBLEM

14-17

ce


da c

.

PROBLEM

14-18

ce


Solution

Let

:

spectively

as

Then

system. surface

OZ

which

and

OX

is the

fast directions

OY

and

of

direction

the

light

of

is

slow

incident

a

directions coordinate

re-

propagation. may

The be

taken

as

the

plane.

da c

XOY

on

the

choose

us

the

ce


or

Ex

-

B

cos

-

A

cS

[1

-

-

11 l

and

rearranging,

2

cos

(wt

)1/2

( E: we

kz)]1/2

-

sin

sin

cS

cS

have

da c

squaring

-

ce


E

nents, are

A

=

of

out

cos

8

and

0

when

phase

=

sin

A

These

8.

arrive

they

at

the

two

components

analyzer.

Now

"

the

the

and

E

components of

analyzer

are

E

parallel

to

the

principal

transmit ed.

da c

only plane

ce


colloidal

the

ized. viewed

is

s

(

)

max

this

from

given

be

the

Nicol

the

through

tated

this

particles, I

Let

of

prism.

position,

the

wil

beam

intensity As

intensity

the

be

this

plane beam

Nicol transmit ed

polarwhen

prism

is

ro-

by

it

by I

=

I

s

(

max

)

co

s

2

8

da c

by

ce


the

Hence,

I

-

20 1

20

30

x

100

=

A

2

in 2

cas

A

2

20 cos

A

-

2

intensity 2

cas

is 2

30

x

100

20

da c

I

reduction

percentage

ce


da c

.

PROBLEM

14-24

ce


da c

or

ce


Solving

for

t

-

21T(n in

t. .'I1e

-

E

given

we

have,

-

E

n

0

,

)

da

ta,

gives

us

da c

substituting

(1)

equation A(

t

and

in

ce


At are

(c)

the

amplitude E

sin

parallel

components 20°

and

0

cos

20°

to

the

E*

.

da c

(c). axis

ce


figure

4,

it

can

be

seen

that

da c

From

ce


da c

where

ce


Solution

internal transmit ed

to

of

beam. at

birefringent

perpendicular

faces

The

the

the

index

of of

angle

the

lit le

perpendicularly

the

ordinary

of

whereas

with

cut

refraction

cut,

reflection,

the

of

plane are

the

sections

two

The

crystal.

crystal

the

of

consists

prism

Glan-type

figure. the

The

cement wave

of

axis

optic ends

to

the

is

such

suffers

extraordinary

or

light that total

wave

is

loss.

da c

is

The

:

uniaxial

natural

ce


Solution

:

by

da c

given

ce


=

1.48641.

These

values

are

obtained

da c

idex, Therefore,

from

a

table.

ce


Solution

in

emergent is the

the

is

wavelets

of

e-

parallel

and

do

rays

is

obeyed

that

clear

depend

not

by

both

the

direction. and

about

for

plane

the

optic Realizing

revolution

sections this

on e-

Since

incidence. of

In

and

deviation.

of

prism

incident

the

edge,

the

circular.

both

0-

law

is

it are

are

the

through between of

angle refracting plane figures

the

wavelets

e

axis,

the the

to to

and

0

angle

as

perpendicular

optic

e

known

refracted c

are

rays The

the

the

0

and

velocities Thus

0-

rays.

da c

axis

e

1.

is

axis

that

and

0

figure

rays

optic

Snell's

The

:

shown

as

ce


is

a

minimum

when

do dS

I

r

da c

o

ce


collecting

after

(w h Thus

only .

lC

h 8

-

i

h

1 y

on

equation

this

way

0

ld

ln

2)sin28i

a

vacuum

)

,

(1

=

or

f

2)sin28t

-

is

satisfied

be

can

·

s

-

·

or

S1n

2

8

for -

i

.

to

.

Sln

2

8

-

1

t.

8t

da c

The

(1

terms,

ce


the

into

Substitutuing A,

o

,

and

oe

=

it 20°50'

is

equations

above

the

given

values

for

that

found ,

da c

8

,

ce


advantage

stant

for

that all

positions

the of

interference the

pattern

remains

con-

specimen.

da c

great

ce


I

da c

FIG.

ce


Solving

equations H

'

respectively,

4,

5,

yields

6,

and the

7

fol owing

for

om

a

,

om results:

e

,

om

H

'

and

a

S

da c

om

ce


15

da c

CHAPTER

ce


the

through normal The to

point

of

center

of

point locate

has

the been

retraces

intersection

image found,

of of

the the

strikes its

C,

curvature

hence

and

incidence

paths

of

two

any

head

of of

mirror

the

original these

the

arrow.

any

other

at

path. suffices this

rays Once

rays

may

be

da c

traced.

ce


line

is

W

drawn

draw

and focal axis.

parallel

which the

second

in

can

used are

directed

lines

can

be

paths

show

to are

directed

and

through

is

QI

when

QI,

one

ray and

F

also

through

Q

lines

dotted of

virtual,

backward

how

and

the

incident dotted

virtual,

extensions

of

the

QI.

da c

emerging

Q

and the

directed

extensions

through used

is

Q forward

how

Q

through

system Q

is

space

the

to

located the

Q

at

primary

the

line

the

through

object

in

parallel

crosses

Having through

When

show

to

paths

W

drawn

directed

in

axis.

the

to

Q'.

again

start

can

locate

to

is pi

trace

is

space

One F

and

at to

which

ray

be

axis

the

object

parallel

a

VI

pencil

red

a

W

through

to use

and line

Q

Through line

The

F

and

through

plane.

can

VI

through line

a

ce


axis

emerge

parallel

and

directed

toward

da c

the

Q'.

ce


da c

o

ce


Image-forming important

image

mirrors characteristics

are

either

concave are

or

the

convex.

fol owing:

da c

The

ce


da c

FIG.4

ce


8

da c

FIG.

ce


direction tion

backward,

from

is

same

ray

wil

be

reflection.

of it

backward

projected The

after

reflection

found

to

direction

its

and

MV, law

the

after the

When wil

be

found from

pass

through

to

M'V,

pass

when

reflecreflected

through projected

P".

da c

ray Pl.

mirror

the

to

found

ce


practice,

In

cisely departure

from is

not

the

A

image

The

object

is

the viewed

is the

generally magnifying is

point

focal

the

with

together

of

point

great. 1.

Figure

shown

when

focal

situation formed final

by through

the

this of the

the

is

il ustrated is

lens

retinal

magnifying

prethe

intro-

error

type

magnifying

the

size

If

lens.

small, of

located

not

da c

duced in

object

the the

at

image lens.

ce


second lens.

not

deviated

ray

This

by

passes ray the

directly travels

the

through in

a

straight

of

center

line

and

is

lens.

da c

The the

ce


da c

.

PROBLEM

15-7

ce


the

principal

from

the

Point.

lens.

a

located

conjugate

Rays

focal

axis,

lens

focal

the

is

point to

an

object

point by

are

(Fig. at

brought 3). that

stating

located

infinity

axis

the

on

object, to

a

Another

optical

the

by of

point

focal the

also

focus

way

the on

defined.

be

can

distant

infinitely

an

PI'

is

point

focal

The

plane,

in

brought

are

rays

da c

lens

the

at

principal

from

the

on

parallel

the

If

.

2

another

right,

Focal of

P

plane

defining is

axis.

ce


da c

.

PROBLEM

15-8

ce


this

Ordinarily located

point

is

point.

in

means

that

air.

In

that

on

the

superimposed nodal The

point ordinary

is

thick

on

is

and

point the

second

il ustrated

are

nodal

first

principal

first lens

the

the

superimposed

equal. image

are

space and

situation,

refraction

of

indices

the

image object

and the

object

the

da c

both second

of

media

the

situations,

optical

most

In

of

in

principal Figure

the

4.

ce


One

point

extra-axial

an

surface

refracting Conversely,

single

locate

be

the

object

principal object

traced

planes

point

backward

to

through image

conjugate

the

locate

to can

rays to

and

points

from

rays

from

the

point.

da c

a

point. image

focal

the

use

can

certain

trace

ce


point to

after the

refraction axis

or

after

that

the

ray

QF

proceeds

refraction.

da c

focal

parallel

ce


F. /'/

I

I

/

/

/(

I I

f'/

I

#: r

I I

I I I I

I

14

1 I

: I

I I

e

:

.

/,

/

4

D

da c

I

S

.

A/

I I I I

ce


da c

TELESCOPES

ce


da c

.

PROBLEM

15-12

ce


since

the

the

image

seen

image by

formed the

by eye

is

the

ocular

is

not

reinverted,

inverted.

da c

and

ce


subtended

by

small, of

the The

telescope angles is

the

the :

object 8.

apparent the

to are

given

il ustrated

at

the

eye the

Therefore, of

angle apparent in

an

angle Figure

since

magnifying object without 1.

8

when

is

power at

the

eye

the

The

very is

the

with

telescope.

magnifying

by

da c

ratio

power

8

tan

ce


unit, successive reinverts viewed

which

consist

may

of

conjunction the

through

or

image the

formed

single

two

by

the

terrestrial

The

is

in

lens,

convex

objective. telescope

used

lenses

convex

double

simple

a

image

therefore

da c

upright.

ce


that

light

has

in been

the

present

case

the

direction

of

travel

of

reversed.

da c

bered

ce


axis

Therefore, A

system. is

formed

real

image outside

which

is can

the

parallel

not

path

not

viewed of

so

centered

a

be

tilted to

the

da c

eyepiece

is it

is

mirror

the

telescope, reflection

after

reflection.

tical an

Herschel

the

In

the

that before

axis

the op-

directly incoming

with beam.

ce


da c

.

PROBLEM

15-16

ce


Since

the

subtended

position point

the

object converge the

fore

aft

and

parallel

by and

the

long

of

form

formed the

eye

independent

movement

each

eye

can

image

an

the

on or

from

rays the

as

to

rays

image

is the

an

visual

the

rays, as

as

parallel of

rotation direction.

long

from

emerge

parallel points

image As

eye.

these

of

set

of

pair

size

the

affected

be

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of

set

a

a

a

a

da c

in

by of emerge

retina,

not

as

point

object

each

telescope

angle of

from

rays

astronomical

on

retina

wil of

it

ce


da c

MICROSCOPES

ce


therefore

and

the

above

the

equation

lateral

magnification

angular

the

magnification

of of

the

the

ocular.

becomes

da c

Thus,

of

product

the

objective,

ce


optics objective object image

of

tance

from

acts

as

the

objective

as

il ustrated

of

usually regard the a

by

located

the

therefore

is

magnifier.

simple of

the

to

the formed

its

near

image

It

by

the

focal located

of is contributes

magnifying objective

and

the The

at

with

viewed

microscope overall

microscope length point.

The

focal

short

very

is This

figure.

the

a

objective

image

large

further

is

it.

magnification The

in has

dis-

some

which

ocular

an

interest

that

note

to

significant of

power

is

enlarged

the

system. even

ocular.

da c

OBJECTIVE

ce


I

SPHERICAL

ABERRATION.

da c

FIG.

ce


and the

it

because

3).

power The

tral

rays

70% as

the

on

to

correct

both

spherical

to

attempts to

infrequently,

Not

zone.

axis

the

than

more

rays

passing

rays

the

from

it

aberration,

spherical

peripheral designer

optical

distance

has

too

the

of

through periphery. the aberrations

the

reduce

(Fig. the

superimpose the

doublet

wil central 70%

of called is calculated

lens This

is

so

and

cen-

the

the

chromatic

da c

aberration.

ce


(b)

a

find

to

axial fol ow

point that also.

imaged aspherical

that

such

imaged

are

at

from

rays In

other

mirror

revolution all second other

words,

for

anyone

pair

of

point. point proper

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from

axial axial the

always or

diverging

rays

by

spherical

is

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can

conjugate

da c

a

a

some

surface,

it

elim-

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course,

However,

of

surface

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of

cannot, mirror.

spherical

form,

aspherical

from

aberration

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inated sible

wil choice be

given

any

does

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not

sharply

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of

an

eliminated

points.

ce


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at

ray

the

image

distance

and the

quality,

interval

of

the vertical circle

horizontal

the

between the

of

surface

midway

Approximately

lens.

images of and

least

is confusion.

vertical

an

images

of

area

is

The known

Sturm.

da c

as

the horizontal

ce


would and

reflected shown

be as

c,

(or

plate,

plane

one

central

region

portion portion

as

2,

and

called)

often

3

as

converging

a

are

is surface

other in

concave

functions

thus

lines.

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a

the

by

ful

points

a,

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lens is

at

It

in

convex

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portion.

outer

lens

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at

correcting is

diverging

lens.

shown

axis The shown

the

cross

dotted

the

and

corrected

da c

1,

is

it surface.

as on

to

as

so

the

by

paths

b,

the outer

central of

rays

ce


light

omit ed, would

on

portion the

of

strike the

the

through

passing not

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slide

lens

near

its

center

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of

portions

outer

projecting

and

would

only

a

be

imaged

screen.

da c

slide small

ce


The determine diameter

that the

the

the

of formed

source

projecting

the

by

The

evidently

be

slide

largest

projecting

from

lens of

least

as

the

by its

image,

the lens

the

of

focal

the

and

fil

diagonal

the

as

detetermined and

just

condensing

while

lens the

of

should the

great

projected,

is slide the

(the

condensing image

length

of

magnification and

the

distance

screen.

da c

the

the

the

lens

conditions

the

explained of

diameter

at

be

to

lens between lantern

has

length

projecting condensing

lens).

must

the desired of

discussion focal

preceding

ce


(c)

All

glass.

Burning at

one

point

where

parallel

rays

the

burning

to

takes

da c

focused

the

axis

are

place.

ce


real

P'Q',

image from

the

therefore, field

lens

of

should the

be

placed

at

this

eyepiece.

da c

The distance

ce


traced

cipal

from

are

plane conjugate

ray WI.

directed The

W,

then

0'

as

plane

as

must

at

cross

rays

O.

to

plane

W

M'

on

is

conjugate

the

the

from

to

M,

in

lies

which which

lies

a

through

system 0',

point

image

axis

thus

and

magnification,

emerge the

WI

Wand

axis.

the

to

prin-

primary

the

to

F,

unit

having

toward two

through parallel

path

a

and

which the

is

im-

same

in

the

same

Q.

da c

age

along

at

points

conjugate object

0

ce


+IOD

+40

1\

VERGENCE

\

I

,

, \

DISTANCE

-4D

,

,

I

V I

FIG.

2

(b)

LI

0.58

M

L2

da c

.

PROBLEM

15-29

ce


C

1.0

,

1

/

da c

METER

F

0

0 8 0.5

ce


pressed pressed

the

strength in in

meters,

of

power

or

then

the

diopters.

a

power

lens, (lens

.

strength)

da c

find

If

F

is

ex-

is

ex-

ce


tions

posed

problems the

axis

optical below

the

to

convention, the

and

diopteral the

figures

by

distances,

1-7, are

powers

in

meters,

respectively. above

expressed are

the

indicated

axis.

da c

By

ce


da c

(g)

ce


rays

verge

tion

they

P' of which refraction. coma

a

aberration

ray

The ra

of in

P'

general

chief from

in

some

the the

meridian

this

P',

consider

parachief

in

ys

refrac-

after

but

than

crosses

displacement the

point.

us

ray

of

space,

common

other

Let

marginal

given

object marginal

points

behind.

some

a

the

but at

ray

The or

at

P',

at

chief

the and

s

in

P

ray P

'

represents

front in

case

at

P'

after the

question:

da c

os

point

to_verge

converge

rays cr

the

at

fail

ce


18

da

c

CHAPTER

ce


Solution eral and surface

tion, incident reflection.

are

as

passing the

A, lies

Hence

the

normal

A,

Band

C.

in C

must

C so

in

be

mirror

the

in

the

x

mirror reflec-

specular at

gen-

any that

determined

plane be

the

to Now

the

the

let

diagram normals

the

to

Band

and

B, the

Draw

ray and

and

A

Orient

through reflected

ray

be

mirror. shown.

the

on

y-axes

points

the

Let

:

point

the

by the same

point plane.

of

da

c

'Y

ce


y

y

h

+

Y the

h

2

(1)

+

(y

-

0

y)

2

figure,

da

c

From

y

-

0

-

ce


Solution

:

mirrors,

M

I

the

describes

figure

The

and

M

2

,

are

separated

given by

a

situation. distance

The 4d.

da

c

two

ce


Similarly, duce called an

angle

of

opposite

COD

of

the

image

of

PIon

point

on

the

all

equidistant

It

fol ows circumference

P

,

PI

line from the

a

are

0 of

circle

OM

since

P

from

Thus, where both whose

perpen-

from

equidistant S, and

OM

series center

every

is

2

the

every

and

PI

probe

with dihedral the

equidistant

are

Similarly,

ON.

point images of

Since

PI

terminates the is

in

MON.

and

S

OM.

and

2

straight that

is

I line

straight ON,

mirrors

ON can

P

2

are

intersect.

ON

arranged

are

is

0

and

on

whose

OS.

da

c

the radius

angle SP

series

these

both

the

bisector on

Q2'

mirror which

on

etc.,

...

of

behind

lies

first

fal

QI' Each

which

image

which

rays

images

Q-series.

the

dicular

point

the

series

a

ce


depend

on

the

the

distance

the

mirror

is x

that

h/2, the

and

person

this is

mirror.

does

standing

not

away

da

c

from

of

length

m1n1murn

ce


OF

REFLECTION INCLINED

BY

MmRORS

da

c

TWO

LIGHT

ce


through

an

angle

8.

da

c

tation

ce


angle with

each

other.

The

incoming

beam

makes

an

angle

MI.

da

c

\J;

with

8

ce


lie

on

a

circle

of

radius

OS,

as

shown

in

the

figure.

da

c

all

ce


8.

The

solving

side,

2Ky

2Ky

+

integer

smallest for

2

a

°

180

>

is

S

-

derived

-

y

the

on

+

a,

from

into

fal

wil

which

180°

>

OC

2Ky

or

the

and

side; +

second

2

a

>

180

0

equa-

K:

da

c

tion,

OD

which

for

P-series

the

in

image

that

the

on -

first

The is

COD

ce


17

da

c

CHAPTER

ce


so,

da

c

and

ce


this

is, the

object.

image,

due It

is

to not

mirror

magnified,

lands

C, but

right it

is

on

top

inverted.

da

c

That of

ce


.".-

.".""

.".-

--

da

c

--

----

ce


da

c

so

ce


da

c

Thus,

ce


PROBLEM

17-6

da

c

.

ce


Solution

radius

(a)

:

distances

to

of

curvature

If the

and

u

v

mirror, of

the

object

the

represent

respectively, mirror,

and

the

and r

mirror

=

21fl

image is

the

equation

da

c

states:

ce


da

c

thus,

ce


If

the

Applying

to

equation =

32.65

be

mirror,

the

and

the

distance

a

image

the

(1), em,

moved

using same

of

distance the

radius

15

image

new

of

further

cm

would

be

32.65

cm.

distance

curvature,

gives

da

c

q'

is

image from

away

ce


must

be

real

image

considered;

produces the

virtual

a

limits cm

and

in

as

distance 60

em

the

each

the

produces that 2.

figure of

for

mirror

second,

and

1

image the

when

20

that

figure

image

a

mirror

the We

want

is

at

find

to

the

two

case.

da

c

limits

first, in

as

ce


=

11.46

cm

and

78.54

cm.

da

c

U2

ce


These

three

fol owing

equations

can

be

added

to

one

another,

yielding

result:

da

c

the

ce


PROBLEM

17-13

da

c

.

ce


AND

CONCAVE

MffiRORS

da

c

CONVEX

ce


The

is

image the

size

of

and

erect

the

second

is

image

1

3

size

the

of

the

object.

is

da

c

Thus,

first

ce


PROBLEM

17-16

da

c

.

ce


SCM ...

,.

da

c

ce


so,

da

c

and

ce


da

c

AX'S

ce


gives

da

c

(3)

ce


PROBLEM

17

-19

da

c

.

ce


For

which

mirror,

concave

(exit

eye

mirror

the

focal

of wil

length pupil mirror

the as

wil

determine

as

in

the

second the mirror. be virtual in shown convex mirror field the

in

For

and

figure

the

view

the outside

lat er

2.

We

inside

the the

focal

proceed the

en-

length draw

to

again,

object

of

length

case,

the

and

in

is

eye

in

one

cases,

focal

the

case

of

two

are

outside which

angle

8

space.

da

c

rays

there

is

pupil)

the

and

trance

of

a

the

ce


the

axis,

renders

with

the

its of

direction the

beam

emergent of

incident

travel

parallel just

to

the

reverse

the

lens of

the

beam.

da

c

direction

lens

ce


da

c

or

ce


da

c

Then,

ce


18

da c

CHAPTER

ce


the

power

of

a

lens

we

have

da c

for

ce


da c

0'

ce


da c

.

PROBLEM

18-5

ce


in is

magnification

,

diameter 1.35

m

=

cm

_

in

-0.091 0 is

.

=

33

magnified

diameter. is

0

.

27

by It

positive.

da c

hence

ce


Thus, 2000

f

em

-

37 -

54

-

f

em

and

fIt

x

1.33

(equation

1

(54

x

1.33)

(1)

)

em

da c

-

ce


da c

.

PROBLEM

18-8

ce


da c

.

PROBLEM

18-9

ce


the

quadratic

equation,

da c

Using

ce


the

light,

focal

Now

again,

wil

we

plane

;. 1:

find

let ing =

(1.

f,

faces

surface

but

Rl 50

surface is

convex

length,

-

1)

focal

the

light.

f1 L

1

_

30

of

Using R

mm

lens

faces

the

mm.

length

and

00

=

·

the

the

of 60

-

2

]

30

rom

the

equation yields

lens

when

(1)

·

'

da c

the

the

when

Therefore,

ce


of

n

the

both

X,

L

the

ratio

medium. mediums

so

is

equation

Since we

can

(1)

of

the

index

we

are

working

represent becomes

of

the

lens

with 1

[ Rl

_

1 R

2

J

the in

equation

da c

where

to same

index

the

lens

(1)

for

by

ce


The

focal

length

of

the

lens

applying

in

water,

fw'

da c

by

can

be

found

ce


n2

-

(n

-

n1)/R

1

+

(n2

-

n)/R2

da c

f2

ce


Solution

:

ipulate

the

lems.

There

or

Newtonian

determine

to

a

length. ray

the

axis

Let

is

that at

the

lens.

for

thin formulation. dif erent look

us

not

optics lenses

of

1,

figure

at

by

by

the

lens

the

either

use

problem

this

method

prob-

handling

can

one

man-

can

person

of In

deviated Then

a

ease

where

the we

and

are

we

expressing passes

look

at

through

geometry,

da c

focal the

for

example,

For

Gaussian asked

which

in

ways

many of

are

geometry

ce


Substituting

the

value

for

U2

determined

in

the

preceding

da c

step,

ce


The

vergence

equation

in

general

form

da c

Solution:

is

ce


using

Again

111 J.

+

=

-20

30

da c

S.

(1),

equation

ce


Then S

-

65

em

-

15

em

0

f 1

+-1.. S.

65

S.

predicted.

the

object

15

1

19.5

em

is

S.

-

1

da c

So as

-

1

1 -

ce


Solution

:

We

use

1

+ 1

1

r

=

1

1

lJ

da c

0-

ce


PROBLEM

18-20

da

c

.

ce


da c

.

PROBLEM

18-22

ce


da c

.

PROBLEM

18-23

ce


da c

.

PROBLEM

18-25

ce


lens 1 -

00

1

+

-

ql -

£1

1

,

da c

q1

l/f

ce


da c

so

ce


(b)

r2

or

(c)

r2

00

=

is

(pIano

positive

convex) but

I r 21

>

rl

·

da c

or

ce


and

have

CAI,we

.

sin sin

a -

cp P

r

-0-

p

sin sin

+

I

a

l

r

-

-

b

l

(2)

da c

and

cp

ce


da c

adding:

ce


L=

fix

y

6 -

x

10 -

-

12

5.0

da

c

y'

ce


the

triangle

OAI,

we

can

write

da c

From

ce


values,

da c

Substituting

ce


da c

T

ce


equation

p

also

=

s'

(3),

da c

or

ce


or,

1 -

r2

[1

+

q

r4 -

-

1

J

1 -

fen

Therefore, -

2f(n

1)

-

(q

-

1)

da c

r2

1)

-

(9)

ce


PROBLEM

18-32

da

c

.

ce


multiplying

Then,

1

the +

0

-

0.2

+

matrices

right-hand

two

1

0

-

first

gives

0.2

-

+

0

0.12 1.5

+

1

0.4

0.92

da c

0.6 1.5

ce


by

da c

given

ce


da c

(1.02 \0.4

ce


In

the

(assuming

if

part, the

units

thickness

te are

the

same

as

is for

changed r

to

and

f),

1 we

:

da c

have

second that

ce


for

Hence,

first

the

surface, n'

n'

n -

n

-

-

f'

£1

r

1

1

or

1

1.48

1

1.48

-

-

-

f

f'

1

1

1

Therefore, f

1.48 -.48

-

1

-

-3.083

nun

-2.083

nun

and 1

f'

-

-

-.48

1

the

second

surface, n' ft

nil

nil

-

n'

-

-

f"

2

r2

2

or

1.48

1.48

1

1

-

-

-

f'

da c

For

.

f"

-1

NON-ACTIVATED VERSION www.avs4you.com 2

2

Therefore,

and

1

-

-

-

.48

2

-2.083

rom

ce

fl

f"

1.48

-

-

2

-.48

-3.083

nun

da c

f1

ce


as

(Rl to

first

the its =

the

22

is

11.4

mm

of

the

to

the

image

the

left

another

therefore,

rom); left

taking

surface,

object,

of of

focal

second

the

the

first

point,

lens.

da c

by

FI'

surface

surface is

11.4

rom

ce


da c

.

PROBLEM

18-38

ce


da c

.

PROBLEM

18-39

ce


da c

or

ce


center

axis, to

the

of

because

any

other

(Fig.

Such

curvature.

normal

ray

to

a

one

lens surface

has

unique

no

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First

observe

figure

from

=

j

75

2

+

125

tha

t 2

=

145.77

da c

u

1

ce


10

da c

36.44

ce


of y,

1

and

surface,

the

p

,

P

2

y,

of

surfaces

'

and an

to

the

so

that

gmatic

z,l

lens

the lens

rbfracting is

total and

=

P

2

y,l

of

the in

power +

P

y,2

of

powers the ,

=

z

the

a

two

principal

two P

of

P

power

P

z,l

+

P

the

two

xz-plane,

and

xy-plane

the

of

powers

lens

thin

equal

is

surfaces

of

the

sections

of

an

lens, asti-

.

z,2

da c

y

in

meters).

(in

curvature

refracting

the

refracting P

P

of

denote

2

z,

astigmatic

P

sum,

radius

the

P

,

The

respectively.

is

r

P

ce


r

10

=

1 p

=

z

=

2

16

() 2/3

=

3

cm.

da c

r

cm.

ce


da c

.

PROBLEM

21-13

ce


the

diameter

to

which

the

iris

diaphragm

must

be

da c

Hence,

stopped

down

is

ce


(c) What

before.

the

away,

original everything

again changed

once

glass

is What

is

the

else

with to

(a),

part

remaining

the

suppose

the

same

as

is

stop before.

now?

starting astigmatic

of

data

dif erence

astigmatic

Now, the

the

far

as

is

(d) of

with

Working times

1.61,

the

everything dif erence

original

data, else

remaining

the

suppose

the

index same

as

now?

da c

5

ce


the

of

vertex

()

of

inclination -

By

the

surface; incoming

surface).

OPA

Angle P

angle ray

to

=

the

180

optical

the

is -

e,

8

axis;

of

angle the

is and

incidence

of

angle angle

POA

is

.

the

law

of

sines,

da c

e

first

the

first

the

at

ce


n

is

in

air,

so

n

=

1.0.

da c

b.

ce


da c

(a)

ce


da c

n

ce


of

Location

AD

stop -16.87

4.7

-16.61

4.8

-16.31

409

-16.00

500

-15.65

5.1

-15.29

da c

4.6

stop

AD

ce


da c

.

PROBLEM

21-17

ce


(.u :1.0)

da c

AIR

ce


22

da c

CHAPTER

ce


+

and

angle

-

0

=

angle

CDB CDB

180°,

+ =

and

a'

180°

-

-

a'.

0

=

180°.

Hence,

Angle a

-

BCD

a

+

8'

=

-

a'

a

a

-

+

180°

so,

da c

BCD

ce


equation,

of

plane into which

prism

the

first

the

is

angle, of

used

the in

one

equations the

obtains

by (1)

determination

da c

sector

tution

the

substi-

fol owing of

n:

ce


in

the

the

light line

to BA

segment to

false,

of

parallel

under for

is of

this

in

the

be

cannot

and

incidence,

the

position

a

and

a

Thus,

case.

of

minimum

to

being

Therefore, deviation

the

exists

experi-

by

However,

E.

inci-

stil

minimum.

minimum

E,

imagis,

parallel

emergent altered,

not

stil

a

Then,

E.

that

and

is

therefore

direction;

,

that

angles

a

in

deviation

assumption

two

deviation,

segment The

.

and

0,

least reversed

of

beam

equal

ment

position

the

is

hypothesis deviation,

da c

is ine dent line

a

=

E.

ce


o 1

sin

+ m

y =

2

n

2

sin

da c

n

ce


the

trigonometric

identity

for

the

sine

of

the

sum

of

angles,

da c

By two

ce


(1)

becomes

da c

equation

ce


da c

.

PROBLEM

22-6

ce


than

greater

prism

90°,

second

there

is

surface,

light is

which il ustrated

in

as

has

the

angle in

total

the

reflection

internal

this of

path

The

case.

minimum

deviation

at

of

a

for

ray

this

figure.

da c

the of

ce


the

above

equation

gives

da c

into

ce


da c

.

PROBLEM

22-9

ce


A

=

0.590

satisfied

be

for

some

average

wavelength

such

as

1J.

da c

only

can

ce


da c

1m

ce


this

problem,

the

fol owing

values

are

given:

da c

In

ce


solution P

of

The

:

100

=

the

of 0

power where

0,

tan

prism.

In

this

a

prism

is

represents

problem,

P

given the is

relation deviation

the of

by angle given

to

5

be

prism-

Thus,

diopters. =

tan-

1

( lO )

=

tan-

1

( lO)

=

tan-

1

(.05)

da c

o

-

2.86°

ce


power

of

a

prism

is

given

by

the

relation

da c

The

ce


da c

by

ce


-

cas

(

°

.

m1n

2

)

=

da c

n

ce


refraction:

da c

of

ce


Snell's

law

of

refraction

states:

da c

Solution:

ce


for

Substituting (3»,

sin

(from

r

Sf

r

(from

sin

(6»,

equation

equation

(5»,

da c

equation becomes

e

cos

and

e

equation

r

(from (4)

ce


23

da c

CHAPTER

ce


J..4*<

da c

t1--

ce


spectrum

show

how

you

included

calculate

could between

these

the

linear

two

wavelengths.

length

da c

to

of

the

ce


sin

i

sin

i

i

and

-1

. -

lA

Sl.n

i

2

nlsl.n i

-

-1(

,

lA

.

n

l

A)2"

.

sl.n'2.

A)

da c

Sl.n

A

2A

. -

2A

rIA

nIsin

-

lA

since

i

nIsin

-

lA

ce


=

i

-

2B

i

2A

da c

a

ce


-

l

54.20°

da c

i

ce


da c

-

ce


Substituting for

in

the

cD

gives

given the

value result

for

wand

the

da c

computed

value

just

ce


(1)

glass

prisms,

and

(2)

refer

to

the

crown-glass

and

the

flint-

respectively.

da c

where

ce


da c

or

ce


(400)

01

°2(400)

°2(700) o

=

1

01

=

A

=

A

(700)

1

(n

1

2

(n

2

1

(n

1

2

(n

2

-

(400)

-

(400)

-

(700)

-

(700)

01

-

(400)

1) 1) 1) 1). =

A

1

(n

1

(700)

da c

Then

A

-

(700)

01

A

-

-

n

1

(400)

)

ce


,

o

=

(n

'

0

-

l)A'

da c

Q

ce


DISPERSION

da c

ANOMALOUS

ce


of

proportional

length,

undeviated violet

the

and

in

(from

and

the

hence, is

beam

the

red, longest

.

d 1rect

1S

dA

colors

of

order

yellow,

orange, to

the

shortest

I y

away green, wave-

order).

da c

from

blue,

to

de.

prism,

the

given

Hence

prism.

the

ce


reen

A,

so

light

has

from

equation

A

wavelength

equal

to

approximately

(1),

da c

(b) 5500

ce


v

da c

g

ce


ce da c

Thus,


for

n

in

equation

(1)

gives

da c

Substituting

ce


)..

dn -

d)"

da c

Thus,

ce


and

from

sin

45°

-

1.653(sin

e)

sin

45°

-

1.614(sin

e

)

which, e

8

-1

sin

-

1.653

.

' -

(Sin 45°)

Sln

-l(Sin

1.614

45°)

.

da c

and

I

ce


to

the

the

the beam

incident

involved.

angles of

light

blue

light

beam.

is

By

The

angle

45°

beeause

between

the

it

is

plate

parallel

trigonometry,

da c

clarify and

ce


da c

or

ce


Thus,

the

dispersive

dn

power

-

(

1 -

dispersive

power

v

n-f) -

(n-l) dn

da c

Now

ce


Substituting given

f

in

2

for

expression

C

using

(1),

equation

light

and

in

given F

light,

indices

the we

equation

(2) of

for

f

l

refraction

have

da c

and

the

ce


for

the

flint

glass

lens,

we

have

da c

Similarly,

ce


da c

or

ce


da c

or

ce


1 -

(1.5436

-

l)p

da c

20

ce


24

ce da

c

CHAPTER


LENS

(D=8cm)

ce da

c

IMAGE


Substituting

the

values

for

F

and

A

into

this

gives

ce da

c

equation


(c)

The

luminous

of

efficiency

luminous

flux

source.

So

to

the

power

source

a

(radiant

is

the

flux)

ratio

applied

of to

ce da

c

the the


Corresponding 0.265. x

10

-3

lumens

flux

luminous =

0.09

is

given

rn,

approximately

is

table

the

633

wavelength

laser

He-Ne

from

the

Hence

the

by

685

x

0.265

x

lumens.

ce da

c

.5

to

luminosity

relative

the


-

Luminance

ce da

c

L


or,

if

source

candlepower

use

energy square

per

mIl imeters

as

unit

time, is

given

our

measure

the

il umination

of

emit ed in

light lumens

by

ce da

c

per

we


where the

is

I

the

source

being

il uminated,

source

to

intensity,

source

and

surface

r

is

the

is

the

distance

between

angle

normal

vector

the

to

in

feet

surface from

the

il uminated.

ce da

c

the

e

unit

the

and

rays


the

light

rays

are

perpendicular

to

the

surface.

ce da

c

provided


PROBLEM

24-10

ce da

c

.


from

the

lamp, for

a

which

was

right

triangle.

found

using

by The

the

il uminance

trigonometric is

ce da

c

relation


of

get

table, the

and

maximum

Then,

is

e

shown

il umination from

in

the the

at

the

figure,

we

figure.

We

edge see

of that

a

for

want

to

circular any

h,

ce da

c

table.

the


em

ce da

c

300


equation x

=

(1), -

135

(1).

equation

the

is

right,

x

=

and

the

1335

135

-

as

the same cm

the

x

cm.

obtained

from

which

at

distances

left,

1335

=

that

points

two

at

are

to

and

cm

result

same

Therefore,

il umination the

get

we

is

em

of of

the

the

135 20

cm

to

candlepower

ce da

c

source.


=

R

tan

8

ce da

c

a


PROBLEM

24-14

ce da

c

.


ce da

c

Solution:


I

is

the

intensity

of

normal

luminance

to B

and

disk

the the the

surface area

of

normal is the

to

equal glass

the

surface. the

to A.

The

of

product So

we

have

ce da

c

the

intensity


25

da c

CHAPTER

ce


first

Thus,

of upon

filters

used

ito

together of

transmit ance

resulting

curve

transmits

30%

transmits

(.9)(.3)

this

In

A

filters shown

in

for

figure

incident

light incident

light X

by B

and

the

the

100

=

27%

of

the

multiplying values

various

it.

the values

of

light for

curve

the

the

and

upon

transmit ance

the found

be

of of

manner,

can

is

90%

transmits

glass

glass

combination

the

incident

The

of

piece piece

both

for

the

wavelength.

3.

da c

the second

ce


(3) r'

=

allr(R+G+B) (R+G+B) allr

that

result

the

gives

+

a

l2

[(all+a21+a3l)r +

l2

g

+

a

(R+G+B)

+

+

13 +

a

13

b(R+G+B)

(a12+a22+a32)g+(a13+a23+a33)

b

(a12+a22+a32)g

+

(a13+a23+a33)b

da c

(a1l+a2l+a3l)r

a

g

ce


the

color-matching

h

A

3

and

functions

A

.

two

be

must

of

the

color-matching

functions,

linear

independent trichromat,

denoted

combinatons_of

da c

by

dichromat's

the

trichromat,

particular denoted

by

r

A

,

gA'

ce


from

ference

the

the

In for

second

specimen is

match

meric is

what of

the

at

three

a

duplicate

lengths

of

the

noticed

as

a

As

shown

visible

an a

isomeric metameric

The

the

dif erence

curves

hue, are

while

those

it

labeled

as

labeled

as

2

and

1

wave-

generally

is

saturation. I

and

original

the

other

at

dif erence

metameric

which

curve

of

curve

from

lightness,

involving

1,

the

widely

deviate

exhibit

reflectance

intersect

spectrum.

dif erence

figure

in

and

meta-

generally

not

original

and

spectral

The

usually

wil

wil

trial

duplicate

dif erence.

metameric

A

observer.

and

first

used

color

same

and

3

dif erence.

da

c

represent represent

then

and

specimen wavelengths

more

or

again,

but,

result,

flux the

those the

produce

to

incident

from

dif erent

are

compounded

given

a

attempted,

called

be

may

for

desired

the

produce

it)

resembling

most

which

colors

values)

(tristimulus

perception

same.

procedure

original

the

color

achromatic the

about

remaining

ce


quired

dominant

section

fal s

such

S2(x,y)

dominant drawn

from

ever,

the

point the

of

the

e

line

defining

the

spectrum

band

or

"e"

a

complementary

is

when

which

wavelength

f ter

fact

that the the

To

-

1.

of

t

Thus

0

-530

,

nm

as

sown

or

530

h

I.n

C

the

a

wavelength matches

complementary a

f1

at

color,

assigned -

Howlocus

is

the is

no

line

locus.

given

distinguish

nm), is

straight

spectrum

the

former

the

the

which with

(770 there

spectrum

meets

mixed

color.

a

the

wavelength,

wavelength, P laced

to

the

nm),

gure

sign

negative 1,

8

2

has

a

nm.

da c

the

purple

e

complementary

(380 colors,

(e),

the

to

S2

inter-

of

region

of intersect

not

achromatic dominant

due

does

from

drawn

specified

from

S2

to

is

this

Sl

triangular corners

region

This

wavelength.

the

the

by the

as

For

nm.

in

defined

described

color.

given 583

as

diagram

chrot icity somet1.mes

the

approximately

at

colors

For

of

wavelength

ce


"

SLIT

da c

ENTRANCE

ce


prism,

quantity and

b

ac

The

ce d


580

nm.

to

wavelength

flux

The

but its

retains

When

the

the

generation from

the

shape

and

location

the

incident

of

about

(exciting)

600 This

on.

of

wavelength

is

accordance

The the

tristimulus

but

throughout flux

sharply

is

data

in

com-

is

that can

matrix

recorded

that

law;

(X,Y,Z)

values

spectrophotometric

band

vanishes

flux than

height.

fluorescent

Stokes's

longer

fluorescent in

and

reflected with

wavelength the

varies

the

enters

drops

Only flux

fluorescent

the

fluxo

arranging

upward.

in

from of

distribution

light nm

somewhat

varies

spectral

form

the

is,

of

the

be

computed

incident as

by fol ows:

da

c

first

much;

fluorescent

of

then

flux

reflected

not

wavelength

pletely from

of

amount

ce


-

=

X

P A

H

A

ZA

A

da c

z

ce


The

trichromatic

coefficients

(1)

equation x

20

=

=

100

20+60+20 =

by

making

use

of

20

=

20+10+20

20 50

=

002;

Y

004;

y'

--

60

=

100 10

50

=

0

002;

0

6;

z

z

20

=

=

002

(2)

=

004

(3)

100 '

20 =

50

da

c

x'

20

=

calculated

be

can

:

ce


PROBLEM

25-9

ce da

c

.


white

yield

light. values

x,y,z

=

x

This

second

E

sunlight"o

complement

necessary

to

To

the

of

490nm.

get

chromatic table

an

for

the

coefficient

yields

values the

fol owing

are

presented

single

American

Institute

be

can

at

where

used, 5

nm

is

intervals.

' white'

the to

appears

and

mono-

it

through

then the

observer

standard

a

desired,

Cadmium-red

which

curve,

is

C

"average

C

and

if

that

color

from

line

where

point

seen

red

spectrum colors

energy

be

can

complementary

spectrum

equal

determination,

accurate

table

Handbook

it

color

spectrum

the

the the

Cadmium

the

straight

a

for

curve

figure to

the

has

.

called

then

E

this

draw

0.3333

=

have

must

light

resultant

be

in

the

of

Physics tri-

the This

data:

da c

range

to

the

points",

From

chromatic

point

shows

"white

the

are

that z:

is

color 1

Figure

z

=

y

spectrum

color.

and

x,y,

color

spectrum the

second

such

for

values

fol owing

the

Clearly, intensities

and

ce


=

644

0.3809

-

comp

0.1427

X

644

comp

da c

Y

ce


004789

-

4.05x

589

comp

da c

=

ce


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tJf -

f we

find tJf

-

f

c -

tJA A

ce da

c

A

tJA A


p" 5

ce da

c

p'


8.4

x

10-

2

rom

ce da

c

-


PROBLEM

26-17

ce da

c

.


CAVITIES

ce da

c

LASER


27

ce da

c

CHAPTER


AR, A Brackett

-

s

1875.62m]J

n

820.58m Series

n

is

-

=

given

4

00

.

by

ce da

c

The

-


On

other is

hand obtained

by

v

and

using

vHg

for

a

deuterium

(H

2

)

spec-

ce da

c

trum

the


cm-

439002.7

1

(n

Then

the

fol owing of

ionized

helium:

the

n

table

first

2

be

can

five

members

giving

constructed, of

the

spectrum

the

emit ed

wave-

by

ce da

c

numbers

)

l


for

h,

c,

e,

and

V,

ce da

c

Substituting


N,

H,

and

c

ce da

c

for


bers

the

only mech

and

h is

the

is

where Planck's

J

is

angular the

constant.

a

The

needed. the

depends

state

is

linear correct

quantum

momenta

to

rotational the

given

be

number

quantum Thus

on

and

molecule

angular

momen-

ce da

c

represents

1)

rotational HC!

allows

t reatment +

the

specify molecule.

number

quantum

one

I J(J

to

of

anical

by tum

required

are

geometry


for

I

into

equation

(4),

ce da

c

puted


The

total kinetic

of

energy

energies.

the

orbit

is

the

sum

of

the

potential

Therefore,

ce da

c

and


RADIATION

ce da

c

BLACKBODY


A

for

determined:

be

can

max

A

T

T

A

=

579.4

K

2.8971

x

A

-

max

mll

555

mll

1060K ,

max

mll,

ce da

c

for

0

5000

(b)

so

5794

SOooK

(a)

max


PROBLEM

27-12

ce da

c

.


4

aT

-

2

aT

gain

wil

but

4 2

so

the

energy net

heat

from

the

lost

wil

surface

other

at

a

rate

be

ce da

c

W

l


Optics given two

closely e

1

and

the

spaced

surfaces, the

other

edition,

3rd

Wesley, relates

net

one

at

1949,

heat

a

(H 2

transfer at

temperature

temperature

derivation

is

Tl T

2

and

between

HI)

-

emissivity

and

emise

2

ce da

c

sivity

Addision

,

which


is of

a

the

energy

black per

x

in

body unit

wavelength

10-

2

m

Kelvin.

degrees interval.

deg

K);

T

is

This We

the

temperature

wil also

yield have

the

the

law

ce da

c

Stefan-Boltzmann

(1.4388

constant


=

500

nrn

6.

63

X

500

10

-34 x

3

X

10-

x

10

8

9

ce da

c

E


N

-

:

( )3 21T

3

KT

h

ce da

c

1T

2


01)

2

-

f

-

-y

dy

01)

o

-

e

Y

f

2

de

-y

y

o

de-

Y

-

dV

and

Y

-

u.

Then

du

-

dy

and

V

-

e-Yi

ce da

c

Let


co

o -

ys «

u

CJ)

«(f) V

UO 02

<{«

en LtJ

0 -

CD

o -

...

W

ce da c

N

CD

0 -

I

o

CJ);>

«

0::>

...


lJ. .

cr

o

cr

V

0

lOa::

W

ftQ: Z -

(f)

ce da c

280


ce da c

0.900


ce da c

0.8


10.0

a::

ce da c

z

o

NON-ACTIVATED VERSION www.avs4you.com c

>C) a:: w

z

w

-I

a:: u w

a.. CJ) w

>

-I

W

a::

1.0

ce da c

0.9


." c

oel.

e

-

da

c

U

&:. U C

ce


UNITS

ce da c

ILLUMINATION


PROPERTIES

OF

SELECTED

CLEAR

MATERIALS

ce da c

OPTICAL


Level

of

Fineness of

Color

ce da c

Designation


da

c

INDEX

ce


on

this

page

refer

to

PROBLEM

NUMBERS,

not

page

numbers

da

c

Numbers

ce


on

this

page

refer

to

PROBLEM

NUMBERS,

not

page

numbers

da

c

Numbers

ce


on

this

page

refer

to

PROBLEM

NUMBERS,

not

page

numbers

da

c

Numbers

ce


on

this

page

refer

to

PROBLEM

NUMBERS,

not

page

numbers

da

c

Numbers

ce


on

this

page

refer

to

PROBLEM

NUMBERS,

not

page

numbers

da

c

Numbers

ce


on

this

page

refer

to

PROBLEM

NUMBERS,

not

page

numbers

da

c

Numbers

ce


on

this

page

refer

to

NUMBERS,

PROBLEM

not

page

numbers

da

c

Numbers

ce


on

this

page

refer

to

NUMBERS,

PROBLEM

not

page

numbers

da

c

Numbers

ce


on

this

page

refer

to

PROBLEM

NUMBERS,

not

page

numbers

da

c

Numbers

ce


on

this

page

refer

to

PROBLEM

NUMBERS,

not

page

numbers

da

c

Numbers

ce


on

this

page

refer

to

PROBLEM

NUMBERS,

not

page

numbers

da

c

Numbers

ce


on

this

page

refer

to

PROBLEM

NUMBERS,

not

page

numbers

da

c

Numbers

ce


da c

ce


The Optics Problem Solver, Fogiel - REA Publishers

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