matriculation physics geometrical optics.pdf

8/12/2019 Matriculation Physics Geometrical Optics.pdf

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1

PHYSICS CHAPTER 1

The study of light based on

the assumption that lightlight

travels in straight linestravels in straight lines

and is concerned ith the

laws controlling thelaws controlling the

reflection and refractionreflection and refraction

of rays of lightlight!

CHAPTER 1"CHAPTER 1"#eometrical optics#eometrical optics

$% Hours&$% Hours&


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PHYSICS CHAPTER 1

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At the end of this chapter, students should be able to:At the end of this chapter, students should be able to:

StateStatelaws of reflection.laws of reflection.

StateStatethe characteristics of image formed by a planethe characteristics of image formed by a plane

mirror.mirror. SketchSketch ray diagrams with minimum two rays.ray diagrams with minimum two rays.

Learning Outcome:

1!1 Reflection at a plane surface $1 hour&

!'mp

h!m

atri'

!edu

!my

(ph

ys

ics

!'mp

h!m

atri'

!edu

!my

(ph

ys

ics


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PHYSICS CHAPTER 1

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Figure .Figure .

. !eflection at a plane surface

.. !eflection of light is defined as the return of all or part of a beam of light whenthe return of all or part of a beam of light when

it encounters the boundary between two mediait encounters the boundary between two media!

There are to types of reflection due to the plane surface

Specular "regular# reflectionSpecular "regular# reflectionis the reflection of light fromreflection of light from

a smooth shiny surfacea smooth shiny surfaceas shon in )igure 1!1!


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Figure .$Figure .$

All the reflected rays are parallel to each another or mo*ein the same direction!

%iffuse reflection%iffuse reflectionis the reflection of light from a roughreflection of light from a roughsurfacesurfacesuch as papers+ floers+ people as shon in )igure1!,!

The reflected rays is sent out in a *ariety of directions!

)or both types of reflection+ the las of reflection are obeyed!


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Laws of reflectionLaws of reflectionstate "

The incident ray, the reflected ray and the normal all lieincident ray, the reflected ray and the normal all lie

in the same planein the same plane! The angle of incidence,angle of incidence, iie&uals the angle of reflection,e&uals the angle of reflection, rr

as shon in )igure 1!-!

i r

'lane surface'lane surface

ri=

Stimulation 1!1Figure .(Figure .(

Picture 1!1
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PHYSICS CHAPTER 1

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Image formation by a plane mirror as shon in )igures 1!.a and1!.b!

Point ob/ect

..$ !eflection at a plane mirror

Figure .)aFigure .)a

A 'A

u v

i

i

r

i

distanceobject:uhere

distanceimage:v

g

gangleglancing:g


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PHYSICS CHAPTER 1

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0ertical $etended& ob/ect

Stimulation 1!,

Figure .)bFigure .)b

Object

vu

ir

i

rImage

ihoh

here heightobject:ohheightimage:ih
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The characteristics of the image formed by the plane mirror are

virtual imagevirtual image

is seem to form by light coming from the image butseem to form by light coming from the image butlight does not actually pass through the imagelight does not actually pass through the image!

ould not appear on paper+ screen or film placed at the

location of the image! upright or erect imageupright or erect image

laterally reverselaterally reverse

right2hand side of the ob/ect becomes the left2hand sideof the image!

the ob*ect distance,ob*ect distance, uue&uals the image distance,e&uals the image distance, vv

the same si+esame si+ehere the linear magnification+ mis gi*en by

obey the laws of reflectionobey the laws of reflection!

1height,Object

height,Image

o

i ==h

hm

Picture 1!,
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A omen is 1!34 m tall and her eyes are 14 cm belo the top

of her head! She ishes to see the hole length of her body

in a *ertical plane mirror hilst she herself is standing

*ertically!

a! S'etch and label a ray diagram to sho the formation ofomen5s image!

b! 6hat is the minimum length of mirror that ma'es this

possible7

c! Ho far abo*e the ground is the bottom of the mirror7

Eample 1 "


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PHYSICS CHAPTER 1

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A

B

L

Solution :Solution :a! The ray diagram to sho the formation of the omen5s image is

HE2

1AL=

EF

2

1LB =

)feetF

)e!esE)headH

h

y

m"#$1

m1#$#


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Solution :Solution :

b! The minimum *ertical length of the mirror is gi*en by

b! The mirror can be placed on the all ith the bottom of the

mirror is hal*ed of the distance beteen the eyes and feet of the

omen! Therefore

LBAL+=hEF

2

1HE

2

1+=h

( )EFHE

2

1+=h

Height of the omen

( ) m%#$#"#$12

1==h

( )1#$#"#$12

1=y

m&$#=y


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u v

m##$1

x

A rose in a *ase is placed 4!-%4 m in front of a plane mirror!

Ahmad loo's into the mirror from 1!44 m in front of it! Ho far aayfrom Ahmad is the image of the rose7


)rom the characteristic of the image formed by the plane mirror+thus

Therefore+

Eample , "

m(#$#=vuv=

vx += ##$1m('#$1=x

m('#$#=u

(#$###$1 +=x


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Eercise 1!1 "1!

The to mirrors in )igure 1!% meet at a right angle! The beam

of light in the *ertical plane P stri'es mirror 1 as shon!

a! 8etermine the distance of the reflected light beam tra*els

before stri'ing mirror ,!

b! Calculate the angle of reflection for the light beam after

being reflected from mirror ,!

AS. :AS. : .- m.- m / )0/ )0to the mirror $.to the mirror $.

Figure .Figure .


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Eercise 1!1 ",!

A person hose eyes are 1!%. m abo*e the floor stands ,!-4

m in front of a *ertical plane mirror hose bottom edge is .4

cm abo*e the floor as shon in )igure 1!3! 8eterminex!

AS. :AS. : 0.1 m0.1 m

Figure .2Figure .2


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Eercise 1!1 "-! Standing ,!44 m in front of a small *ertical mirror+ you see the

reflection of your belt buc'le+ hich is 4!94 m belo youreyes!

a! 6hat is the *ertical location of the mirror relati*e to the

le*el of your eyes7

b! 6hat is the angle do your eyes ma'e ith the hori:ontal

hen you loo' at the buc'le7c! If you no mo*e bac'ard until you are 3!4 m from the

mirror+ ill you still see the buc'le7 Eplain!

AS. :AS. : ( cm below/ -.-( cm below/ -.-/ 3 think/ 3 think.! You are 1!;4 m tall and stand -!44 m from a plane mirror that

etends *ertically upard from the floor!


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Sketch and useSketch and useray diagrams toray diagrams to determinedeterminethethe

characteristics of image formed by spherical mirrors.characteristics of image formed by spherical mirrors.

3se3se

for real ob*ect only.for real ob*ect only.

3se3sesign convention for focal length:sign convention for focal length:

44 ff for concave mirror and 5for concave mirror and 5 ff for conve6 mirror.for conve6 mirror. SketchSketchray diagrams with minimum two rays.ray diagrams with minimum two rays.

rr7 $7 $ff only applies to spherical mirror.only applies to spherical mirror.

Learning Outcome:

1!, Reflection at a spherical surface $1 hour&

!'mp

h!m

atri'

!ed

u!m

y(ph

ys

ics

!'mp

h!m

atri'

!ed

u!m

y(ph

ysics

rvuf

2111=+=


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PHYSICS CHAPTER 1

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88

AA

99

r''88

AA

99

r''

Figure .aFigure .a

. !eflection at a spherical surface

.$. Spherical mirror is defined as a reflecting surface that is part of a spherea reflecting surface that is part of a sphere!

There are to types of spherical mirror! It is conve6conve6$cur*ingoutards& and concaveconcave$cur*ing inards& mirror!

)igures 1!9a and 1!9b sho the shape of conca*e and con*e

mirrors!

reflecting surface

imaginary sphere

sil*er layer

Figure .bFigure .b

$a&Conca*e $8onverging8onverging&

mirror

$b& Con*e $%iverging%iverging& mirror

Picture 1!-
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;erms of spherical mirror;erms of spherical mirror

8entre of curvature "point 8#8entre of curvature "point 8#

is defined as

the centre of the sphere of which a curvedthe centre of the sphere of which a curvedmirror forms a partmirror forms a part!

!adius of curvature,!adius of curvature, rr is defined as the radius of the sphere of which a curvedthe radius of the sphere of which a curved

mirror forms a partmirror forms a part!

'ole or verte6 "point '#'ole or verte6 "point '#

is defined as the point at the centre of the mirrorthe point at the centre of the mirror!

'rincipal a6is'rincipal a6is

is defined as the straight line through the centre ofthe straight line through the centre of

curvature 8 and pole ' of the mirrorcurvature 8 and pole ' of the mirror.

A=is called the apertureapertureof the mirror!


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Consider the ray diagram for a conca*e and con*e mirrors asshon in )igures 1!;a and 1!;b!

Point FFrepresents the focal pointfocal pointor focusfocusof the mirrors!

8istanceffrepresents the focal lengthfocal lengthof the mirrors! The parallel incident raysparallel incident raysrepresent the ob*ect infinitely farob*ect infinitely far

awayawayfrom the spherical mirror e!g! the sun!

88''88 ''

.$.$ Focal point and focal length,f

Figure .1aFigure .1a

FF f

FFf


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Focal point or focus, FFocal point or focus, F

)or conca*e mirror > is defined as a point where the incidenta point where the incident

parallel rays converge after reflection on the mirrorparallel rays converge after reflection on the mirror! Its focal point is real "principal#real "principal#!

)or con*e mirror > is defined as a point where the incidenta point where the incident

parallel rays seem to diverge from a point behind the mirrorparallel rays seem to diverge from a point behind the mirror

after reflectionafter reflection!

Its focal point is virtualvirtual!

Focal length,Focal length,ff is defined as the distance between the focal point "focus# Fthe distance between the focal point "focus# F

and pole ' of the spherical mirrorand pole ' of the spherical mirror!

The para6ial rayspara6ial raysis defined as the rays that are near to andthe rays that are near to and

almost parallel to the principal a6isalmost parallel to the principal a6is!


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Consider a ray A= parallel to the principal ais of conca*emirror as shon in )igure 1!?!

.$.( !elationship between focal length,fand

radius of curvature, r

Figure .-Figure .-

88

''FF %%

incident rayincident ray99AA

fr

i

ii


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)rom the )igure 1!?+

=C8

=)8

=y using an isosceles triangle C=)+ thus the angle is gi*en by

then

=ecause of A= is paraial ray+ thus point = is too close ith pole

P then

Therefore

ii =

*

B*tan

=F*

B*tan

;aken the angles are ==;aken the angles are ==

small by considering thesmall by considering theray A9 is para6ial ray.ray A9 is para6ial ray.

i2=

r= +* f= F+F*

;his relationship also valid for conve6 mirror.;his relationship also valid for conve6 mirror.

2

rf =

=

*B*2

F*B*


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is defined as the simple graphical method to indicate thethe simple graphical method to indicate thepositions of the ob*ect and image in a system of mirrors orpositions of the ob*ect and image in a system of mirrors or

lenseslenses!

)igures 1!14a and 1!14b sho the graphical method of locatingan image formed by conca*e and con*e mirror!

.$.) !ay diagrams for spherical mirrors

Figure .0aFigure .0a Figure .0bFigure .0b

$a& Conca*e mirror $b& Con*e mirror

88 ''

FF

((

((

I 88

FF

''

$$

$$O

O I

$$

((

$$


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!ay !ay 2 Parallel to principal ais+ after reflection+ passesthrough the focal point $focus& ) of a conca*e

mirror or appears to come from the focal point )

of a con*e mirror! !ay $!ay $ 2 Passes or directed toards focal point ) reflected

parallel to principal ais!

!ay (!ay ( 2 Passes or directed toards centre of cur*ature C+reflected bac' along the same path!


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The characteristics of the image formed are

virtualvirtual

uprightupright diminished "smaller than the ob*ect#diminished "smaller than the ob*ect#

formed at the back of the mirror "behind the mirror#formed at the back of the mirror "behind the mirror#


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PHYSICS CHAPTER 1

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Ob*ect

distance, u!ay diagram ru > r

u = ru = r

OI

O

Real

In*erted

8iminished

)ormedbeteen pointC and )!

Real

In*erted

Same si:e

)ormed at pointC!

88

FF

''

FrontFront backback


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Ob*ect

distance, u!ay diagram


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inear $lateral& magnification of the spherical mirror+ m is definedas the ratio between image height,the ratio between image height, hhiiand ob*ect height,and ob*ect height, hhoo

Ob*ect



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)igure 1!1, shos an ob/ect Oat a distance uand on theprincipal ais of a conca*e mirror! A ray from the ob/ect Oisincident at a point = hich is close to the pole P of the mirror!

.$. %erivation of Spherical mirror e&uation

Figure .$Figure .$

O 88 ''Iv

u

99

%%

)rom the figure+

=


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=y considering point = *ery close to the pole P+ hence

then

therefore

vru === I+I*.+*.O+O*

tan.tan.tan

v

BD

r

BD

u

BD=== .. Substituting thisSubstituting this

value in e&. "(#value in e&. "(#

fr 2=

=+

rvu

B*2

B*

B*

rvu

21

1=+ here

rvuf

21

11 =+= Spherical mirror>sSpherical mirror>se&uatione&uation


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Table 1!, shos the sign con*ention for spherical mirror5seuation !

Dote"

Real image is formed by the actual light rays that passformed by the actual light rays that passthrough the imagethrough the image!

Real image can be pro*ected on the screenpro*ected on the screen!

'hysical ?uantity 'ositive sign"4# egative sign"@#

Ob*ect distance,u


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A dentist uses a small mirror attached to a thin rod to eamine one

of your teeth! 6hen the tooth is 1!,4 cm in front of the mirror+ theimage it forms is ?!,% cm behind the mirror! 8etermine

a! the focal length of the mirror and state the type of the mirror

used+

b! the magnification of the image!


a! =y applying the mirror5s euation+ thus

b! =y using the magnification formula+ thus

Eample - "

cm(%$1+=fvuf

111 +=

u

vm=

cm/$2cm.2#$1 =+= vu

&1$&

2#$1

2$/==m

( )2'$/

1

2#$1

11

+=

f

"8oncave mirror#"8oncave mirror#


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An upright image is formed ,4!% cm from the real ob/ect by usingthe spherical mirror! The image5s height is one fourth of ob/ect5sheight!

a! 6here should the mirror be placed relati*e to the ob/ect7

b! Calculate the radius of cur*ature of the mirror and describe the

type of mirror reuired!

c! S'etch and label a ray diagram to sho the formation of theimage!


Eample . "

oi 2'$# hh =

O Icm#$2

SphericalSpherical

mirrormirroru v


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a! )rom the figure+

=y using the euation of linear magnification+ thus

=y substituting e! $,& into e! $1&+ hence

The mirror should be placed2.) cm in front of the ob*ect2.) cm in front of the ob*ect!

oi 2'$# hh =

$2#=+ vu

u

v

h

hm ==

o

i

"#"#

u

v

h

h=

o

o2$#

uv 2'$#= "$#"$#

'$2#2'$# =+ uucm0$1"=u


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b! =y using the mirror5s euation+ thus

The type of spherical mirror is conve6conve6because the negati*e

*alue of focal length!

oi 2'$# hh =

cm0&$

=f

vuf111 +=

( )uuf 2'$#

111

+=

( )( )0$1"2'$#

1

0$1"

11

+=

f

and

2

r

f=( ) cm/$1#0&$'2 ==r


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PHYSICS CHAPTER 1

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c! The ray diagram is shon belo!oi 2'$# hh =

FF'' 88

O I

frontfront backback

PHYSICS CHAPTER 1


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A person of 1!34 m height stands 4!34 m from a surface of ahanging shiny globe in a garden!

a! If the diameter of the globe is 1; cm+ here is the image of the

person relati*e to the surface of the globe7

b! Ho large is the person5s image7

c! State the characteristics of the person5s image!


Eample % "

m#$"#m."#$1o == uh

u

oh

PHYSICS CHAPTER 1


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a! #i*en

The radius of cur*ature of the globe5s surface $con*e surface&is gi*en by

=y applying the mirror5s euation+ hence

m1%$#=d

m#/$#2

1%$#==r

vur112 +=

"behind the globe>s surface#"behind the globe>s surface#m#02$#=v

m#$"#m."#$1o == uh

v

1

"#$#

1

#/$#

2 +=

PHYSICS CHAPTER 1


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b! =y applying the magnification formula+ thus

c! The characteristics of the person5s image are

virtualvirtual

uprightupright diminisheddiminished

formed behind the reflecting surface.formed behind the reflecting surface.

u

v

h

hm ==

o

i

m#$"#m."#$1o == uh

"#$#

#02$#

"#$1

i =h

m112$#i=h


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88 FF

''

A sha*ing or ma'eup mirror forms an image of a light bulb on aall of a bathroom that is -!%4 m from the mirror! The height of thebulb is ;!4 mm and the height of its image is .4 cm!

a! S'etch a labeled ray diagram to sho the formation of the bulb5s

image!

b! Calculate

i! the position of the bulb from the pole of the mirror+

ii! the focal length of the mirror!


a! The ray diagram of the bulb is

Eample 3 "

I

O

cm0#

mm#$%

u

m1#0#m.1##$%m.($'# 2i(

o

=== hhv

m#$(

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

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b! i! =y applying the magnification formula+ thus

;he position of the bulb is .0 cm in front of the mirror.;he position of the bulb is .0 cm in front of the mirror.

ii! =y applying the mirror5s euation+ thus

u

v

h

hm ==

o

i

u

#$(

1##$%

1#0#

(

2

=

m#&$#=u


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Eercise 1!, "1! a! A conca*e mirror forms an in*erted image four times larger

than the ob/ect! Calculate the focal length of the mirror+assuming the distance beteen ob/ect and image is

4!344 m!

b! A con*e mirror forms a *irtual image half the si:e of theob/ect! Assuming the distance beteen image and ob/ectis ,4!4 cm+ determine the radius of cur*ature of the mirror!

AS. :AS. : 20 mm20 mm / $2 mm/ $2 mm

,! a! A 1!9. m tall shopper in a department store is %!1? m froma security mirror! The shopper notices that his image in themirror appears to be only 13!- cm tall!

i! Is the shopper5s image upright or in*erted7 Eplain!ii! 8etermine the radius of cur*ature of the mirror!

b! A conca*e mirror of a focal length -3 cm produces animage hose distance from the mirror is one third of theob/ect distance! Calculate the ob/ect and image distances!

AS. :AS. : u think, .0 mu think, .0 m / )) cm, )1 cm/ )) cm, )1 cm

PHYSICS CHAPTER 1


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State and useState and usethe laws of refraction "Snell>s Law# forthe laws of refraction "Snell>s Law# for

layers of materials with different densities.layers of materials with different densities.

ApplyApply

for spherical surface.for spherical surface.

Learning Outcome:

1!- Refraction at a plane and spherical surfaces $1

hour&

!'mp

h!m

atri'

!ed

u!m

y(ph

ysic

s

!'mp

h!m

atri'

!ed

u!m

y(ph

ysic

s

( )

r

nn

v

n

u

n 1221 =+

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

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.( !efraction at a plane and spherical

surfaces.(. !efraction at a plane surface !efraction!efractionis defined as the changing of direction of a lightthe changing of direction of a light

ray and its speed of propagation as it passes from oneray and its speed of propagation as it passes from onemedium into anothermedium into another!

Laws of refractionLaws of refractionstate " The incident ray, the refracted ray and the normal all lieincident ray, the refracted ray and the normal all lie

in the same planein the same plane! )or to gi*en media+ Snell>s lawSnell>s law states

constant

sin

sin

1

2 ==n

n

r

irnin sinsin 21 =


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PHYSICS CHAPTER 1

45

The light ray is bent toward thebent toward the

normalnormal+ thus

The light ray is bent away frombent away from

the normalthe normal+ thus

Eamples for refraction of light ray tra*els from one medium toanother medium can be shon in )igures 1!1-a and 1!1-b!

21 nn


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!efractive inde6 "inde6 of refraction#,!efractive inde6 "inde6 of refraction#, nn is defined as the constant ratio for the two givenconstant ratio for the two given

mediamedia!

The *alue of refracti*e inde depends on the type of mediumtype of mediumand the colour of the lightcolour of the light!

It is dimensionlessdimensionlessand its *alue greater than greater than !

Consider the light ray tra*els from medium 1 into medium ,+ therefracti*e inde can be denoted by

r

i

sin

sin

2

121

2mediminlightof3elocit!

1mediminlightof3elocit!

v

vn ==

"Bedium containing"Bedium containing

the incident ray#the incident ray#

"Bedium containing the"Bedium containing the

refracted ray#refracted ray#

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

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Absolute refracti*e inde+ n $for the incident ray tra*els fromvacuum or airinto the mediummedium& is gi*en by

Table 1!- shos the refracti*e indices for common substances!

vcn == mediminlightof3elocit!

in 3acmlightof3elocit!

SubstanceSubstance !efractive inde6,!efractive inde6, nn

SolidsSolids8iamond)lint glassCron glass)used uart: $glass&Ice

Li&uidsLi&uids=en:eneEthyl alcohol6ater

CasesCasesCarbon dioide

Air

,!.,1!331!%,1!.31!-1

1!%41!-31!--

1!444.%1!444,?-

;able .(;able .(

$If the densitydensity

of medium isof medium is

greatergreaterhence

the refractiverefractive

inde6 is alsoinde6 is also

greatergreater&

PHYSICS CHAPTER 1


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!elationship between refractive inde6 and the wavelength of!elationship between refractive inde6 and the wavelength of

lightlight

As light tra*els from one medium to another+ its wavelength,wavelength,changeschangesbut its fre&uency,fre&uency,ffremains constantremains constant!

The a*elength changes because of different materialdifferent material! Thefreuency remains constant because the number of wavenumber of wave

cycles arriving per unit time must e&ual the number leavingcycles arriving per unit time must e&ual the number leaving

per unit timeper unit timeso that the boundary surface cannot create orcannot create ordestroy wavesdestroy waves!

=y considering a light tra*els from medium 1 $n1& into medium ,

$n2&+ the *elocity of light in each medium is gi*en by

then

11 fv

= 22 fv

=and

2

1

2

1

f

f

v

v= here

1

1n

cv =

2

2n

cv =and

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

49

If medium 1 is *acuum or air+ then n1 1! Therefore the

refracti*e inde for any medium+ ncan be epressedas

2

1

2

1

=

n

cn

c

2211 nn =

"!efractive inde6 is inversely"!efractive inde6 is inversely

proportional to the wavelength#proportional to the wavelength#

here

#

=n

in 3acmlightofh5a3elengt:#

mediminlightofh5a3elengt:

Picture 1!% Picture 1!3

PHYSICS CHAPTER 1
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PHYSICS CHAPTER 1

50

A fifty cent coin is at the bottom of a simming pool of depth

-!44 m! The refracti*e inde of air and ater are 1!44 and 1!--

respecti*ely! 8etermine the apparent depth of the coin!


Eample 9 "

(($1.1$## 5a == nn

heredethaa-ent:AB

m($##dethactal:A =

A

i

Air $na

&

r

B

6ater $n5

&

i

r

m##$(

*

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

51


)rom the diagram+

A=8

AC8

=y considering only small angles of rand i+ thus

(($1.1$## 5a == nn

ABA*tan =r

A

A*tan =i

andrr sintan ii sintan

A

AB

AB

A*

A

A*

sin

sin

tan

tan=

==r

i

r

i

then

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

52

ote :ote : "


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PHYSICS CHAPTER 1

53

A pond ith a total depth $ice F ater& of .!44 m is co*ered by a

transparent layer of ice of thic'ness 4!-, m! 8etermine the time

reuired for light to tra*el *ertically from the surface of the ice to

the bottom of the pond! The refracti*e inde of ice and ater are

1!-1 and 1!-- respecti*ely!

$#i*en the speed of light in *acuum is -!44 14;m s21!&


Eample ; "

(($1.1$(1 5i == nn


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PHYSICS CHAPTER 1

54


The speed of light in ice and ater are

Since the light propagates in ice and ater at constant speed thus

Therefore the time reuired is gi*en byt

sv=

v

st=

i

ivcn =

1%

i sm1#2/$2 =v

(($1.1$(1 5i == nn

i

%

1###$((1$1v=

55 v

c

n =1%

5 sm1#2"$2 =v

5

%1###$(

(($1 v

=

5i ttt +=

+

=+=

%%

5

5

i

i

1#2"$2

"%$(

1#2/$2

(2$#

v

h

v

ht

s1#&&$1%

=t

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

55

)igure 1!1. shos a spherical surface ith radius+ rforms aninterface beteen to media ith refracti*e indices n

1

and n2

!

The surface forms an image Iof a point ob/ect O$

The incident ray OBma'ing an angle iith the normal and is

refracted to ray BIma'ing an angle here n16 n

2!

Point )is the centre of cur*ature of the spherical surface and

B)is normal!

.(.$ !efraction at a spherical surface

Figure .)Figure .)

+

B

O I)*

1n

v

r

u

2ni

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

56

)rom the figure+

BO)

BI)

)rom the Snell5s la

=y using =


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PHYSICS CHAPTER 1

57

=y substituting e! $1& and $,& into e! $-&+ thus

then

)) 21 =+ nn ) 1221 nnnn =+

=

+

rnn

vn

un

B*)

B*B*1221

r

nn

v

n

u

n ) 1221 =+

here olef-omdistanceimage:volef-omdistanceobject:u

1medimofinde-ef-acti3e:1n-a!)incidentthecontaining4edim

2medimofinde-ef-acti3e:2n-a!)-ef-actedthecontaining4edim

E&uation of sphericalE&uation of spherical

refracting surfacerefracting surface

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

58

Dote "

If the refracting surface is flat "plane#flat "plane#"

then

The euation $formula& of linear magnification for refractionby the spherical surface is gi*en by

#21 =+v

n

u

n=r

un

vn

h

hm

2

1

o

i==

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

59

Table 1!. shos the sign con*ention for refraction or thinrefraction or thin

lenseslenses"

'hysical ?uantity 'ositive sign "4# egative sign "@#

Ob*ect distance,u


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60

A cylindrical glass rod in air has a refracti*e inde of 1!%,!


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PHYSICS CHAPTER 1

61


a! =y using the euation of spherical refracting surface+ thus

;he image is $0. cm at the back of the conve6 surface.;he image is $0. cm at the back of the conve6 surface.

b! The linear magnification of the image is gi*en by

( )r

nn

v

n

u

n agga =+

cm($##cm.#$1#.1$'2g +=== run

un

vn

m g

a

=

cm&$2#+=v( )

##$(

##$1'2$1'2$1

#$1#

##$1

+

=+

v

un

vn

m 2

1

=( )( )

( )( )#$1#2$1

&$2###$1=m

("$1=m PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

62

)igure 1!13 shos an ob/ect < placed at a distance ,4!4 cm from

the surface P of a glass sphere of radius %!4 cm and refracti*e

inde of 1!3-!

8eterminea! the position of the image formed by the surface P of the glass

sphere+

b! the position of the final image formed by the glass sphere!

$#i*en the refracti*e inde of air + na 1!44&

Eample 14 "

Figure .2Figure .2

O

+

cm#$2#

#lass sphere

air

cm#$'

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

63


a! =y using the euation of spherical refracting surface+ thus

;he image is $. cm at the back of the first surface '.;he image is $. cm at the back of the first surface '.


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PHYSICS CHAPTER 1

64


b!

)rom the figure abo*e+ the image I1formed by the first surface P

is in the glass and 11!% cm from the second surface G! I1acts

as a virtual ob*ectvirtual ob*ectfor the second surface and

O

2I cm1$2

+

gnan

)irst surface

1I

an

7

cm1$1

Second surface

cm.'$111$##..1$"( a2g1 ===== unnnncm'$##+=r8entre of curvature is located in8entre of curvature is located in

more dense mediummore dense medium

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

65


b! =y using

;he image is real and (.) cm at the back of the second;he image is real and (.) cm at the back of the second

surface ?.surface ?.

( )r

nn

v

n

u

n gaag =+

cm&0$(+=v

( )

#$'

"($1##$1##$1

'$11

"($1

+

=+

v

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

66

Eercise 1!- "1! A student ishes to determine the depth of a simming pool

filled ith ater by measuring the idth $x %!%4 m& and thennoting that the bottom edge of the pool is /ust *isible at anangle of 1.!4abo*e the hori:ontal as shon in )igure 1!19!

Calculate the depth of the pool!

$#i*en nater

1!-- and nair 1!44&

AS. :AS. : .2 m.2 m

Figure .Figure .


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PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

68


Sketch and useSketch and useray diagrams toray diagrams to determinedeterminethethe

characteristics of image formed by diverging andcharacteristics of image formed by diverging and

converging lenses.converging lenses.

3se3see&uation stated in .( toe&uation stated in .( to derivederivethin lens formula,thin lens formula,

for real ob*ect only.for real ob*ect only.

3se3selensmaker>s e&uation:lensmaker>s e&uation:

3se3sethe thin lens formula for a combination ofthe thin lens formula for a combination of

converging lenses.converging lenses.

Learning Outcome:

1!. Thin lenses $, hours&

!'mp

h!m

atri'

!ed

u!m

y(ph

ysic

s

!'mp

h!m

atri'

!ed

u!m

y(ph

ysic

s

fvu

111=+

( )

+=

21

111

1

rrn

f

PHYSICS CHAPTER 1


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69

.) ;hin lenses is defined as a transparent material with two sphericala transparent material with two spherical

refracting surfaces whose thickness is thin compared torefracting surfaces whose thickness is thin compared tothe radii of curvature of the two refracting surfacesthe radii of curvature of the two refracting surfaces!

There are to types of thin lenses! It is convergingconvergingand

divergingdiverginglenses!

)igures 1!1;a and 1!1;b sho the *arious types of thin lenses+

both con*erging and di*erging!$a& 8onverging "8onve6# lenses8onverging "8onve6# lenses

9iconve69iconve6 'lano@conve6'lano@conve6 8onve6 meniscus8onve6 meniscus

Figure .1aFigure .1a

rr11(+ve)(+ve)

rr22(+ve)(+ve)

rr11(+ve)(+ve)

rr22(( ))

rr11(+ve)(+ve)

rr22(( ve)ve)

PHYSICS CHAPTER 1


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70

.). ;erms of thin lenses )igures 1!1? sho the shape of con*erging $con*e& and

di*erging $conca*e& lenses!

$b& %iverging "8oncave# lenses%iverging "8oncave# lenses

9iconcave9iconcave 'lano@concave'lano@concave 8oncave meniscus8oncave meniscusFigure .1bFigure .1b

$a& Con*erging lens $b& 8i*erging lens

88 88$$

rr11

rr22OO 88 88$$

rr11

rr22OO

Figure .-Figure .-

rr11((ve)ve)

rr22((ve)ve)

rr11((ve)ve)

rr22(( ))

rr11(+ve)(+ve)

rr22(( ve)ve)

PHYSICS CHAPTER 1


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71

8entre of curvature "point 88entre of curvature "point 8and 8and 8

$$##

is defined as the centre of the sphere of which the surfacethe centre of the sphere of which the surface

of the lens is a partof the lens is a part!

!adius of curvature "r!adius of curvature "rand rand r

$$##

is defined as the radius of the sphere of which the surfacethe radius of the sphere of which the surface

of the lens is a partof the lens is a part!

'rincipal "Optical# a6is'rincipal "Optical# a6is is defined as the line *oining the two centres of curvaturethe line *oining the two centres of curvature

of a lensof a lens!

Optical centre "point O#Optical centre "point O#

is defined as the point at which any rays entering the lensthe point at which any rays entering the lens

pass without deviationpass without deviation!

PHYSICS CHAPTER 1


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72

Consider the ray diagrams for con*erging and di*erging lenses

as shon in )igures 1!,4a and 1!,4b!

)rom the figures+ Points )

1and )

,represent the focus of the lenses!

8istancefrepresents the focal length of the lenses!

.).$ Focal point and focal length,f

FF FF$$

OO

ffff

Figure .$0aFigure .$0a Figure .$0bFigure .$0b

FF F

F$$

OO

PHYSICS CHAPTER 1


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73

Focus "point FFocus "point Fand Fand F$$##

)or converging "conve6#converging "conve6#lens > is defined as the point on thethe point on the

principal a6is where rays which are parallel and close to theprincipal a6is where rays which are parallel and close to theprincipal a6is converges after passing through the lensprincipal a6is converges after passing through the lens!

Its focus is real $principal&!

)or diverging "concave#diverging "concave#lens > is defined as the point on thethe point on the

principal a6is where rays which are parallel to the principalprincipal a6is where rays which are parallel to the principal

a6is seem to diverge from after passing through the lensa6is seem to diverge from after passing through the lens! Its focus is *irtual!

Focal length "Focal length "ff## is defined as the distance between the focus F and the opticalthe distance between the focus F and the optical

centre O of the lenscentre O of the lens!

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74

)igures 1!,1a and 1!,1b sho the graphical method of locating

an image formed by a con*erging $con*e& and di*erging$conca*e& lenses!

.).( !ay diagram for thin lenses

Figure .$aFigure .$a

FF

FF$$

$a& Con*erging $con*e& lens

$$

$$

OO

((

((

II

u v

PHYSICS CHAPTER 1


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75

!ay !ay 2 Parallel to the principal ais+ after refraction by the lens+passes through the focal point $focus& ),of a

con*erging lens or appears to come from the focal point),of a di*erging lens!

!ay $!ay $ 2 Passes through the optical centre of the lens isunde*iated!

!ay (!ay ( 2 Passes through the focus )1of a con*erging lens or

appears to con*erge toards the focus )1of a di*erging

lens+ after refraction by the lens the ray parallel to the principalais!

$b& 8i*erging $conca*e& lens

OO FF$$ FF

$$

$$

((

((

II

v

u Figure .$bFigure .$b

At leastAt least

any twoany two

rays forrays for

drawingdrawingthe raythe ray

diagram.diagram.

PHYSICS CHAPTER 1


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76


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77

Ob*ect



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78

Ob*ect



79/99

79

distance, u

OOFF FF$$ $F$F$$$F$F

u = fu = f

Real or *irtual

)ormed at infinity!

FrontFront backback

u < fu < f

0irtual

Bpright

@agnified

)ormed in front

of the lens!

OOFF FF$$ $F$F$$$F$FFrontFront backback

I

;able .;able .

Stimulation 1!%

PHYSICS CHAPTER 1
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;hin lens formula and lens maker>s e&uation;hin lens formula and lens maker>s e&uation Considering the ray diagram of refraction for to spherical

surfaces as shon in )igure 1!,-!

.).) ;hin lens formula, lens maker>s and linear

magnification e&uations

Figure .$(Figure .$(

OO

8888$$II11

II22'' ''$$

EE99

AA %%

1u 1v 2v1r 2r

t

1n

12 vtu =

1n2n


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PHYSICS CHAPTER 1


82/99

82

=y substituting e! $,& into e! $1&+ thus

If u18 and v

28f thus e! $-& becomes

1

12

2

12

2

1

1

1 )

r

nn

r

nn

v

n

u

n =

+

2

12

1

12

2

1

1

1 ))

r

nn

r

nn

v

n

u

n +=+

"(#"(#

+

=+ 2112

21

11

1

11

rrn

n

vu

+

=

211

2 1111

rrn

n

f

Lens maker>sLens maker>s

e&uatione&uation

here lengthfocal:fs-face-ef-acting1fo-c-3at-eof-adis: st1r

medimtheofinde-ef-acti3e:1nmate-iallenstheofinde-ef-acti3e:2n

s-face-ef-acting2fo-c-3at-eof-adis: nd2r

PHYSICS CHAPTER 1


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83

=y euating e! $-& and the lens ma'er5s euation+ thus

therefore in general+

Dote " If the medium is airair$n

18n

ai-81& thus the lens ma'er5s

euation can be ritten as

)or thin lenses and lens ma'er5s euations+ use the signsignconventionconventionfor refractionrefraction!

fvu

111

21

=+

vuf

111 += ;hin lens formula;hin lens formula

here mate-iallenstheofinde-ef-acti3e:n

( )

+= 21

111

1

rrn

f

PHYSICS CHAPTER 1


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84

Linear magnification,Linear magnification, mm

is defined as thetheratio between image height,ratio between image height, hhiiand ob*ectand ob*ect

height,height, hhoo!

Since the linear magnification euation can be

ritten as

u

v

h

hm ==

o

i

here cent-eoticalf-omdistanceimage:v

cent-eoticalf-omdistanceobject:u

vuf

111 +=

vvuf

+=

111

1+=u

v

f

v1=

f

vm

PHYSICS CHAPTER 1


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85

A person of height 1!9% m is standing ,!%4 m in from of a camera!

The camera uses a thin bicon*e lens of radii of cur*ature

9!3? mm! The lens made from the cron glass of refracti*e inde1!%,!

a! Calculate the focal length of the lens!

b! S'etch a labelled ray diagram to sho the formation of the

image!c! 8etermine the position of the image and its height!

d! State the characteristics of the image!


a! =y applying the lens ma'er5s euation in air+ thus

Eample 11 "

.'2$1m.'#$2m.1$&'o === nuhm1#"/$& (21

+==rr

( )

+=

21

111

1

rrn

f

PHYSICS CHAPTER 1


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86


a!

b! The ray diagram for the case is

.'2$1m.'#$2m.1$&'o === nuhm1#"/$& (21

+==rr

( )

+

= (( 1#"/$&

11#"/$&

11'2$11f

m1#(/$& (+=f

FF FF$$ $F$F$$$F$F

FrontFront backback

OO

I

PHYSICS CHAPTER 1


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87


c! The position of the image formed is

=y using the linear magnification euation+ thus

d! The characteristics of the image are

realreal

invertedinverted

diminisheddiminished

formed at the back of the lensformed at the back of the lens

vuf111 +=

m1#01$& (=vv1

'#$21

1#(/$&1 ( +=+

"at the back of the lens#"at the back of the lens#

u

v

h

hm ==o

i

'#$2

1#01$&

&'$1

(i

=h

m1#1/$' (i=h


88/99

PHYSICS CHAPTER 1


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89


a!

b! Since the thin lens is plano2con*e thus

Therefore

1#cm.1#1$"". === mun

=2r

( )

+=

21

111

1

rrn

f

( )

+= 111""$1

1$11

1

1r

cm(($&1 +=r

( )( )1#1#

1

1#

11

+=fcm1$11+=f

PHYSICS CHAPTER 1


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90

The radii of cur*ature of the faces of a thin conca*e meniscus lens

of material of refracti*e inde -(, are ,4 cm and 14 cm! 6hat is

the focal length of lens

a! in air+

b! hen completely immersed in ater of refracti*e inde .(-7


a! =y applying the lens ma'er5s euation in air+

Eample 1- "

29(2=n

( )

+=21

111

1

rrn

f

cm2#1 +=r cm1#2 =r

29(2 ==nnand

PHYSICS CHAPTER 1


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91


a!

b! #i*en

=y using the general lens ma'er5s euation+ therefore

(901=n

+

=

211

2 1111

rrn

n

f

cm1"#=f

cm0#=f

29(2=n

( )

++

= 1#1

2#

1

12

(1

f

( )( ) ( )

++

= 1#

1

2#

11

1

(02

(

f

PHYSICS CHAPTER 1


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92

@any optical instruments+ such as microscopes andmicroscopes and

telescopestelescopes+ use two converging lensestwo converging lensestogether to producean image!

In both instruments+ the 1stlens $closest to the ob*ectclosest to the ob*ect&is calledthe ob*ectiveob*ectiveand the ,ndlens $closest to the eyeclosest to the eye& is referred toas the eyepieceeyepiece orocularocular!

The image formedimage formed by theststlenslensis treatedtreatedas the ob*ect forob*ect for

the $the $ndndlenslensand the final imagefinal imageis the image formed by the $$ndndlenslens!

The position of the final imageposition of the final imagein a to lenses system can bedetermined by applying the thin lens formula to each lensthin lens formula to each lens

separatelyseparately!

The overall magnification of a two lenses systemoverall magnification of a two lenses systemis theproduct of the magnifications of the separate lensesproduct of the magnifications of the separate lenses!

.). 8ombination of lenses

21mmm=here

ionmagnificato3e-all:mlens1thetodeionmagnificat: st1mlens2thetodeionmagnificat: nd

2

m

Picture 1!9

Picture 1!;

PHYSICS CHAPTER 1
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The ob/ecti*e and eyepiece of the compound microscope are both

con*erging lenses and ha*e focal lengths of 1%!4 mm and ,%!%

mm respecti*ely! A distance of 31!4 mm separates the lenses! Themicroscope is being used to eamine a sample placed ,.!1 mm in

front of the ob/ecti*e!

a! 8etermine

i! the position of the final image+

ii! the o*erall magnification of the microscope!

b! State the characteristics of the final image!


Eample 1. "

mm."1$#mm.'$2'mm.#$1' 21 =+=+= dffmm20$11=u

d

1u

1f1f 2f2f

FF FF FF$$ FF$$O

ob*ective "ob*ective "stst ## eyepiece"$eyepiece"$ndnd ##

PHYSICS CHAPTER 1


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94

d

1u

1f1f 2f2f

FF FF FF$$ FF$$O


a! i! =y applying the thin lens formula for the 1stlens $ob/ecti*e&+

mm&$(/1 +=v111

111

vuf+=

mm."1$#mm.'$2'mm.#$1' 21 =+=+= dffmm20$11=u

1

1

1$20

1

#$1'

1

v+=

+"real#"real#

1I

1v 2u12 vdu =

&$(/#$"12 =umm($212=u

PHYSICS CHAPTER 1


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95


a! i! and the position of the final image formed by the ,ndlens

$eyepiece& is

mm."1$#mm.'$2'mm.#$1' 21 =+=+= dffmm20$11=u

222

111

vuf+=

2

1

($21

1

'$2'

1

v+=

+mm12/2 =v

"in front of the $"in front of the $

ndnd

lens#lens#

2I

mm12/2 =v

d

1u

1f1f 2f2f

FF FF FF$$ FF$$O1I

1v 2u

PHYSICS CHAPTER 1


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96


a! ii! The o*erall $total& magnification of the microscope is gi*en by

b! The characteristics of the final image are virtualvirtual

invertedinverted

magnifiedmagnified

formed in front of the formed in front of the ststand $and $ndndlenseslenses!

mm."1$#mm.'$2'mm.#$1' 21 =+=+= dffmm20$11=u

21mmm= here1

11

u

vm =

2

22

u

vm =and

2

2

1

1

u

v

u

vm =

( )

($21

12/

1$20

&$(/ =m /%$/=m

PHYSICS CHAPTER 1


97/99

97

Eercise 1!. "1! a! A glass of refracti*e inde 1!%4 plano2conca*e lens has a

focal length of ,1!% cm! Calculate the radius of the

conca*e surface!

b! A rod of length 1%!4 cm is placed hori:ontally along the

principal ais of a con*erging lens of focal length 14!4 cm! If

the closest end of the rod is ,4!4 cm from the lens calculate

the length of the image formed!

AS. :AS. : 0.1 cm0.1 cm/ 2.00 cm/ 2.00 cm

,! An ob/ect is placed 13!4 cm to the left of a lens! The lens

forms an image hich is -3!4 cm to the right of the lens!

a! Calculate the focal length of the lens and state the type of

the lens!b! If the ob/ect is ;!44 mm tall+ calculate the height of the

image!

c! S'etch a labelled ray diagram for the case abo*e!

AS. :AS. : . cm/ .1 cm. cm/ .1 cm

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-! 6hen a small light bulb is placed on the left side of acon*erging lens+ a sharp image is formed on a screen placed-4!4 cm on the right side of the lens! 6hen the lens is mo*ed

%!4 cm to the right+ the screen has to be mo*ed %!4 cm to theleft so that a sharp image is again formed on the screen! 6hatis the focal length of the lens7

AS. :AS. : 0.0 cm0.0 cm

.! A con*erging lens of focal length ;!44 cm is ,4!4 cm to the left

of a con*erging lens of focal length 3!44 cm! A coin is placed14!4 cm to the left of the 1st lens! Calculate

a! the distance of the final image from the 1st lens+

b! the total magnification of the system!

AS. :AS. : $).2 cm/ 0.-$)$).2 cm/ 0.-$)

%! A con*erging lens ith a focal length of .!4 cm is to the left of asecond identical lens! 6hen a feather is placed 1, cm to theleft of the first lens+ the final image is the same si:e andorientation as the feather itself! Calculate the separationbeteen the lenses!

AS. :AS. : $.0 cm$.0 cm

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