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1

The study of light based onthe assumption that lightlighttravels in straight linestravels in straight linesand is concerned with the

laws controlling thelaws controlling thereflection and refractionreflection and refraction

of rays of lightlight.

CHAPTER 22:CHAPTER 22:Geometrical opticsGeometrical optics

(4 Hours)(4 Hours)

Learning Outcome:

At the end of this chapter, students should be able to:At the end of this chapter, students should be able to: StateState laws of reflection.laws of reflection.

Sketch and useSketch and use ray diagrams to determine theray diagrams to determine thecharacteristics of image formed by spherical mirrors.characteristics of image formed by spherical mirrors.

UseUse

for real object only.for real object only.

UseUse sign convention for focal length:sign convention for focal length:

++ff for concave mirror andfor concave mirror and ff for convex mirror.for convex mirror.

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

rr = 2= 2ff only applies to spherical mirror.only applies to spherical mirror.

2

22.1 Reflection at a spherical surface (1 hour)

rvuf

2111=+=

22.1 Reflection at a spherical surface22.1.1 Reflection 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 two types of reflection due to the plane surface Specular (regular) reflectionSpecular (regular) reflection is the reflection of light fromreflection of light from

a smooth shiny surfacea smooth shiny surface as shown in Figure 22.1.

3

Figure 22.1Figure 22.1

All the reflected rays are parallel to each another or movein the same direction.

Diffuse reflectionDiffuse reflection is the reflection of light from a roughreflection of light from a roughsurfacesurface such as papers, flowers, people as shown in Figure22.2.

The reflected rays is sent out in a variety of directions.

For both types of reflection, the laws of reflection are obeyed.

4


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Laws of reflectionLaws of reflection state :

The incident ray, the reflected ray and the normal all lie inincident ray, the reflected ray and the normal all lie inthe same planethe same plane.

The angle of incidence,angle of incidence, ii equals the angle of reflection,equals the angle of reflection,rr

as shown in Figure 22.3.

5

i r

Plane surfacePlane surface

ri =


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

There are two types of spherical mirror. It is convexconvex (curvingoutwards) and concaveconcave (curving inwards) mirror.

Figures 22.4a and 22.4b show the shape of concave andconvex mirrors.

6

CC

AA

BB

rPPCC

AA

BB

rPP

Figure 22.4aFigure 22.4areflecting surface

imaginary sphere

silver layer

Figure 22.4bFigure 22.4b

(a)Concave (ConvergingConverging)mirror

(b) Convex (DivergingDiverging) mirror

Terms of spherical mirrorTerms of spherical mirror

Centre of curvature (point C)Centre of curvature (point C)

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.

Radius of curvature,Radius of curvature,rr

is defined as the radius of the sphere of which a curvedthe radius of the sphere of which a curvedmirror forms a partmirror forms a part.

Pole or vertex (point P)Pole or vertex (point P)

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

Principal axisPrincipal axis

is defined as the straight line through the centre ofthe straight line through the centre ofcurvature C and pole P of the mirrorcurvature C and pole P of the mirror.

AB is called the apertureaperture of the mirror.

7

22.1.3 Focal point and focal length,f Consider the ray diagram for a concave and convex mirrors as

shown in Figures 22.5a and 22.5b.

Point FF represents the focal pointfocal point or focusfocus of the mirrors.

Distanceffrepresents the focal lengthfocal length of the mirrors.

The parallel incident raysparallel incident rays represent the object infinitely farobject infinitely farawayaway from the spherical mirror e.g. the sun.

8

CCPPCC PP

Figure 22.5aFigure 22.5a

FF f

FFf

IncidentIncidentraysrays


IncidentIncidentraysrays

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Focal point or focus, FFocal point or focus, F

For concave mirror is defined as a point where the incidenta point where the incidentparallel rays converge after reflection on the mirrorparallel rays converge after reflection on the mirror.

Its focal point is real (principal)real (principal).

For convex mirror is defined as a point where the incidenta point where the incidentparallel rays seem to diverge from a point behind the mirrorparallel rays seem to diverge from a point behind the mirrorafter 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) Fand pole P of the spherical mirrorand pole P of the spherical mirror.

The paraxial raysparaxial rays is defined as the rays that are near to andthe rays that are near to and

almost parallel to the principal axisalmost parallel to the principal axis.

9

22.1.4 Relationship between focal length,fand

radius of curvature,r Consider a ray AB parallel to the principal axis of concave

mirror as shown in Figure 22.6.

10


CC

PPFF DD

incident rayincident rayBBAA

f

r

ii

i

From the Figure 22.6,

BCD

BFD

By using an isosceles triangle CBF, thus the angle is given by

then

Because of AB is paraxial ray, thus point B is too close with poleP then

Therefore

11

ii =CD

BDtan

=FD

BDtan

Taken the angles are

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Ray 1Ray 1 - Parallel to principal axis, after reflection, passesthrough the focal point (focus) F of a concave mirroror appears to come from the focal point F of aconvex mirror.

Ray 2Ray 2 - Passes or directed towards focal point F reflectedparallel to principal axis.

Ray 3Ray 3 - Passes or directed towards centre of curvature C,reflected back along the same path.

Images formed by a convex mirrorImages formed by a convex mirror

Figure 22.8 shows the graphical method of locating an imageformed by a convex mirror.

13

At least anyAt least anytwo raystwo raysfor drawingfor drawing

the raythe raydiagram.diagram.

CC

FF

PP

O I

u v

frontfront backbackFigure 22.8Figure 22.8

The characteristics of the image formed are

VirtualVirtual

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.

would not appear on paper, screen or film placed at thelocation of the image.

uprightupright

diminished (smaller than the object)diminished (smaller than the object)

formed at the back of the mirror (behind the mirror)formed at the back of the mirror (behind the mirror)

Object position any positionany position in front of the convex mirror.

Convex mirror always being used as a driving mirrordriving mirror because ithas a wide field of viewwide field of view and providing an upright imageupright image.

Images formed by a concave mirrorImages formed by a concave mirror Concave mirror can be used as a shaving and makeup mirrorsshaving and makeup mirrors

because it provides an upright and virtual imagesupright and virtual images.

Table 22.1 shows the ray diagrams of locating an image formed

by a concave mirror for various object distance, u.14

15

Object

distance, uRay diagram

Imagecharacteristic

I

CC

FrontFront backback

FFPPu > ru > r

u = ru = r

OI

O

Real

Inverted

Diminished

Formedbetween pointC and F.

Real

Inverted

Same size Formed at point

C.

CC

FF

PP

FrontFront backback16

Object


Imagecharacteristic

FFCC PP

FrontFront backback

f < u < rf < u < r

u = fu = f

O

Real

Inverted

Magnified

Formed at adistancegreater thanCP.

Real or virtual

Formed atinfinity.

IO

CC

FF

PP

FrontFront backback

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Linear (lateral) magnification of the spherical mirror, m is definedas the ratio between image height,the ratio between image height,hhii and object height,and object height,hhoo

17

Object


Imagecharacteristic

u < fu < f

O

Virtual

Upright

Magnified Formed at the

back of themirror

IFF

CC PP

FrontFront backback

i

o

h vm

h u= =

wherepolethefromdistanceimage:vpolethefromdistanceobject:u

Table 22.1Table 22.1

22.1.6 Derivation of Spherical mirror equation

Figure 22.9 shows an object O at a distance u and on theprincipal axis of a concave mirror. A ray from the object O is

incident at a point B which is close to the pole P of the mirror.

18


O CC PPIv

u

BB

DD

From the figure,

BOC

BCI

then, eq. (1)(2) :

By usingBOD,BCD andBID

thus

+= (1)(1) += (2)(2)

= 2=+ (3)(3)

ID

BDtan;

CD

BDtan;

OD

BDtan ===

By considering point B very close to the pole P, hence

then

therefore

19

vru === IPID;CPCD;OPOD

tan;tan;tan

v

BD

r

BD

u

BD=== ;;

Substituting thisSubstituting thisvalue in eq. (3)value in eq. (3)

fr 2=

=+

rvu

BD2

BD

BD

rvu

21

1=+ where

rvuf

21

11

=+=

Spherical mirrorsSpherical mirrorsequationequation

Table 22.2 shows the sign convention for spherical mirrorsequation .

Note:

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 projected on the screenprojected on the screen.

20

Physical Quantity Positive sign (+) Negative sign (-)

Object distance, u

Image distance, v

Focal length,f

Real object Virtual object

Real image Virtual image

Concave mirror Convex mirror

(same side of the object) (opposite side of the object)

(in front of the mirror) (at the back of the mirror)


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21

A dentist uses a small mirror attached to a thin rod to examine oneof your teeth. When the tooth is 1.20 cm in front of the mirror, theimage it forms is 9.25 cm behind the mirror. Determine

a. the focal length of the mirror and state the type of the mirrorused,

b. the magnification of the image.

Solution :Solution :

a. By applying the mirrors equation, thus

b. By using the magnification formula, thus

Example 22.1 :

cm38.1+=f

vuf

111+=

vm

u=

cm9.25cm;20.1 =+= vu

( 9.25)7.71

1.20m

= =

( )25.91

20.1

11

+=

f

(Concave mirror)(Concave mirror)

22

An upright image is formed 20.5 cm from the real object by usingthe spherical mirror. The images height is one fourth of objectsheight.

a. Where should the mirror be placed relative to the object?b. Calculate the radius of curvature of the mirror and describe the

type of mirror required.

c. Sketch and label a ray diagram to show the formation of the

image.


Example 22.2 :

oi 25.0 hh =

O Icm0.52

SphericalSpherical

mirrormirror

u v

23


a. From the figure,

By using the equation of linear magnification, thus

By substituting eq. (2) into eq. (1), hence

The mirror should be placed 16.4 cm in front of the object16.4 cm in front of the object.

oi 25.0 hh =

20.5u v+ =

i

o

h vmh u

= =

(1)(1)

o

o

0.25h vh u

=

uv 25.0= (2)(2)

5.2025.0 =+ uucm4.16=u

24


b. By using the mirrors equation, thus

The type of spherical mirror is convexconvex because the negative

value of focal length.

oi 25.0 hh =

cm47.5=f

vuf

111+=

( )uuf 25.0111

+=

( )( )4.1625.01

4.16

11

+=

f

and

2

rf =

( ) cm9.1047.52 ==r

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c. The ray diagram is shown below.oi 25.0 hh =

FF

PP CC

O I

frontfront backback

26

A person of 1.60 m height stands 0.60 m from a surface of ahanging shiny globe in a garden.

a. If the diameter of the globe is 18 cm, where is the image of the

person relative to the surface of the globe?b. How tall is the persons image?

c. State the characteristics of the persons image.


Example 22.3 :

m0.60m;60.1o == uh

u

oh

27


a. GivenThe radius of curvature of the globes surface (convex surface)is given by

By applying the mirrors equation, hence

m18.0=d

m09.0

2

18.0==r

vur

112+=

(behind the globes surface)(behind the globes surface)m042.0=v

m0.60m;60.1o == uh

v

1

60.0

1

09.0

2+=

28


b. By applying the magnification formula, thus

c. The characteristics of the persons image are

virtualvirtual

uprightupright

diminisheddiminished

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

i

o

h vm

h u= =

m0.60m;60.1o == uh

i ( 0.042)

1.60 0.60

h =

m112.0i =h OR cm2.11

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CC FFPP

A shaving or makeup mirror forms an image of a light bulb on awall of a bathroom that is 3.50 m from the mirror. The height of thebulb is 8.0 mm and the height of its image is 40 cm.

a. Sketch a labeled ray diagram to show the formation of the bulbs

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

Example 22.4 :

I

O

cm40

mm0.8

u

m1040m;100.8m;3.50 2i3

o

=== hhv

m50.3 30


b. i. By applying the magnification formula, thus

The position of the bulb is 7.0 cm in front of the mirror.The position of the bulb is 7.0 cm in front of the mirror.ii. By applying the mirrors equation, thus

i

o

h vm

h u= =

2

3

40 10 3.50

8.0 10 u

=

0.07 mu = OR 7.0 cm

m1040m;100.8m;3.50 2i3

o

=== hhv

vuf

111+=

m0687.0=f

50.3

1

07.0

11+=

f

OR cm87.6

31

Exercise 22.1 :1. a. A concave mirror forms an inverted image four times larger

than the object. Calculate the focal length of the mirror,assuming the distance between object and image is0.600 m.

b. A convex mirror forms a virtual image half the size of theobject. Assuming the distance between image and objectis 20.0 cm, determine the radius of curvature of the mirror.

ANS. :ANS. : 160 mm160 mm ; 267 mm; 267 mm

2. a. A 1.74 m tall shopper in a department store is 5.19 m froma security mirror. The shopper notices that his image in themirror appears to be only 16.3 cm tall.

i. Is the shoppers image upright or inverted? Explain.

ii. Determine the radius of curvature of the mirror.

b. A concave mirror of a focal length 36 cm produces animage whose distance from the mirror is one third of theobject distance. Calculate the object and image distances.

ANS. :ANS. : u think, 1.07 mu think, 1.07 m ; 144 cm, 48 cm; 144 cm, 48 cm

Learning Outcome:

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

State and useState and use the laws of refraction (Snells Law) forthe laws of refraction (Snells Law) forlayers of materials with different densities.layers of materials with different densities.

ApplyApply

for spherical surface.for spherical surface.

UseUse sign convention for radius of curvature,sign convention for radius of curvature,rr::

++ veve forfor convex surfaceconvex surface andand veve forfor concave surfaceconcave surface

32

22.2 Refraction at a plane and sphericalsurfaces (1 hour)

1 2 2 1n n n n

u v r

+ =

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Relationship between refractive index and the wavelength ofRelationship between refractive index and the wavelength of

lightlight

As light travels from one medium to another, its wavelength,wavelength, changeschanges but its frequency,frequency, ffremains constantremains constant.

The wavelength changes because of different materialdifferent material. Thefrequency remains constant because the number of wavenumber of wavecycles arriving per unit time must equal the number leavingcycles arriving per unit time must equal the number leavingper unit timeper unit time so that the boundary surface cannot create orcannot create ordestroy wavesdestroy waves.

By considering a light travels from medium 1 (n1) into medium 2

(n2), the velocity of light in each medium is given by

then

37

11 fv = 22 fv =and

2

1

2

1

f

f

v

v= where

1

1n

cv =2

2n

cv =and

If medium 1 is vacuum or air, then n1 = 1. Therefore the

refractive index for any medium, n can be expressed as

38

2

1

2

1

=

nc

n

c

2211 nn =

(Refractive index is inversely(Refractive index is inverselyproportional to the wavelength)proportional to the wavelength)

where

0

=n

in vacuumlightofhwavelengt:0

mediuminlightofhwavelengt:

39

A fifty cent coin is at the bottom of a swimming pool of depth3.00 m. The refractive index of air and water are 1.00 and 1.33respectively. Determine the apparent depth of the coin.


Example 22.5 :

33.1;1.00 wa == nn

where depthapparent:ABm3.00depthactual:AC =

A

i

Air (na)

C

r

B

Water (nw)

i

r

m00.3

D

40


From the diagram,

ABD

ACD

By considering only small angles of rand i , thus

33.1;1.00 wa == nn

AB

ADtan =r

ACADtan =i

andrr sintan ii sintan

AC

AB

ABAD

AC

AD

sin

sin

tan

tan=

==

r

i

r

i

then

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41

Note :Note : (Important)(Important)


From the Snells law,33.1;1.00 wa == nn

w

a

AC

AB

n

n=

w

a

1

2

sin

sin

n

n

n

n

r

i==

33.1

00.1

3.00

AB=

m26.2AB =

depthapparent

depthreal

1

2==

n

nn

Other equation for absolute

refractive index in term ofdepth is given by

42

A pond with a total depth (ice + water) of 4.00 m is covered by atransparent layer of ice of thickness 0.32 m. Determine the timerequired for light to travel vertically from the surface of the ice to

the bottom of the pond. The refractive index of ice and water are1.31 and 1.33 respectively.

(Given the speed of light in vacuum is 3.00 108 m s-1.)


Example 22.6 :

33.1;1.31 wi == nn

Ice (Ice (nnii))

Water (Water (nnww))

BottomBottom

m00.4

m32.0i =h

32.000.4w =h

m68.3w =h

43


The speed of light in ice and water are

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

Therefore the time required is given byt

sv =

v

st =

i

iv

cn =

18

ism1029.2 =v

33.1;1.31 wi == nn

i

81000.331.1

v

=

w

wv

cn =

18

w sm1026.2

=vw

81000.333.1

v

=

wi ttt +=

+

=+=88

w

w

i

i

1026.268.3

1029.232.0

vh

vht

s1077.18

=t

22.2.2 Refraction at a spherical surface Figure 22.11 shows a spherical surface with radius, rforms an

interface between two media with refractive indices n1 and n2.

The surface forms an image I of a point object O.

The incident ray OB making an angle i with the normal and isrefracted to ray BI making an angle where n1< n2.

Point C is the centre of curvature of the spherical surface andBC is normal. 44


P

B

O ICD

1n

v

r

u

2ni

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From the Figure 22.11,

BOC

BIC

From the Snells law

By usingBOD,BCD andBID thus

By considering point B very close to the pole P, hence

then Snells law can be written as

45

+=i (1)

+= = (2)

sinsin 21 nin =

ID

BDtan;

CD

BDtan;

OD

BDtan ===

vru === IPID;CPCD;OPOD

21 nin =

tan;tan;tan;sin;sin ii

(3)

By substituting eq. (1) and (2) into eq. (3), thus

then

46

)()( 21 =+ nn )( 1221 nnnn =+

=

+

rnn

vn

un BD)(BDBD 1221

1 2 2 1n n n n

u v r

+ =

where polefromdistanceimage:v

polefromdistanceobject:u 1mediumofindexrefractive:1nray)incidentthecontainingMedium(

2mediumofindexrefractive:2nray)refractedthecontainingMedium(

Equation of sphericalEquation of sphericalrefracting surfacerefracting surface

Note :

If the refracting surface is flat (plane)flat (plane) :

then

The equation (formula) of linear magnification for refractionby the spherical surface is given by

47

021

=+

v

n

u

n

=r

un

vn

h

hm

o

i

2

1==

Table 22.4 shows the sign convention for refraction or thin lensesrefraction or thin lenses:

48

Physical Quantity Positive sign (+) Negative sign (-)

Object distance, u

Image distance, v

Focal length,f

Real object Virtual object

Real image Virtual image

Converging lens Diverging lens

(same side of theobject)

(opposite side of theobject)

(in front of the refracting

surface)

(at the back of the

refracting surface)

Radius of

curvature,r


convex surface concave surface

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49

A cylindrical glass rod in air has a refractive index of 1.52. One end

is ground to a hemispherical surface with radius, r=3.00 cm asshown in Figure 22.12.

Calculate,

a. the position of the image for a small object on the axis of the rod,

10.0 cm to the left of the pole as shown in figure.b. the linear magnification.

(Given the refractive index of air , na= 1.00)

Example 22.7 :

O ICP

cm0.10

glassair


50


a. By using the equation of spherical refracting surface, thus

The image is 20.7 cm at the back of the convex surface.The image is 20.7 cm at the back of the convex surface.

b. The linear magnification of the image is given by

ga g an n nn

u v r

+ =

cm3.00cm;0.10;1.52g +=== run

cm7.20+=v1.00 1.52 1.52 1.00

10.0 3.00v

+ =

+

36.1=m

un

vnm

2

1=

un

vnm

g

a=

)0.10)(52.1(

)7.20)(00.1(=m

51

Figure 22.13 shows an object O placed at a distance 20.0 cm fromthe surface P of a glass sphere of radius 5.0 cm and refractiveindex of 1.63.

Determine

a. 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.

(Given the refractive index of air , na= 1.00)

Example 22.8 :


O

P

cm0.20

Glass sphere

air

cm0.5

52


a. By using the equation of spherical refracting surface, thus

The image is 21.5 cm at the back of the first surface P.The image is 21.5 cm at the back of the first surface P.

OR

ga g an n nn

u v r

+ =

cm.05cm;0.20;1.63g +=== run

cm5.21+=v1.00 1.63 1.63 1.0020.0 5.0v

+ =

+

O C 1I

cm020u .= cm5.21=v

P

gnan

r

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53


b.

From the figure above, the image I1 formed by the first surface P

is in the glass and 11.5 cm from the second surface Q. I1 acts

as a virtual objectvirtual object for the second surface and

O C2

Icm1.52

P

gnan

First surface

1I

an

Q

cm1.51

Second surface

cm;5.111.00;;1.63 a2g1 ===== unnnn

cm5.00=r

convex surfaceconvex surface54


b. By using

The image is real and 3.74 cm at the back of the secondThe image is real and 3.74 cm at the back of the second

surface Q.surface Q.

g a a gn n nn

u v r

+ =

cm74.3+=v

0.5

63.100.100.1

5.11

63.1

=+

v

55

Exercise 22.2 :1. A student wishes to determine the depth of a swimming pool

filled with water by measuring the width (x = 5.50 m) and thennoting that the bottom edge of the pool is just visible at anangle of 14.0 above the horizontal as shown in Figure 22.14.

Calculate the depth of the pool.

(Given nwater = 1.33 and nair = 1.00)ANS. :ANS. : 5.16 m5.16 m


56

2. A small strip of paper is pasted on one side of a glass sphereof radius 5 cm. The paper is then view from the oppositesurface of the sphere. Determine the position of the image.

(Given the refractive index of glass =1.52 and the refractive

index of air =1.00)ANS. :ANS. : 20.83 cm in front of the 220.83 cm in front of the 2ndnd refracting surface.refracting surface.

3. A point source of light is placed at a distance of 25.0 cm fromthe centre of a glass sphere of radius 10 cm. Determine theimage position of the source.

(Given the refractive index of glass =1.52 and the refractiveindex of air =1.00)

ANS. :ANS. : 25.2 cm at the back of the 225.2 cm at the back of the 2ndnd refracting surface.refracting surface.

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Learning Outcome:


Sketch and useSketch and use ray diagrams toray diagrams to determinedetermine thethecharacteristics of image formed by diverging andcharacteristics of image formed by diverging andconverging lenses.converging lenses.

UseUse thin lens equation,thin lens equation,

for real object only.for real object only.

UseUse sign convention for focal length,sign convention for focal length,ff::

++ff forfor converging lensconverging lens andand ff forfor diverging lensdiverging lens..

57

22.3 Thin lenses (2 hours)

fvu

111=+

Learning Outcome:


UseUse lensmakers equation:lensmakers equation:

if the medium is air.if the medium is air.

UseUse the thin lens formula for a combination ofthe thin lens formula for a combination ofconverging lenses.converging lenses.

58

22.3 Thin lenses (2 hours)

( )1 2

1 1 11n

f r r

=

material

medium 1 2

1 1 11

n

f n r r

=

22.3 Thin lenses is defined as a transparent material with two sphericala transparent material with two spherical

refracting surfaces whose thickness is thin compared to therefracting surfaces whose thickness is thin compared to theradii of curvature of the two refracting surfacesradii of curvature of the two refracting surfaces.

There are two types of thin lenses. It is convergingconverging anddivergingdiverging lenses.

Figures 22.15a and 22.15b show the various types of thinlenses, both converging and diverging.

59

(a) Converging (Convex) lensesConverging (Convex) lenses

BiconvexBiconvex PlanoPlano--convexconvex Convex meniscusConvex meniscus


22.3.1 Terms of thin lenses Figures 22.16 show the shape of converging (convex) and

diverging (concave) lenses.

60

(b) Diverging (Concave) lensesDiverging (Concave) lenses

BiconcaveBiconcave PlanoPlano--concaveconcave Concave meniscusConcave meniscusFigure 22.15bFigure 22.15b

(a) Converging lens (b) Diverging lens

CC11 CC22

rr11

rr22OO CC11 CC22

rr11

rr22OO


rr22(())

rr11(+ve)(+ve)

rr22((veve))

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Focus (point FFocus (point F11 and Fand F22))

For converging (convex)converging (convex) lens is defined as the point on thethe point on theprincipal axis where rays which are parallel and close to theprincipal axis where rays which are parallel and close to theprincipal axis converges after passing through the lensprincipal axis converges after passing through the lens.

Its focus is real (principal).

For diverging (concave)diverging (concave) lens is defined as the point on thethe point on theprincipal axis where rays which are parallel to the principalprincipal axis where rays which are parallel to the principalaxis seem to diverge from after passing through the lensaxis seem to diverge from after passing through the lens.

Its focus is virtual.

Focal length (Focal length (ff))

is defined as the distance between the focus F and thethe distance between the focus F and theoptical centre O of the lensoptical centre O of the lens.

65

22.3.3 Ray diagram for thin lenses

Figures 22.18a and 22.18b show the graphical method oflocating an image formed by a converging (convex) anddiverging (concave) lenses.

66


FF11

FF22

(a) Converging (convex) lens

11

11

22

22

OO

33

33

II

u v

Ray 1Ray 1 - Parallel to the principal axis, after refraction by the lens,passes through the focal point (focus) F2 of aconverging lens or appears to come from the focal pointF2 of a diverging lens.

Ray 2Ray 2 - Passes through the optical centre of the lens isundeviated.

Ray 3Ray 3 - Passes through the focus F1 of a converging lens orappears to converge towards the focus F1 of a diverginglens, after refraction by the lens the ray parallel to theprincipal axis. 67

(b) Diverging (concave) lens

OO FF22 FF11

11

11

22

22

33

33

II

vu


At leastAt leastany twoany tworays forrays fordrawingdrawingthe raythe ray

diagram.diagram.

Images formed by a diverging lensImages formed by a diverging lens

Figure 22.19 shows the graphical method of locating an imageformed by a diverging lens.

The characteristics of the image formed are

virtualvirtual

uprightupright

diminished (smaller than the object)diminished (smaller than the object) formed in front of the lensformed in front of the lens.

Object position any positionany position in front of the diverging lens.68

FrontFront backback

OO FF22 FF11II


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Images formed by a converging lensImages formed by a converging lens

Table 22.5 shows the ray diagrams of locating an image formed

by a converging lens for various object distance, u.

69

Objectdistance, u Ray diagram Imagecharacteristic

FF11 FF22 2F2F222F2F11

FrontFront backback

u >u > 22ff

Real

Inverted

Diminished

Formed betweenpoint F2 and 2F2.

(at the back of

the lens)

OOI

70

Object


Imagecharacteristic

OOFF11 FF22

2F2F22

2F2F11u =u = 22ff

Real

Inverted Same size

Formed at point2F2. (at the backof the lens)FrontFront backback

I

f < u

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By using the equation of spherical refracting surface, therefraction by first surface AB and second surface DE are givenby

Surface AB ((rr == +r+r11))

Surface DE ((rr == --rr22))

Assuming the lens is very thin thusvery thin thustt = 0= 0,

73

1

12

1

2

1

1 )(r

nnvn

un

+

=+

( ) 2

21

2

1

1

2 )(

r

nn

v

n

vt

n

=+

(1)(1)

=

2

21

2

1

1

2

r

nn

v

n

v

n2

21

2

1

1

2 )(

r

nn

v

n

v

n

=+

(2)(2)

By substituting eq. (2) into eq. (1), thus

If u1= and v2= f thus eq. (3) becomes

74

1

12

2

21

2

1

1

1 )(

r

nn

r

nn

v

n

u

n =

+

2

21

1

12

2

1

1

1 )()(

r

nn

r

nn

v

n

u

n

+

=+

(3)(3)2

1 2 1 1 2

1 1 1 11

n

u v n r r

+ =

2

1 1 2

1 1 11

n

f n r r

=

Lens makersLens makersequationequation

where lengthfocal:fsurfacerefracting1forcurvatureofradius:

st

1r

mediumtheofindexrefractive:1nmateriallenstheofindexrefractive:2n

surfacerefracting2forcurvatureofradius: nd2r

By equating eq. (3) and the lens makers equation, thus

therefore in general,

Note :

If the medium is airair (n1=nair=1) thus the lens makersequation can be written as

For thin lenses and lens makers equations, use the signsignconventionconvention for refractionrefraction.

75

fvu

111

21

=+

vuf

111+=

Thin lens formulaThin lens formula

where materiallenstheofindexrefractive:n

( )1 2

1 1 11n

f r r

=

vuf

111+=

76

Linear magnification,Linear magnification,mm

is defined as thethe ratio between image height,ratio between image height,hhii and objectand object

height,height,hhoo.

Since the linear magnification equation can be

written as

i

o

h vm

h u= =

where centreopticalfromdistanceimage:vcentreopticalfromdistanceobject:u

vvuf

+=

111

1+=uv

fv 1=

fvm

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( )1 2

1 1 11n

f r r

=

77

A person of height 1.75 m is standing 2.50 m in from of a camera.The camera uses a thin biconvex lens of radii of curvature7.69 mm. The lens made from the crown glass of refractive index1.52.

a. Calculate the focal length of the lens.

b. Sketch a labelled ray diagram to show the formation of the

image.

c. Determine the position of the image and its height.

d. State the characteristics of the image.


a. By applying the lens makers equation in air, thus

Example 22.9 :

;52.1m;50.2m;1.75o === nuhm1069.7 321

+==rr

78


a.

b. The ray diagram for the case is

;52.1m;50.2m;1.75o === nuhm1069.7 321

+==rr

( ) 3 31 1 1

1.52 1

7.69 10 ( 7.69 10 )f

=

m1039.7 3+=f

FF11 FF22 2F2F222F2F11

FrontFront backback

OOI

79


c. The position of the image formed is

By using the linear magnification equation, thus

d. The characteristics of the image are

realreal

Inverted(Inverted(--ve)ve)

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

vuf

111+=

m1041.73

=vv

1

50.2

1

1039.7

13

+=

+

(at the back of the lens)(at the back of the lens)

i

o

h vm

h u= =

3

i 7.41 10

1.75 2.50

h =

3

i 5.19 10 mh

= OR mm19.5 uv 10=

80

A thin plano-convex lens is made of glass of refractive index 1.66.When an object is set up 10 cm from the lens, a virtual image tentimes its size is formed. Determine

a. the focal length of the lens,

b. the radius of curvature of the convex surface.


a. By applying the linear magnification equation for thin lens, thus

By using the thin lens formula, thus

Example 22.10 :

10cm;101.66; === mun

( )10

v vm

u u

= = =

vuf

111+=

( )uuf 10111

+=

Virtual imageVirtual image

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81


a.

b. Since the thin lens is plano-convex thus

Therefore

10cm;101.66; === mun

=2r

( )1 2

1 1 11n

f r r

=

( )1

1 1 11.66 1

11.1 r

=

cm33.71 +=r

( )( )10101

10

11

+=

f

cm1.11+=f

( )1 2

1 1 11nf r r

=

82

The radii of curvature of the faces of a thin concave meniscus lensof material of refractive index 3/2 are 20 cm and 10 cm. What is thefocal length of lens

a. in air,b. when completely immersed in water of refractive index 4/3?


a. By applying the lens makers equation in air,

Example 22.11 :

2/32 =n

cm201 +=r cm102 +=r

2/32 ==nnand

83


a.

b. Given

By using the general lens makers equation, therefore3/41 =n

2

1 1 2

1 1 11

n

f n r r

=

cm160=f

cm40=f

2/32 =n

1 3 1 11

2 20 10f

=

+

( )

( )

32

43

1 1 11

20 10f

= +

22.3.5 Combination of lenses Many optical instruments, such as microscopes andmicroscopes and

telescopestelescopes, use two converging lensestwo converging lenses together to producean image.

In both instruments, the 1st lens (closest to the objectclosest to the object )is calledthe objectiveobjective and the 2nd lens (closest to the eyeclosest to the eye) is referred to

as the eyepieceeyepiece or ocularocular. The image formedimage formed by the 11stst lenslens is treatedtreated as the object forobject for

the 2the 2ndnd lenslens and the final imagefinal image is the image formed by the 22ndnd

lenslens.

The position of the final imageposition of the final image in a two lenses system can bedetermined by applying the thin lens formula to each lensthin lens formula to each lensseparatelyseparately.

The overall magnification of a two lenses systemoverall magnification of a two lenses system is theproduct of the magnifications of the separate lensesproduct of the magnifications of the separate lenses.

84

21mmm =where ionmagnificatoverall:m

lens1thetodueionmagnificat: st1mlens2thetodueionmagnificat: nd2m

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85

The objective and eyepiece of the compound microscope are bothconverging lenses and have focal lengths of 15.0 mm and 25.5 mmrespectively. A distance of 61.0 mm separates the lenses. Themicroscope is being used to examine a sample placed 24.1 mm infront of the objective.

a. Determine

i. the position of the final image,

ii. the overall magnification of the microscope.

b. State the characteristics of the final image.


Example 22.12 :

mm;61.0mm;5.25mm;0.15 21 =+=+= dffmm24.11 =u

d

1u

1f

1f

2f2f

FF11 FF11 FF22 FF22O

objective (1objective (1stst)) eyepiece(2eyepiece(2ndnd)) 86

d

1u

1f

1f

2f2f

FF11 FF11 FF22 FF22O


a. i. By applying the thin lens formula for the 1st lens (objective),

mm7.391 +=v111

111

vuf+=

mm;61.0mm;5.25mm;0.15 21 =+=+= dffmm24.11 =u

1

1

1.24

1

0.15

1

v+=

+

(real)(real)

1I

1v

2u

12 vdu =7.390.612 =u

mm3.212 =u

87


a. i. and the position of the final image formed by the 2 nd lens

(eyepiece) is

mm;61.0mm;5.25mm;0.15 21 =+=+= dffmm24.11 =u

222

111

vuf+=

2

1

3.21

1

5.25

1

v+=

+

mm1292 =v(in front of the 2(in front of the 2ndnd lens)lens)

2I

mm1292 =v

d

1u

1f

1f

2f

2f

FF11 FF11 FF22 FF22O1

I

1v

2u

88


a. ii. The overall (total) magnification of the microscope is given by

b. The characteristics of the final image are

Virtual (m=Virtual (m=--ve)ve)

invertedinverted magnifiedmagnified

formed in front of the 1formed in front of the 1stst and 2and 2ndnd lenseslenses.

mm;61.0mm;5.25mm;0.15 21 =+=+= dffmm24.11 =u

21mmm = where1

1

1

vm

u=

22

2

vm

u= and

1 2

1 2

v vmu u

= 39.7 129

24.1 21.3m

=

9.98m =

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89

Exercise 22.3 :1. a. A glass of refractive index 1.50 plano-concave lens has a

focal length of 21.5 cm. Calculate the radius of theconcave surface.

b. A rod of length 15.0 cm is placed horizontally along theprincipal axis of a converging lens of focal length 10.0 cm.If the closest end of the rod is 20.0 cm from the lenscalculate the length of the image formed.

ANS. :ANS. : 10.8 cm10.8 cm; 6.00 cm; 6.00 cm

2. An object is placed 16.0 cm to the left of a lens. The lensforms an image which is 36.0 cm to the right of the lens.

a. Calculate the focal length of the lens and state the type ofthe lens.

b. If the object is 8.00 mm tall, calculate the height of theimage.

c. Sketch a labelled ray diagram for the case above.

ANS. :ANS. : 11.1 cm; 1.8 cm11.1 cm; 1.8 cm90

3. When a small light bulb is placed on the left side of aconverging lens, a sharp image is formed on a screen placed30.0 cm on the right side of the lens. When the lens is moved5.0 cm to the right, the screen has to be moved 5.0 cm to theleft so that a sharp image is again formed on the screen. What

is the focal length of the lens?ANS. :ANS. : 10.0 cm10.0 cm

4. A converging lens of focal length 8.00 cm is 20.0 cm to the leftof a converging lens of focal length 6.00 cm. A coin is placed10.0 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.

ANS. :ANS. : 24.6 cm; 0.92424.6 cm; 0.924

5. A converging lens with a focal length of 4.0 cm is to the left ofa second identical lens. When a feather is placed 12 cm to theleft of the first lens, the final image is the same size andorientation as the feather itself. Calculate the separationbetween the lenses.

ANS. :ANS. : 12.0 cm12.0 cm

91

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