pdt unit9 (students)

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UNIT 9:Geometrical OpticsUNIT 9:Geometrical Optics

The study of light based on The study of light based on

the assumption that light the assumption that light

travels in straight lines and travels in straight lines and

is concerned with the laws is concerned with the laws

controlling the reflection controlling the reflection

and refraction of rays of and refraction of rays of

light.light.

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Geometrical Optics

Reflection

Plane surface

Refraction

Spherical surface

Plane surface

- Law of reflection

- Ray diagram (plane mirror)

- Characteristic of image

- Term in used: principal axis. Centre of curvature( c ),radius of curvature( r ), focal point( F ),focal length( f ),pole of the spherical mirror

- Ray diagram (spherical mirror)


- Sign convention( u, v , f )

- Snell’s Law

- Refractive index( n )

- Material of different n

- Convex and concave lenses

- Term in used: principal axis. focal point( F ),focal length( f ),optical centre of lens

- Ray diagram (lenses)


- Sign convention( u, v , f )

- Thin lens formula:

Thin lenses

f

1

v

1

u

1=+

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9.1 Reflection of Plane Mirror 9.1.1 Reflection of Plane Mirror

� Definition – is defined as the return of all or part of a beam of particles or waves when it encounters the boundary between twomedia.

� Laws of reflection state :

� The incident ray, the reflected ray and the normal all lie in the same plane.

� The angle of incidence, i is equal to the angle of reflection, r as shown in figure below.

i r

Plane mirrorPlane mirror

ri =

Simulation

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A 'A

i

u v

i

r

i

� Image formation by a plane mirror. (ray diagrams)

� Point object

� Vertical (extended) object

Objecti

v

i

rr

u

Image

ihoh

distanceobject :uwhere

distance image :vheightobject :ohheight image :ih

Simulation

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� The properties of image formed are

� virtual

� upright or erect

� laterally reverse

� the object distance, u is equal to the image distance, v� same size as the object where the linear magnification is

given by

� obey the laws of reflection.

� Example 1 :

Find the minimum vertical length of a plane mirror for an observer of

2.0 m height standing upright close to the mirror to see his whole

reflection. How should this minimum length mirror be placed on the

wall?

1h

hM

o

i ==height,Object

height, Image

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Solution: By using the ray diagram as shown in figure below.

HE2

1AL =

EF2

1LB =

The minimum vertical length of the mirror is given by

LBALh +=

EF2

1HE

2

1h +=

( )EFHE2

1h +=

Height of observer

m01h .=

)head(H A)eyes(E

)feet(F

B

L

h

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The mirror can be placed on the wall with the lower end of the mirror

is halved of the distance between the eyes and feet of the observer.

� Example 2 :

A rose in a vase is placed 0.250 m in front of a plane mirror. Nagar

looks into the mirror from 2.00 m in front of it. How far away from

Nagar is the image of the rose?

Solution: u=0.250 m

u v

m002 .

x

From the properties of the

image formed by the plane

mirror, thus

Therefore, the distance between Nagar and the image of the rose is

given by

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9.2 Reflection at a Spherical surface9.2.1 Spherical mirror

� Definition – is defined as a reflecting surface that is part of a sphere.

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

� Figures below show the shape of concave and convex mirrors.

� Some terms of spherical mirror

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

� is defined as the centre of the sphere of which a curved mirror forms a part.

(a) Concave (ConvergingConverging) mirror (b) Convex (DivergingDiverging) mirror

reflecting surface

imaginary sphere

CC CC

AA

BB

AA

BB

silver layer

r rPP PP

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�� Radius of curvature, Radius of curvature, rr� is defined as the radius of the sphere of which a curved

mirror forms a part.

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

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

�� Principal axisPrincipal axis

� is defined as the straight line through the centre of curvature C and pole P of the mirror.

� AB is called the aperture aperture of the mirror.

9.2.2 Focal point and focal length, f� Consider the ray diagram for concave and convex mirror as shown in

figures below.

CC

FFf

Incident Incident

raysrays

PP CC

FFf

Incident Incident

raysrays

PP

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� From the figures,

� Point F represents the focal point or focus of the mirrors.

� Distance f represents the focal length of the mirrors.� The parallel incident rays represent the object infinitely far away

from the spherical mirror e.g. the sun.

�� Focal point or focus, FFocal point or focus, F

� for concave mirror – is defined as a point where the incident parallel rays converge after reflection on the mirror.

� Its focal point is real (principal).

� for convex mirror – is defined as a point where the incident parallel rays seem to diverge from a point behind the mirror after reflection.

� Its focal point is virtual.

�� Focal length, Focal length, ff� Definition – is defined as the distance between the focal point

(focus) F and pole P of the spherical mirror.

� The paraxial raysparaxial rays is defined as the rays that are near to and almost parallel to the principal axis.

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9.2.3 Relationship between focal length, f and radius of curvature, r� Consider a ray AB parallel to the principal axis of concave mirror as

shown in figure below.

� From the figure,

�BCD

�BFD

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

CC

FF

incident rayincident ray

PPDD

BBAA

fr

iiθ

i

iCD

BDi ≈=tan

θθ ≈=FD

BDtan

Taken the angles are << Taken the angles are <<

small by considering the ray small by considering the ray

AB is paraxial ray.AB is paraxial ray.

i2=θ

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then

� Because of AB is paraxial ray, thus point B is too close with pole P then

� Therefore

rCPCD =≈fFPFD =≈

This relationship also valid This relationship also valid

for convex mirror.for convex mirror.

2

rf =

=CD

BD2

FD

BD

or

FD2CD =

f2r =

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9.2.4 Ray Diagrams for Spherical Mirrors

� Definition – is defined as the simple graphical method to indicate the positions of the object and image in a system of mirrors or lenses.

� Ray diagrams below showing the graphical method of locating an image formed by concave and convex mirror.

�� Ray 1Ray 1 - Parallel to principal axis, after reflection, passes through the focal point (focus) F of a concave mirror or appears to come from the focal point F of a convex mirror.

�� Ray 2Ray 2 - Passes or directed towards focal point F reflected parallel to principal axis.

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

(a) Concave mirror (b) Convex mirror

CC PP

FF

11

33

33

11

I CC

FF

PP

11

22

22O O I

22

33

11

22

At least any At least any

two rays two rays

for drawing for drawing

the ray the ray

diagram.diagram.

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9.2.5 Images formed by a convex mirror

� Ray diagrams below showing the graphical method of locating an image formed by a convex mirror.

� Properties of image formed are

� virtual

� upright

� diminished (smaller than the object)

� formed at the back of the mirror

� Object position → any position in front of the convex mirror.

CC

FF

PP

O I

u v

FrontFront backback

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Image propertyRay diagramObject

distance, u

I

9.2.6 Images formed by a concave mirror

� Table below shows the ray diagrams of locating an image formed by a

concave mirror for various object distance, u.

CC

FrontFront backback

FFPP

u > ru > r

u = ru = r

OI

O

� Real

� Inverted

� Diminished

� Formed between point C and F.

� Real

� Inverted

� Same size

� Formed at point C.CC

FF

PP

FrontFront backback

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distance, u

FFCC PP

FrontFront backback

f < u < rf < u < r

u = fu = f

O

� Real

� Inverted

� Magnified

� Formed at a distance greater than CP.

� Real

� Formed at infinity.

IO

CC

FF

PP

FrontFront backback

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distance, u

� Linear (lateral) magnification of the spherical mirror, M is defined as the ratio

between image height, hi and object height, ho

Negative sign indicates that the object and image are on opposite sides of the

principal axis (refer to the real image), If ho is positive, hi is negative.

u < fu < f

O

� Virtual

� Upright

� Magnified

� Formed at the back of the mirror

IFF

CC PP

FrontFront backback

i

o

h vM

h u= =

where

pole from distance image :vpole from distanceobject :u

Simulation

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O CC PPIv

u

BB

θθ

α φ βDD

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

then

� From the figure,

�BOC

�BCI

then, eq. (1)-(2) :

By using �BOD, �BCD and �BID thus

9.2.7 Derivation of Spherical mirror equation

� Figure below shows an object O at a distance u and on the principal

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.

θαφ += (1)(1)

θφβ += (2)(2)

φαβφ −=−φβα 2=+ (3)(3)

ID

BD

CD

BD

OD

BD=== βφα tan; tan ; tan

vIPIDrCPCDuOPOD =≈=≈=≈ ; ;

v

BD

r

BD

u

BD=== βφα ; ;

ββφφαα ≈≈≈ tan; tan ; tan

Substituting this Substituting this

value in eq. (3)value in eq. (3)

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therefore

� Table below shows the sign convention for equation of spherical mirror .

f2r =

=+r

BD2

v

BD

u

BD

r

2

v

1

u

1=+ where

v

1

u

1

f

1 += Equation (formula) Equation (formula)

of spherical mirrorof spherical mirror

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

Object distance, u

Image distance, v

Focal length, f

Linear

magnification, M

Real object Virtual object

Real image Virtual image

Concave mirror Convex mirror

Upright (erect)

imageInverted image

(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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� Example 7 :

An object is placed 10 cm in front of a concave mirror whose focal length is 15 cm. Determine

a. the position of the image.

b. the linear magnification and state the properties of the image.

Solution: u=+10 cm, f=+15 cm

a. By applying the equation of spherical mirror, thus

The image is 30 cm from the mirror on the opposite side of the

object (or 30 cm at the back of the mirror).

b. The linear magnification is given by

cm30v −=

3M =

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� Example 8 :

An upright image is formed 30 cm from the real object by using the spherical mirror. The height of image is twice the height of object.

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.

Solution: hi=2ho

a. From the figure above,

By using the equation of linear magnification, thus

O Icm30

Spherical Spherical

mirrormirror

u v

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

b. By using the equation of spherical mirror,

and therefore

The type of spherical mirror is concaveconcave because the positive value

of focal length.

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� Example 9 :

A mirror on the passenger side of your car is convex and has a radius of curvature 20.0 cm. Another car is seen in this side mirror and is 11.0 m behind the mirror. If this car is 1.5 m tall, calculate the height of the car image . (Similar to No. 34.66, pg. 1333, University Physics with Modern Physics,11th edition, Young & Freedman.)

Solution: ho=1.5x102 cm, r=-20.0 cm, u=+11.0x102 cm

By applying the equation of spherical mirror,

From equation of linear magnification,

cm351hi .=

and

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I

O

CC FF

PP

cm035 .

mm05 .

m203 .

u

� Example 10 :

A concave mirror forms an image on a wall 3.20 m from the mirror of the filament of a headlight lamp. If the height of the filament is 5.0 mm and the height of its image is 35.0 cm, calculate

a. the position of the filament from the pole of the mirror.

b. the radius of curvature of the mirror.

Solution: hi=-35.0 cm, v=3.20x102 cm, ho=0.5 cm

a. By applying the equation of linear magnification,

cm574u .=cm019r .=

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b. By applying the equation of spherical mirror, thus

� Example 11 : (exercise)

a. A concave mirror forms an inverted image four times larger than the object. Find the focal length of the mirror, assuming the distance between object and image is 0.600 m.

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

No. 14, pg. 1169,Physics for scientists and engineers with modern physics,

Serway & Jewett,6th edition.

Ans. : 160 mm, -267 mm

and

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9.3 Refraction on plane surface� Definition – is defined as the changing of direction of a light ray and its

speed of propagation as it passes from one medium into another.

� Laws of refraction state :

� The incident ray, the refracted ray and the normal all lie in the same plane.

� For two given media,

incidence of angle :iwhere

refraction of angle :r1 medium theofindex refractive:1nray)incident thecontaining Medium(

constantsin

sin==

1

2

n

n

r

i

rnin 21 sinsin =

Or

2 medium theofindex refractive:2nray) refracted thecontaining Medium(

SnellSnell’’s laws law

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� Examples for refraction of light ray travels from one medium to another medium can be shown in figures below.

21 nn <(a)

ir >

21 nn >(b)

1n

2n

i

r

Incident ray

Refracted ray

1n

2n

i

r

Incident ray

Refracted ray

(Medium 1 is less (Medium 1 is less

dense than medium 2)dense than medium 2)(Medium 1 is denser (Medium 1 is denser

than medium 2)than medium 2)

ir <

The light ray is bent toward the

normal, thus

The light ray is bent away from the

normal, thus

Simulation-1 Simulation-2

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� Refractive index (index of refractionindex of refraction)

� Definition – is defined as the constant ratio for the two given media.

� The value of refractive index depends on the type of medium and the colour of the light.

� It is dimensionless and its value greater than 1.

� Consider the light ray travels from medium 1 into medium 2, the refractive index can be denoted by

� Absolute refractive index, n (for the incident ray is travelling in

vacuumvacuum or airor air and is then refracted into the medium concernedmedium concerned) is given by

r

i

sin

sin

2

121

v

vn ==

2 mediumin light ofvelocity

1 mediumin light ofvelocity

(Medium containing (Medium containing

the incident ray)the incident ray)

(Medium containing (Medium containing

the refracted ray)the refracted ray)

v

cn ==

mediumin light ofvelocity

in vacuumlight ofvelocity

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� Table below shows the indices of refraction for yellow sodium light having a wavelength of 589 nm in vacuum.

(If the density of medium is greater hence the refractive index (If the density of medium is greater hence the refractive index is also greater)is also greater)

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� The relationship between refractive index and the wavelength of light.

� As light travels from one medium to another, its wavelength, wavelength, λλλλλλλλchangeschanges but its frequency, frequency, ff remains constantremains constant.

� The wavelength changes because of different materialdifferent material. The frequency remains constant because the number of wave cycles arriving per unit time must equal the number leaving per unit time so that the boundary surface cannot create or destroy wavescannot create or destroy waves.

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

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

then11 fv λ=

22 fv λ=and

2

1

2

1

f

f

v

v

λλ

= where

1

1n

cv =

2

2n

cv =and

2

1

2

1

n

c

n

c

λλ

=

2211 nn λλ =

(Refractive index is inversely (Refractive index is inversely

proportional to the wavelength)proportional to the wavelength)

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� If medium 1 is vacuum or air, then n1 = 1. Hence the refractive

index for any medium, n can be expressed as

� Example 3 :

A fifty cent coin is at the bottom of a swimming pool of depth 2.00 m.

The refractive index of air and water are 1.00 and 1.33 respectively.

What is the apparent depth of the coin?

Solution: na=1.00, nw=1.33

where

λλ0n = in vacuumlight ofh wavelengt:0λ

mediumin light ofh wavelengt:λ

wheredepthapparent :AB

m 2.00 depth actual : =AC

A

i

Air (na)

C

r

B

Water (nw)

i

r

m002 .

D

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From the diagram,

�ABD

�ACD

By considering only small angles of r and i , hence

From the Snell’s law,

and

AB

ADr =tan

w

a

n

n

AC

AB=

m501AB .=

rr sintan ≈

AC

ADi =tan

depthapparent

depth real=n

ii sintan ≈

AC

AB

AB

AD

AC

AD

r

i

r

i=

==sin

sin

tan

tan

w

a

1

2

n

n

n

n

r

i==

sin

sin

then

Note : Note : (Important)(Important)

Other equation for absolute

refractive index in term of

depth is given by

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� Example 4 :

A light beam travels at 1.94 x 108 m s-1 in quartz. The wavelength of

the light in quartz is 355 nm.

a. Find the index of refraction of quartz at this wavelength.

b. If this same light travels through air, what is its wavelength there?

(Given the speed of light in vacuum, c = 3.00 x 108 m s-1)

No. 33.3, pg. 1278, University Physics with Modern Physics,11th edition, Young &

Freedman.

Solution: v=1.94 x 108 m s-1, λ=355 x 10-9 ma. By applying the equation of absolute refractive index, hence

b. By using the equation below, thus

551n .=

nm550m10x505 7

0 @. −=λ

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We wish to determine the depth of a swimming pool filled with water by measuring the width (x = 5.50 m) and then noting that the bottom edge of the pool is just visible at an angle of 14.0° above the horizontal as shown in figure below. (Gc.835.60)

Calculate the depth of the pool. (Given nwater = 1.33 and nair = 1.00)

Ans. : 5.16 m


A person whose eyes are 1.54 m above the floor stands 2.30 m in front of a vertical plane mirror whose bottom edge is 40 cm above the floor as shown in figure below. (Gc.832.10)

Find x.

Ans. : 0.81 m

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9.4 Thin Lenses� Definition – is defined as a transparent material with two spherical

refracting surfaces whose thickness is thin compared to the radii of curvature of the two refracting surfaces.

� There are two types of thin lens. It is convergingconverging and diverging diverging lens.

� Figures below show the various types of thin lenses, both converging and diverging.

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

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

BiconvexBiconvex PlanoPlano--convexconvex Convex meniscusConvex meniscus

BiconcaveBiconcave PlanoPlano--concaveconcave Concave meniscusConcave meniscus

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9.4.1 Terms of lens

� Figures below show the shape of converging (convex) and diverging (concave) lenses.

�� Centre of curvature (point CCentre of curvature (point C11 and Cand C22))

� is defined as the centre of the sphere of which the surface of the lens is a part.

�� Radius of curvature (rRadius of curvature (r11 and rand r22))

� is defined as the radius of the sphere of which the surface of the lens is a part.

�� Principal (Optical) axisPrincipal (Optical) axis

� is defined as the line joining the two centre's of curvature of a lens.

�� Optical centre (point O)Optical centre (point O)

� is defined as the point at which any rays entering the lens pass without deviation.

(a) Converging lens (b) Diverging lens

CC11 CC22

rr11

rr22OO CC11 CC22

rr11

rr22

OO

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FF11 FF22OO

ff

FF11 FF22OO

ff

9.4.2 Focus (Focal point) and focal length

� Consider the ray diagrams for converging and diverging lens as shown in figures below.

� From the figures,

� Point F1 and F2 represent the focus of the lens.

� Distance f represents the focal length of the lens.�� Focus (point FFocus (point F11 and Fand F22))

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

� Its focus is real (principal).

� For diverging (concave)diverging (concave) lens – is defined as the point on the principal axis where rays which are parallel to the principal axis seem to diverge from after passing through the lens.

� Its focus is virtual.

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FF11

FF22

�� Focal length ( Focal length ( f f ))� Definition – is defined as the distance between the focus F and the

optical centre O of the lens.

9.4.3 Ray Diagrams for Lenses

� Ray diagrams below showing the graphical method of locating an image formed by converging (convex) and diverging (concave) lenses.

(a) Converging (convex) lens

11

11

22

22

OO

33

33

II

u v

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(b) Diverging (concave) lens

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

�� Ray 2Ray 2 - Passes through the optical centre of the lens is not deviated.

�� Ray 3Ray 3 - Passes through the focus point F1 of a converging lens or appears to converge towards the focus F1 of a diverging lens, after refraction by the lens the ray parallel to the principal axis.

OO FF22FF11

11

11

22

22

33

33

II

vu

At least any At least any

two rays two rays

for drawing for drawing

the ray the ray

diagram.diagram.

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9.4.4 Images formed by a diverging (concave) lens

� Ray diagrams below showing the graphical method of locating an image formed by a diverging lens.

� Properties of image formed are

� virtual

� upright

� diminished (smaller than the object)

� formed in front of the lens.

� Object position → any position in front of the diverging lens.

FrontFront backback

OO FF22FF11II

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distance, u

FF11FF22

OO 2F2F22

2F2F11

9.4.5 Images formed by a converging lens

� Table below shows the ray diagrams of locating an image formed by a

converging lens for various object distance, u.

FrontFront backback

u > 2fu > 2f

u = 2fu = 2f

� Real

� Inverted

� Diminished

� Formed between point F2 and 2F2.

(at the back of the lens)

� Real

� Inverted

� Same size

� Formed at point 2F2. (at the back of the lens)

FF11FF22OO 2F2F222F2F11

I

FrontFront backback

I

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distance, u

FF11FF22

OO

2F2F222F2F11

f < u < 2ff < u < 2f

u = fu = f

� Real

� Inverted

� Magnified

� Formed at a distance greater

than 2f at the back of the lens.

� Real

� Formed at infinity.

FF11FF22OO 2F2F222F2F11

FrontFront backback

I

FrontFront backback

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distance, u

� Linear (lateral) magnification of the thin lenses, M is defined as the ratio

between image height, hi and object height, ho

Negative sign indicates that when u and v are both positive, the image is

inverted and ho and hi have opposite signs.

u < fu < f

� Virtual

� Upright

� Magnified

� Formed in front of the lens.

u

v

h

hM

o

i −==where

centre optical from distance image :vcentre optical from distanceobject :u

Simulation

FF11 FF22OO 2F2F222F2F11

FrontFront backback

I

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Thin Lenses Formula and Lens maker’s

Equation

u

O

vo

v

Io

n1

n2

I

first surfacesecond surface

A ray from an object O in medium of refractive index n1 passes

through the first surface of thin lens with refractive index n2. An

image is formed (Io).

Io acts as the virtual object for the second surface of the lens and

finally form a final image

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� By using the equation of spherical refracting surface, the refraction by first surface and second surface are given by

�� Convex surface (Convex surface (r = +rr = +r11))

�� Concave surface (Concave surface (r = r = --rr22))

adding up eq. (1) and eq. (2):

1

12

o

21

r

)nn(

v

n

u

n −=+

2

211

o

2

r

)nn(

v

n

v

n

−−

=+−

(1)(1)

2

21

1

121

o

2

o

21

r

)nn(

r

nn

v

n

v

n

v

n

u

n

−−

+−

=+−

++

(2)(2)

2

12

1

1211

r

nn

r

nn

v

n

u

n −+

−=+

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+−=

+21

121r

1

r

1)nn(

v

1

u

1n

+

−=

211

2

r

1

r

11

n

n

f

1

+

−=+

211

12

r

1

r

1

n

nn

v

1

u

1

Lens makerLens maker’’s s

equationequation

where

length focal :fsurface refractingfirst of curvature of radius :1r

medium theofindex refractive :1nmaterial lens theofindex refractive :2n

surface refracting second of curvature of radius :2r

�If u at infinity then v = f, hence eq. (3) becomes

(3)(3)

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� By equating eq. (3) with lens maker’s equation, hence

therefore in general,

� Note :

� If the medium is airair (n1= nair=1) thus the lens maker’s equation will be

� For thin lens formula and lens maker’s equation, Use the sign sign

conventionconvention for refractionrefraction.

� The radius of curvature of flat refracting surface is infinity, r = r = ∞∞∞∞∞∞∞∞.

f

1

v

1

u

1=+

v

1

u

1

f

1+= Thin lens formulaThin lens formula

where material lens theofindex refractive :n

( )

+−=

21 r

1

r

11n

f

1

Very ImportantVery Important

SF027 48

� Example 16 :

A biconvex lens is made of glass with refractive index 1.52 having the radii of curvature of 20 cm respectively. Calculate the focal length of the lens in

a. water,

b. carbon disulfide.

(Given nw = 1.33 and nc=1.63)

Solution: r1=+20 cm, r2=+20 cm, ng=n2=1.52

a. Given the refractive index of water, nw = n1By using the lens maker’s equation, thus

b. Given the refractive index of carbon disulfide, nc = n1By using the lens maker’s equation, thus

+

−=

21w

g

r

1

r

11

n

n

f

1

cm70f +=

+

−=

21c

g

r

1

r

11

n

n

f

1

cm18148f .−=

SF027 49

� Example 17 :

A converging lens with a focal length of 90.0 cm forms an image of a 3.20 cm tall real object that is to the left of the lens. The image is 4.50 cm tall and inverted. Find

a. the object position from the lens.

b. the image position from the lens. Is the image real or virtual?

No. 34.26, pg. 1331, University Physics with Modern Physics,11th edition,

Young & Freedman.

Solution: f=+90.0 cm, ho=3.20 cm, hi=-4.50 cma. By using the linear magnification equation, hence

By applying the thin lens formula,

u

v

h

hM

o

i −==

u411v .=

v

1

u

1

f

1+=

v

1

u

1

090

1+=

.

(1)(1)

(2)(2)

SF027 50

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

The object is placed 154 cm in front of the lens.

b. By substituting u = 154 cm into eq. (1),therefore

The image forms 217 cm at the back of the lens (at the

opposite side of the object placed) and the image is real.

� Example 18 :

An object is placed 90.0 cm from a glass lens (n=1.56) with one concave surface of radius 22.0 cm and one convex surface of radius 18.5 cm. Determine

a. the image position.

b. the linear magnification. (Gc.862.28)

Solution: u=+90.0 cm, n=1.56, r1=-22.0 cm, r2=+18.5 cma. By applying the lens maker’s equation in air,

cm217v =

cm154u =

( )

+−=

21 r

1

r

11n

f

1

cm208f +=

SF027 51

By applying the thin lens formula, thus

The image forms 159 cm in front of the lens (at the same side

of the object placed)

b. By applying equation of linear magnification for thin lens, thus


A glass (n=1.50) plano-concave lens has a focal length of 21.5 cm. Calculate the radius of the concave surface. (Gc.862.26)

Ans. : -10.8 cm


An object is 16.0 cm to the left of a lens. The lens forms an image 36.0 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 object is 8.00 mm tall, find the height of the image.

c. Sketch the ray diagram for the case above. (UP. 1332.34.34)

Ans. : +11.1 cm, -1.8 cm

cm159v −=

771M .=

v

1

u

1

f

1+=

u

vM −=

pdt unit9 (students)

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