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UNIT 1:Geometrical OpticsUNIT 1: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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1.1 Reflection of Plane Mirror and Refraction1.1.1 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 two media.

Laws of reflection state : The incident ray, the reflected ray and the normal all lie in the

same plane. The angle of incidence, i equals 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 vi

r

i

Image formation by a plane mirror. Point object

Vertical (extended) object

distanceobject :uwhere

distance image :vheightobject :ohheight image :ih

Simulation

A 'A

i

u vi

r

i

Objecti

v

i

rr

u

Image

ihoh

A 'A

i

u vi

r

i

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The properties of image formed are virtual upright or erect laterally reverse

the object distance, u equals the image distance, v the same size 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 byLBALh

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 .

xFrom the properties of the image formed by the plane mirror, thus

m2500v .uv

Therefore, the distance between Nagar and the image of the rose is given by v002x . m252x .

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1.1.2 Refraction 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:1n

ray)incident thecontaining Medium(

constantsin

sin

1

2

n

n

r

i

rnin 21 sinsin Or

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

Snell’s lawSnell’s 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 is also greater)(If the density of medium is greater hence the refractive index 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,

changeschanges but its frequency, frequency, ff remains constant remains 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 22 fv and

2

1

2

1

f

f

v

v

where

11 n

cv

22 n

cv and

2

1

2

1

nc

nc

2211 nn

(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

0n in vacuumlight ofh wavelengt:0

mediumin light ofh wavelengt:

wheredepthapparent :AB

m 2.00 depth actual : AC

A

i

Air (na)

C

r

B

Water (nw)

ir

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 realn

ii sintan

AC

AB

ABADACAD

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, =355 x 10-9 ma. By applying the equation of absolute refractive index, hence

b. By using the equation below, thus

v

cn

551n .

n0 0n

nm550m10x505 70 @.

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Example 5 : (exercise)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

Example 6 : (exercise)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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1.2 Reflection of Spherical Mirrors1.2.1 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.

1.2.2 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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1.2.3 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 is given by

CC

FF

incident rayincident ray

PPDD

BBAA

fr

ii

i

iCD

BDi tan

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

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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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1.2.4 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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1.2.5 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 Iu v

FrontFront backback

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Object

distance, u Ray diagram Image property

I

1.2.6 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.

CCFF

PP

FrontFront backback

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Object

distance, u Ray diagram Image property

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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Object

distance, u Ray diagram Image property

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

u

v

h

hM

o

i where

pole from distance image :vpole from distanceobject :u

Simulation

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O CC PPIv

u

BB

DD

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

then

From the figure, BOC BCIthen, eq. (1)-(2) :

By using BOD, BCD and BID thus

1.2.7 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.

(1)(1)

(2)(2)

2 (3)(3)

ID

BD

CD

BD

OD

BD tan; tan ; tan

vIPIDrCPCDuOPOD ; ;

v

BD

r

BD

u

BD ; ;

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

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

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) image

Inverted 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. Determinea. the position of the image.b. the linear magnification and state the properties of the image.

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

v

1

u

1

f

1

10

30

u

vM

3M

v

1

10

1

15

1

The properties of the image are

Virtual

Upright

Magnified

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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

u

v

h

hM

o

i

cm30vu

u2v

O Icm30

Spherical Spherical

mirrormirroru v

(1)(1)

(2)(2)

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

The mirror should be placed 10 cm in front of the object.

b. By using the equation of spherical mirror,

and therefore

The type of spherical mirror is concaveconcave because the positive value of focal length.

cm40r

v

1

u

1

f

1

cm10u

cm20f u2

1

u

1

f

1

2

rf

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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 cmBy applying the equation of spherical mirror,

From equation of linear magnification,

v

1

u

1

f

1

cm919v .

cm351hi .

2

rf and

v

1

u

1

r

2

u

v

h

hM

o

i

oi hu

vh

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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, calculatea. 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,

The position of the filament is 4.57 cm in front of the concave

mirror.

cm574u .u

10x203

50

035 2.

.

.

u

v

h

hM

o

i

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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

cm019r .v

1

u

1

r

2

v

1

u

1

f

1

2

rf and

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1.3 Refraction of Spherical Surfaces Figure below shows a spherical surface with radius, r forms an

interface between two media with refractive indices n1 and n2.

The surface forms an image I of a point object O as shown in figure above.

The incident ray OB making an angle i with the normal and is

refracted to ray BI making an angle where n1<n2.

Point C is the centre of curvature of the spherical surface and BC is normal.

P

B

O ICD

1n

vr

u

2ni

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From the figure,

BOC

BIC

From the Snell’s law

By using BOD, BCD and BID thus

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

then Snell’s law can be written as

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

then

sinsin 21 nin

)()( 21 nn

i (1)(1)

vIPIDrCPCDuOPOD ; ;

21 nin

tan; tan ; tan ; sin ; sin ii

(3)(3)

ID

BD

CD

BD

OD

BD tan; tan ; tan

(2)(2)

)( 1221 nnnn

r

BDnn

v

BDn

u

BDn 1221 )(

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Note : If the refraction surface is flat (plane) flat (plane) :

then

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

0v

n

u

n 21 r

un

vn

h

hM

2

1

o

i

r

nn

v

n

u

n 1221 )(

wherepole from distance image :vpole from distanceobject :u

1 medium ofindex refractive :1nray)incident thecontaining Medium(

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

Equation of spherical Equation of spherical refracting surfacerefracting surface

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O IC

P

cm008 .

2.00 cm

glassair

Sign convention for refraction refraction :

Example 12 :A cylindrical glass rod in air has refractive index of 1.52. One endis ground to a hemispherical surface with radius, r =2.00 cm as shown in figure below.

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

Object distance, u

Image distance, v

Focal length, f

Linear magnification, M

Real object Virtual object

Real image Virtual image

Convex surface Concave surface

Upright (erect) image

Inverted image

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

(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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Find,a. the position of the image for a small object on the axis of the rod, 8.00 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 34.5, pg. 1302, University Physics with Modern Physics,11th edition, Young & Freedman.

Solution: ng=1.52, u=8.00 cm, r=+2.00 cma. By applying the equation of spherical refracting surface,

The image is 11.26 cm at the back of the convex surface.

b. By using the equation of linear magnification for refracting surface,

930M .

cm2611v .

r

nn

v

n

u

n 1221 )(

un

vn

h

hM

2

1

o

i

r

nn

v

n

u

n agga)(

un

vnM

g

a Negative sign indicates the image is inverted.

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Example 13 :A point object is 25.0 cm from the centre of a glass sphere of radius 5.0 cm. The refractive index of glass is 1.50. Find the position of the image formed due to refraction bya. the first spherical glass surface.b. the first and second refractive surfaces of spheres.

Solution: a. Given na=n1=1.00, ng=n2=1.50, u=20.0 cm, r=5.0 cm By using the equation of spherical refracting surface, thus

r

nn

v

n

u

n agga)(

cm05r .andConvex surfaceConvex surface

(first surface)(first surface)

05

001501

v

501

020

001

.

)..(.

.

.

cm30v The image is real and 30 cm at the back

of the convex surface.

O C 1I

cm020u . r cm30v

P

gnan

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b.

From the figure above, the image I1 formed by the first surface is in

glass and 20 cm from the point Q of the second surface.I1 acts as a virtual objectvirtual object for the second refraction surface and

ng=n1=1.50, na=n2=1.00, u=-20.0 cm, r=-5.0 cmUsing

r

nn

v

n

u

n gaag )(

).(

)..(.

).(

.

05

501001

v

001

020

501

cm715v .

O C

2I cm30

P

gnan

First surfaceFirst surface

1I

an

Q

cm20

Second surfaceSecond surface

Concave surfaceConcave surface

(second surface)(second surface)

The image is real and 5.71cm at the back of the concave surface

(5.71 cm from point Q as shown in figure above).

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Example 14 : (exercise)

A small strip of paper is pasted on one side of a glass sphere of radius 5 cm. The paper is then view from the opposite surface of the sphere. Find the position of the image.

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

Ans. : 20.83 cm in front of the concave surface (second refracting

surface) Example 15 : (exercise)

A point source of light is placed at a distance of 25.0 cm from the centre of a glass sphere of radius 10 cm. Find the image position of the source. (Gc.830.Exam.33-11)

(Given refractive index of glass =1.50 and refractive index of air=1.00)

Ans. : 28 cm at the back of the concave surface (second refracting

surface).

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1.4 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) lensesBiconvexBiconvex Plano-convexPlano-convex Convex meniscusConvex meniscus

BiconcaveBiconcave Plano-concavePlano-concave Concave meniscusConcave meniscus

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1.4.1 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 C and 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 r and 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 centres 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

rr22

OO CC11 CC22

rr11

rr22

OO

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FF11 FF22OO

ff

FF11 FF22OO

ff

1.4.2 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 F and 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.1.4.3 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 undeviated.

Ray 3Ray 3 - Passes through the focus 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 FF22 FF11

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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1.4.4 Images formed by a diverging 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 FF22 FF11II

SF027 48

Object

distance, u Ray diagram Image property

FF11FF22

OO 2F2F22

2F2F11

1.4.5 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

SF027 49

Object

distance, u Ray diagram Image property

FF11FF22

OO2F2F222F2F11

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

SF027 50

Object

distance, u Ray diagram Image property

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

SF027 51

1.5 Thin Lenses Formula and Lens maker’s Equation Considering the ray diagram of refraction for 2 spherical surfaces as

shown in figure below.

OOCC11

CC22II11

II22PP11

PP22

EEBB

AA DD

1u 1v2v

1r 2r

t

1n

1vt

1n2n

SF027 52

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

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

Concave surface DE (Concave surface DE (r = -rr = -r22))

Assuming the lens is very thin thus t = 0,

1

12

1

2

1

1

r

nn

v

n

u

n )(

2

1

2

21

1

2

v

n

r

nn

v

n

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

n

)(

2

12

2

1

1

2

r

nn

v

n

v

n(2)(2)

SF027 53

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

then

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

1

12

2

12

2

1

1

1

r

nn

r

nn

v

n

u

n )(

211

2

r

1

r

11

n

n

f

1

2

12

1

12

2

1

1

1

r

nn

r

nn

v

n

u

n )()(

(3)(3)

211

2

21 r

1

r

11

n

n

v

1

u

1

Lens maker’s Lens maker’s equationequation

wherelength focal :f

surface refractingfirst of curvature of radius :1r

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

surface refracting second of curvature of radius :2r

SF027 54

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

21

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 55

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 = n1

By using the lens maker’s equation, thus

b. Given the refractive index of carbon disulfide, nc = n1

By 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 56

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. Finda. 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 57

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. Determinea. 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 58

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

Example 19 : (exercise)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 Example 20 : (exercise)

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

SF027 59

There are 3 optical devices that extend human vision. It is magnifier, compound microscope magnifier, compound microscope and telescopetelescope.

1.6.1 Angular magnification (magnifying power), Ma

The angular magnification of an optical device is defined

as the ratio of the angle subtended at the eye by the image , to the angle subtended at the unaided eye by the object (without

lens), .

In order to determine the angle it is necessary to specify the position of the object. For microscopemicroscope, the best object position is at the near pointnear point. For telescope, the object position is not meaningful because the

telescope is used for viewing distant object. Near point is defined as the nearest point at which an object is seen

most clearly by the human eye. The distance between the near point to the eye is 25 cm25 cm and is

known as distance of distinct vision (D).

aM

1.6 Optical Devices

SF027 60

1.6.2 Magnifier It also known as magnifying glassmagnifying glass or simple microscopesimple microscope. It is an optical device used for viewing near object. It consists of single converging (biconvex) lens. Suppose a leaf is viewed at near point of the human eye as shown in

figure below.

From the figure,

By making small angle approximation, we get

D

oh

D

hotan

D

hotan

SF027 61

To increase the apparent size of the leaf, a converging lens can be placed in front of the eye as shown in figure below.

The apparent size of the leaf is maximummaximum when the image is at the near point where

From the figure above,

By making small angle approximation, we get

The properties of the image are Virtual, upright and magnified

u

h

D

h oi tan

u

h

D

h oi tan

u < fu < f

cm25Dv

oh

u

ih

vFI O

SF027 62

The angular magnification in terms of D and f can be evaluated by derivation below. By applying the thin lens formula,

From the definition of angular magnification,

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

v

1

u

1

f

1

fD

Dfu

where Dv

u

DM a

Dhuh

Mo

o

a

(1)(1)

(2)(2)

1f

DM a

wherelength focal :f

cm 25isiondistinct v of distance : D

SF027 63

The relationship between linear magnification, M with angular

magnification, Ma From the definition of angular magnification,

then

Note: If the object placed at the focal point of the converging lens, the

image formed at infinityimage formed at infinity. Thus

Therefore, since then

Mh

hM

o

ia

DhDh

Mo

i

a

f

ho

aMf

DM a

D

h

f

h

Mo

o

a

The eye is relax.The eye is relax.

SF027 64

1.6.3 Compound Microscope Because it makes use of two lenses, the magnifying power of the

compound microscope is much greater than that of the magnifier. The two lenses are converging lens and is known as objective lensobjective lens

(close to the object) and eyepiece lens eyepiece lens (close to the eye). The figure below shows the schematic diagram of the compound

microscope.

oFeF'

oF

efu

of 1I

L

O

2I

Objective lensObjective lens

Eyepiece lensEyepiece lens

The properties of first image are Real, inverted and magnified

v >2fo

The properties of final image are Virtual, inverted and

magnified

v >(fo+ fe)

acts as a magnifier.acts as a magnifier.

SF027 65

The properties of the compound microscope are The distance between two lenses, L > (fo+fe) fo < fe

The final image is I2. The angular magnification formula is given by

The negative sign indicates that the image is inverted. It is used for viewing small objects that are very close to the objective

lens.

1.6.4 Astronomical (refracting) Telescope This telescope consists of two converging lenses. Like compound microscope, the two lenses are objective objective and eyepiece eyepiece

lens. It is used to magnify objects that are very far away (considered to be at

infinity).

eoa f

D

f

LM

wherelens eyepiece theoflength focal :ef

cm 25isiondistinct v of distance : Dlens objective theoflength focal :of

SF027 66

The figure below shows the schematic diagram of the telescope.

oF eF

of

'eF

efef

2I

1I

Eyepiece lensEyepiece lensObjective lensObjective lens

Parallel rays Parallel rays from object at from object at infinityinfinity

acts as a magnifier.acts as a magnifier.

The properties of first image are Real, inverted and diminished

v =fo

The properties of final image are Virtual, inverted and magnified

v >(fo+ fe)

SF027 67

The properties of the telescope are The distance between two lenses, L <(fo+fe) fo > fe

The final image is I2. The angular magnification formula is given by

The negative sign indicates that the image is inverted.

e

oa f

fM