matriculation physics ( geometrical optics )

1

PHYSICS CHAPTER 1The study of light based on the assumption that light light travels in straight linestravels in straight lines and is concerned with the

laws controlling the laws controlling the reflection and refractionreflection and refraction

of rays of lightlight.

CHAPTER 1: CHAPTER 1: Geometrical opticsGeometrical optics

(5 Hours)(5 Hours)

PHYSICS CHAPTER 1

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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: StateState laws of reflection. laws of reflection. StateState the characteristics of image formed by a plane the characteristics of image formed by a plane

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

Learning Outcome:

1.1 Reflection at a plane surface (1 hour)

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Figure 1.1Figure 1.1

1.1 Reflection at a plane surface1.1.1 Reflection of light is defined as the return of all or part of a beam of light when the 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 from reflection of light from a smooth shiny surfacea smooth shiny surface as shown in Figure 1.1.

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

Diffuse reflectionDiffuse reflection is the reflection of light from a rough reflection of light from a rough surfacesurface such as papers, flowers, people as shown in Figure 1.2.

The reflected rays is sent out in a variety of directions. For both types of reflection, the laws of reflection are obeyed.

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

in the same planein the 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 1.3.

i r

Plane surfacePlane surface

ri

Stimulation 1.1Figure 1.3Figure 1.3

Picture 1.1

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Image formation by a plane mirror as shown in Figures 1.4a and 1.4b.

Point object

1.1.2 Reflection at a plane mirror

Figure 1.4aFigure 1.4a

A 'Au v

i

i

r

i

distanceobject :uwhere

distance image :v

g

gangle glancing :g

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Vertical (extended) object

Stimulation 1.2

Figure 1.4bFigure 1.4b

Object

vu

ir

i

rImage

ihoh

where heightobject :ohheight image :ih

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The characteristics of the image formed by the plane mirror are virtual imagevirtual image

is seem to form by light coming from the image but seem to form by light coming from the image but light does not actually pass through the imagelight does not actually pass through the image.

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

upright or erect imageupright or erect image laterally reverselaterally reverse

right-hand side of the object becomes the left-hand side of the image.

the object distance, object distance, uu equals the image distance, equals the image distance, vv the same sizesame size where the linear magnification, m is given by

obey the laws of reflectionobey the laws of reflection.

1height,Object

height, Image

o

i h

hm

Picture 1.2

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A women is 1.60 m tall and her eyes are 10 cm below the top of her head. She wishes to see the whole length of her body in a vertical plane mirror whilst she herself is standing vertically.

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

women’s image.

b. What is the minimum length of mirror that makes this

possible?

c. How far above the ground is the bottom of the mirror?

Example 1 :

PHYSICS CHAPTER 1

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A

B

L

Solution :Solution :a. The ray diagram to show the formation of the women’s image is

HE2

1AL

EF2

1LB

)feet(F

)eyes(E)head(H

h

y

m 60.1

m 10.0

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Solution :Solution :b. The minimum vertical length of the mirror is given by

b. The mirror can be placed on the wall with the bottom of the mirror is halved of the distance between the eyes and feet of the women. Therefore

LBAL h

EF2

1HE

2

1h

EFHE2

1h

Height of the women

m 80.060.12

1h

10.060.12

1y

m 75.0y

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u v

m 00.1

x

A rose in a vase is placed 0.350 m in front of a plane mirror. Ahmad looks into the mirror from 1.00 m in front of it. How far away from Ahmad is the image of the rose?

Solution :Solution :

From the characteristic of the image formed by the plane mirror, thus

Therefore,

Example 2 :

m 350.0vuv

vx 00.1m 350.1x

m 350.0u

350.000.1 x

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Exercise 1.1 :1.

The two mirrors in Figure 1.5 meet at a right angle. The beam of light in the vertical plane P strikes mirror 1 as shown.

a. Determine the distance of the reflected light beam travels

before striking mirror 2.

b. Calculate the angle of reflection for the light beam after being reflected from mirror 2.

ANS. :ANS. : 1.95 m 1.95 m ; 40; 40 to the mirror 2. to the mirror 2.


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Exercise 1.1 :2.

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 1.6. Determine x.

ANS. :ANS. : 0.81 m0.81 m


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Exercise 1.1 :3. Standing 2.00 m in front of a small vertical mirror, you see the

reflection of your belt buckle, which is 0.70 m below your eyes.

a. What is the vertical location of the mirror relative to the level of your eyes?

b. What is the angle do your eyes make with the horizontal when you look at the buckle?

c. If you now move backward until you are 6.0 m from the mirror, will you still see the buckle? Explain.

ANS. :ANS. : 35 cm below; 9.935 cm below; 9.9; U think; U think4. You are 1.80 m tall and stand 3.00 m from a plane mirror that

extends vertically upward from the floor. On the floor 1.50 m in front of the mirror is a small table, 0.80 high. What is the minimum height the mirror must have for you to be able to see the top of the table in the mirror?

ANS. :ANS. : 11.13 m.13 m

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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Sketch and useSketch and use ray diagrams to ray diagrams to determinedetermine the the

characteristics 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 and – for 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.

Learning Outcome:

1.2 Reflection at a spherical surface (1 hour)

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rvuf

2111

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CC

AA

BB

rPPCC

AA

BB

rPP


1.1 Reflection at a spherical surface1.2.1 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 (curving

outwards) and concaveconcave (curving inwards) mirror. Figures 1.7a and 1.7b show the shape of concave and convex

mirrors.

reflecting surface

imaginary sphere

silver layer


(a)Concave (ConvergingConverging) mirror

(b) Convex (DivergingDiverging) mirror

Picture 1.3

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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 curved the centre of the sphere of which a curved mirror forms a partmirror forms a part.

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

mirror 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 of the straight line through the centre of curvature C and pole P of the mirrorcurvature C and pole P of the mirror.

AB is called the apertureaperture of the mirror.

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Consider the ray diagram for a concave and convex mirrors as shown in Figures 1.8a and 1.8b.

Point FF represents the focal pointfocal point or focusfocus of the mirrors. Distance ff represents the focal lengthfocal length of the mirrors. The parallel incident raysparallel incident rays represent the object infinitely far object infinitely far

awayaway from the spherical mirror e.g. the sun.

CCPPCC PP

1.2.2 Focal point and focal length, f


FFf

FFf

Incident Incident raysrays


Incident Incident raysrays

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Focal point or focus, FFocal point or focus, F For concave mirror – is defined as a point where the incident a point where the incident

parallel rays converge after reflection on the mirrorparallel rays converge after reflection on the mirror. Its focal point is real (principal)real (principal).

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

Its focal point is virtualvirtual.

Focal length, Focal length, ff is defined as the distance between the focal point (focus) F the distance between the focal point (focus) F

and 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 and the rays that are near to and

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

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Consider a ray AB parallel to the principal axis of concave mirror as shown in Figure 1.9.

1.2.3 Relationship between focal length, f and

radius of curvature, r


CC

PPFF DD

incident rayincident rayBBAA

fr

ii

i

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From the Figure 1.9,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 pole P then

Therefore

ii CD

BDtan

FD

BDtan

Taken the angles are << Taken the angles are << small by considering the small by considering the ray AB is paraxial ray.ray AB is paraxial ray.

i2

rCPCD

fFPFD

This relationship also valid for convex mirror.This relationship also valid for convex mirror.

2

rf

CD

BD2

FD

BD

OR

FD2CD

f2r

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

Figures 1.10a and 1.10b show the graphical method of locating an image formed by concave and convex mirror.

1.2.4 Ray diagrams for spherical mirrors

Figure 1.10aFigure 1.10a Figure 1.10bFigure 1.10b

(a) Concave mirror (b) Convex mirror

CC PPFF

11

33

33

11

I CC

FFPP

1122

22O O I

2233

11

22

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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.

Images formed by a convex mirrorImages formed by a convex mirror Figure 1.11 shows the graphical method of locating an image

formed by a convex mirror.

At least any At least any two rays two rays for drawing for drawing the ray the ray diagram.diagram.

CC

FF

PP

O Iu v

frontfront backbackFigure 1.11Figure 1.11

Picture 1.4

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The characteristics of the image formed are virtualvirtual 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 it

has 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 1.1 shows the ray diagrams of locating an image formed

by a concave mirror for various object distance, u.

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Object

distance, uRay diagram

Image characteristic

I

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

PHYSICS CHAPTER 1

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Object



FFCC PP

FrontFront backback

f < u < rf < u < r

u = fu = f

O

Real Inverted Magnified Formed at a

distance greater than CP.

Real or virtual Formed at infinity.

IO

CC

FF

PP

FrontFront backback

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Linear (lateral) magnification of the spherical mirror, m is

defined as the ratio between image height, the ratio between image height, hhii and object and object

height, height, hhoo

Object



u < fu < f

O

Virtual Upright Magnified Formed at the

back of the mirror

IFF

CC PP

FrontFront backback

u

v

h

hm

o

iwhere

pole thefrom distance image :vpole thefrom distanceobject :u

Table 1.1Table 1.1

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Figure 1.12 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.

1.2.5 Derivation of Spherical mirror equation


O CC PPIv

u

BB

DD

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

By using BOD, BCD and BID thus

(1)(1) (2)(2)

2 (3)(3)

ID

BD tan;

CD

BDtan ;

OD

BDtan

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

then

therefore

vru IPID; CPCD ; OPOD tan; tan ; tan

v

BD

r

BD

u

BD ; ; Substituting this Substituting this

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

fr 2

rvu

BD2

BD

BD

rvu

21

1 where

rvuf

21

11 Spherical mirror’s Spherical mirror’s

equationequation

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Table 1.2 shows the sign convention for spherical mirror’s equation .

Note: Real image is formed by the actual light rays that pass formed by the actual light rays that pass

through the imagethrough the image. Real image can be projected on the screenprojected on the screen.

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)

Table 1.2Table 1.2

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

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

used,

b. the magnification of the image.


a. By applying the mirror’s equation, thus

b. By using the magnification formula, thus

Example 3 :

cm 38.1fvuf

111

u

vm

cm 9.25 cm; 20.1 vu

71.720.1

25.9m

25.9

1

20.1

11

f(Concave mirror)(Concave mirror)

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An upright image is formed 20.5 cm from the real object by using the spherical mirror. The image’s height is one fourth of object’s height.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 4 :

oi 25.0 hh

O Icm 0.52

Spherical Spherical

mirrormirroru v

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Solution :Solution :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 object 16.4 cm in front of the object.

oi 25.0 hh

5.20 vu

u

v

h

hm

o

i

(1)(1)

u

v

h

h

o

o25.0

uv 25.0 (2)(2)

5.2025.0 uucm 4.16u

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Solution :Solution :b. By using the mirror’s equation, thus

The type of spherical mirror is convexconvex because the negative value of focal length.

oi 25.0 hh

cm 47.5f

vuf

111

uuf 25.0

111

4.1625.0

1

4.16

11

f

and2

rf

cm 9.1047.52 r

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Solution :Solution :c. The ray diagram is shown below.

oi 25.0 hh

FF

PP CC

O I

frontfront backback

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A person of 1.60 m height stands 0.60 m from a surface of a hanging 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 large is the person’s image?c. State the characteristics of the person’s image.


Example 5 :

m 0.60 m; 60.1o uh

u

oh

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Solution :Solution :a. Given The radius of curvature of the globe’s surface (convex surface) is given by

By applying the mirror’s equation, hence

m 18.0d

m 09.02

18.0r

vur

112

(behind the globe’s surface)(behind the globe’s surface)m 042.0v

m 0.60 m; 60.1o uh

v

1

60.0

1

09.0

2

PHYSICS CHAPTER 1

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b. By applying the magnification formula, thus

c. The characteristics of the person’s image are virtualvirtual uprightupright diminisheddiminished formed behind the reflecting surface.formed behind the reflecting surface.

u

v

h

hm

o

i

m 0.60 m; 60.1o uh

60.0

042.0

60.1i

h

m 112.0i h OR cm 2.11

PHYSICS CHAPTER 1

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CC FFPP

A shaving or makeup mirror forms an image of a light bulb on a wall of a bathroom that is 3.50 m from the mirror. The height of the bulb 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 bulb’s image.b. Calculate i. the position of the bulb from the pole of the mirror, ii. the focal length of the mirror.Solution :Solution :a. The ray diagram of the bulb is

Example 6 :

I

O

cm 40

mm 0.8

u

m 10 40 m; 10 0.8m; 3.50 2i

3o

hhv

m 50.3

PHYSICS CHAPTER 1

41


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 mirror’s equation, thus

u

v

h

hm

o

i

u

50.3

100.8

10403

2

m 07.0u OR cm 0.7

m 10 40 m; 10 0.8m; 3.50 2i

3o

hhu

vuf

111

m 0687.0f50.3

1

07.0

11

fOR cm 87.6

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Exercise 1.2 :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 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.

ANS. :ANS. : 160 mm 160 mm ; 267 mm; 267 mm2. a. A 1.74 m tall shopper in a department store is 5.19 m

from a security mirror. The shopper notices that his image in the mirror appears to be only 16.3 cm tall.

i. Is the shopper’s 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 an image whose distance from the mirror is one third of the object distance. Calculate the object and image distances.

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

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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: State and useState and use the laws of refraction (Snell’s Law) for the laws of refraction (Snell’s Law) for

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

for spherical surface.for spherical surface.

Learning Outcome:

1.3 Refraction at a plane and spherical surfaces (1 hour)

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r

nn

v

n

u

n 1221

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1.3 Refraction at a plane and spherical surfaces1.3.1 Refraction at a plane surface RefractionRefraction is defined as the changing of direction of a light the changing of direction of a light

ray and its speed of propagation as it passes from one ray and its speed of propagation as it passes from one medium into anothermedium into another.

Laws of refractionLaws of refraction state : The incident ray, the refracted ray and the normal all lie incident ray, the refracted ray and the normal all lie

in the same planein the same plane. For two given media, Snell’s law Snell’s law states

constantsin

sin

1

2 n

n

r

irnin sinsin 21 OR

where 1 medium theofindex refractive:1nray)incident thecontaining Medium(

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

refraction of angle :r

PHYSICS CHAPTER 1

45

The light ray is bent toward the bent toward the normalnormal, thus

The light ray is bent away from bent away from the normalthe normal, thus

Examples for refraction of light ray travels from one medium to another medium can be shown in Figures 1.13a and 1.13b.

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 dense (Medium 1 is less dense medium 2)medium 2)

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

ir


Stimulation 1.3 Stimulation 1.4

PHYSICS CHAPTER 1

46

Refractive index (index of refraction), Refractive index (index of refraction), nn is defined as the constant ratio for the two given constant ratio for the two given

mediamedia.

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

It is dimensionlessdimensionless and its value greater than 1greater than 1. Consider the light ray travels from medium 1 into medium 2, the

refractive index can be denoted by

r

i

sin

sin

2

121 2 mediumin light ofvelocity

1 mediumin light ofvelocity

v

vn

(Medium containing (Medium containing the incident ray)the incident ray)

(Medium containing the (Medium containing the refracted ray)refracted ray)

PHYSICS CHAPTER 1

47

Absolute refractive index, n (for the incident ray travels from vacuum or air into the mediummedium) is given by

Table 1.3 shows the refractive indices for common substances.

v

cn

mediumin light ofvelocity

in vacuumlight ofvelocity

SubstanceSubstance Refractive index, Refractive index, nnSolidsSolids

DiamondFlint glassCrown glassFused quartz (glass)Ice

LiquidsLiquidsBenzeneEthyl alcoholWater

GasesGasesCarbon dioxideAir

2.421.661.521.461.31

1.501.361.33

1.000451.000293

Table 1.3Table 1.3

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

PHYSICS CHAPTER 1

48

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

light 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 number of wave cycles arriving per unit time must equal the number leaving cycles arriving per unit time must equal the number leaving per unit timeper unit time so that the boundary surface cannot create or cannot create or destroy 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

11 fv 22 fv and

2

1

2

1

f

f

v

v where

11 n

cv

22 n

cv and

PHYSICS CHAPTER 1

49

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

refractive index for any medium, n can be expressed as

2

1

2

1

nc

nc

2211 nn

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

where

0n

in vacuumlight ofh wavelengt:0mediumin light ofh wavelengt:

Picture 1.5 Picture 1.6

PHYSICS CHAPTER 1

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A fifty cent coin is at the bottom of a swimming pool of depth 3.00 m. The refractive index of air and water are 1.00 and 1.33 respectively. Determine the apparent depth of the coin.Solution :Solution :

Example 7 :

33.1; 1.00 wa nn

wheredepthapparent :AB

m 3.00 depth actual :AC

A

i

Air (na)

C

r

B

Water (nw)

ir

m 00.3

D

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Solution :Solution :From the diagram,

ABD

ACD

By considering only small angles of r and i , thus

33.1; 1.00 wa nn

AB

ADtan r

AC

ADtan i

andrr sintan ii sintan

AC

AB

ABADACAD

sin

sin

tan

tan

r

i

r

ithen

PHYSICS CHAPTER 1

52

Note : Note : (Important)(Important)

Solution :Solution :From the Snell’s 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

m 26.2AB

depthapparent

depth real

1

2 n

nn

Other equation for absolute refractive index in term of depth is given by

PHYSICS CHAPTER 1

53

A pond with a total depth (ice + water) of 4.00 m is covered by a transparent layer of ice of thickness 0.32 m. Determine the time required for light to travel vertically from the surface of the ice to the bottom of the pond. The refractive index of ice and water are 1.31 and 1.33 respectively.

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

Example 8 :

33.1; 1.31 wi nn

Ice (Ice (nnii))

Water (Water (nnww))

BottomBottom

m 00.4

m 32.0i h

32.000.4w hm 68.3w h

PHYSICS CHAPTER 1

54

Solution :Solution :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

ii v

cn

18i s m 1029.2 v

33.1; 1.31 wi nn

i

81000.331.1

v

ww v

cn

18w s m 1026.2 v

w

81000.333.1

v

wi ttt

88w

w

i

i

1026.2

68.3

1029.2

32.0

v

h

v

ht

s 1077.1 8t

PHYSICS CHAPTER 1

55

Figure 1.14 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. 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.

1.3.2 Refraction at a spherical surface


P

B

O ICD

1n

vr

u

2ni

PHYSICS CHAPTER 1

56

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

i (1)

(2)

sinsin 21 nin

ID

BD tan;

CD

BDtan ;

OD

BDtan

vru IPID; CPCD ; OPOD

21 nin

tan; tan ; tan ; sin ; sin ii

(3)

PHYSICS CHAPTER 1

57

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

then

)()( 21 nn )( 1221 nnnn

rnn

vn

un

BD)(

BDBD1221

r

nn

v

n

u

n )( 1221

where pole 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

PHYSICS CHAPTER 1

58

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

then

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

021 v

n

u

n

r

un

vn

h

hm

2

1

o

i

PHYSICS CHAPTER 1

59

Table 1.4 shows the sign convention for refraction or thin refraction or thin lenseslenses:

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 the object)

(opposite side of the object)

(in front of the refracting surface)

(at the back of the refracting surface)

Radius of

curvature, r

Centre of curvature is located in more dense medium

Centre of curvature is located in less dense medium

Table 1.4Table 1.4

(convex surface) (concave surface)

PHYSICS CHAPTER 1

60

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 as shown in Figure 1.15.

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 9 :

O ICP

cm 0.10

3.00 cm

glassair


PHYSICS CHAPTER 1

61

Solution :Solution :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

r

nn

v

n

u

n agga

cm 3.00 cm; 0.10; 1.52g run

un

vnm

g

a

cm 7.20v

00.3

00.152.152.1

0.10

00.1

v

un

vnm

2

1

0.1052.1

7.2000.1m

36.1m

PHYSICS CHAPTER 1

62

Figure 1.16 shows an object O placed at a distance 20.0 cm from the surface P of a glass sphere of radius 5.0 cm and refractive index 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 10 :


OP

cm 0.20

Glass sphere

air

cm 0.5

PHYSICS CHAPTER 1

63

Solution :Solution :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

r

nn

v

n

u

n agga

cm .05 cm; 0.20; 1.63g run

cm 5.21v

0.5

00.163.163.1

0.20

00.1

v

O C 1I

cm020u . cm 5.21v

P

gnan

r

PHYSICS CHAPTER 1

64

Solution :Solution :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 C

2I cm 1.52

P

gnan

First surface

1I

an

Q

cm 1.51

Second surface

cm; 5.11 1.00; ; 1.63 a2g1 unnnn

cm 5.00rCentre of curvature is located in Centre of curvature is located in more dense mediummore dense medium

PHYSICS CHAPTER 1

65

Solution :Solution :b. By using

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

surface Q.surface Q.

r

nn

v

n

u

n gaag

cm 74.3v

0.5

63.100.100.1

5.11

63.1

v

PHYSICS CHAPTER 1

66

Exercise 1.3 :1. A student wishes 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 1.17.

Calculate the depth of the pool.

(Given nwater = 1.33 and nair = 1.00)

ANS. :ANS. : 5.16 m5.16 m


PHYSICS CHAPTER 1

67

Exercise 1.3 :

2. 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. 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 from the centre of a glass sphere of radius 10 cm. Determine the image position of the source.

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

PHYSICS CHAPTER 1

68

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Sketch and useSketch and use ray diagrams to ray diagrams to determinedetermine the the

characteristics of image formed by diverging and characteristics of image formed by diverging and converging lenses. converging lenses.

UseUse equation stated in 1.3 to equation stated in 1.3 to derivederive thin lens formula, thin lens formula,

for real object only.for real object only.

UseUse lensmaker’s equation: lensmaker’s equation:

UseUse the thin lens formula for a combination of the thin lens formula for a combination of converging lenses.converging lenses.

Learning Outcome:1.4 Thin lenses (2 hours)

ww

w.k

mp

h.m

atri

k.ed

u.m

y/p

hys

ics

ww

w.k

mp

h.m

atri

k.ed

u.m

y/p

hys

ics

fvu

111

21

111

1

rrn

f

PHYSICS CHAPTER 1

69

1.4 Thin lenses is defined as a transparent material with two spherical a transparent material with two spherical

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

There are two types of thin lenses. It is convergingconverging and divergingdiverging lenses.

Figures 1.18a and 1.18b show the various types of thin lenses, both converging and diverging.

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

BiconvexBiconvex Plano-convexPlano-convex Convex meniscusConvex meniscus


rr11

(+ve)(+ve)rr22

(+ve)(+ve)rr11

(+ve)(+ve)rr22

(())rr11

(+ve)(+ve)rr22

((ve)ve)

PHYSICS CHAPTER 1

70

1.4.1 Terms of thin lenses Figures 1.19 show the shape of converging (convex) and

diverging (concave) lenses.

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

BiconcaveBiconcave Plano-concavePlano-concave Concave meniscusConcave meniscusFigure 1.18bFigure 1.18b

(a) Converging lens (b) Diverging lens

CC11 CC22

rr11

rr22OO CC11 CC22

rr11

rr22

OO


rr11

((ve)ve)rr22

((ve)ve)rr11

((ve)ve)rr22

(())rr11

(+ve)(+ve)rr22

((ve)ve)

PHYSICS CHAPTER 1

71

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 the centre of the sphere of which the surface

of the lens is a partof 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 the radius of the sphere of which the surface of the lens is a partof the lens is a part.

Principal (Optical) axisPrincipal (Optical) axis is defined as the line joining the two centres of curvature the line joining the two centres of curvature

of a lensof a lens. Optical centre (point O)Optical centre (point O)

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

PHYSICS CHAPTER 1

72

Consider the ray diagrams for converging and diverging lenses as shown in Figures 1.20a and 1.20b.

From the figures, Points F1 and F2 represent the focus of the lenses. Distance f represents the focal length of the lenses.

1.4.2 Focal point and focal length, f

FF11 FF22OO

ffff


FF11 FF22OO

PHYSICS CHAPTER 1

73

Focus (point FFocus (point F11 and F and F22)) For converging (convex)converging (convex) lens – is defined as the point on the the point on the

principal axis where rays which are parallel and close to the principal axis where rays which are parallel and close to the principal 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 the the point on the

principal axis where rays which are parallel to the principal principal axis where rays which are parallel to the principal axis 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 the the distance between the focus F and the

optical centre O of the lensoptical centre O of the lens.

PHYSICS CHAPTER 1

74

Figures 1.21a and 1.21b show the graphical method of locating an image formed by a converging (convex) and diverging (concave) lenses.

1.4.3 Ray diagram for thin lenses


FF11

FF22

(a) Converging (convex) lens

11

11

22

22

OO

33

33

II

u v

PHYSICS CHAPTER 1

75

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.

(b) Diverging (concave) lens

OO FF22 FF11

11

11

22

22

33

33

II

vu


At least At least any two any two rays for rays for drawing drawing the ray the ray diagram.diagram.

PHYSICS CHAPTER 1

76

Images formed by a diverging lensImages formed by a diverging lens Figure 1.22 shows the graphical method of locating an image

formed 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.

FrontFront backback

OO FF22 FF11II


PHYSICS CHAPTER 1

77

Object



FF11FF22 2F2F222F2F11

Images formed by a converging lensImages formed by a converging lens Table 1.5 shows the ray diagrams of locating an image formed

by a converging lens for various object distance, u.

FrontFront backback

u > u > 22ff

Real Inverted Diminished Formed between

point F2 and 2F2.

(at the back of the lens)

OOI

PHYSICS CHAPTER 1

78

Object



OO

FF11FF22

2F2F22

2F2F11u = u = 22ff

Real Inverted Same size Formed at point

2F2. (at the back of the lens)FrontFront backback

I

f < u < f < u < 22ff

Real Inverted Magnified Formed at a

distance greater than 2f at the back of the lens.

OO FF11FF22 2F2F222F2F11

FrontFront backback

I

PHYSICS CHAPTER 1

79

Object



OOFF11

FF22 2F2F222F2F11u = fu = f

Real or virtual Formed at infinity.

FrontFront backback

u < fu < f

Virtual Upright Magnified Formed in front

of the lens.

OOFF11 FF22 2F2F222F2F11

FrontFront backback

I

Table 1.5Table 1.5

Stimulation 1.5

PHYSICS CHAPTER 1

80

Thin lens formula and lens maker’s equationThin lens formula and lens maker’s equation Considering the ray diagram of refraction for two spherical

surfaces as shown in Figure 1.23.

1.4.4 Thin lens formula, lens maker’s and linear magnification equations


OOCC11

CC22II11

II22PP11 PP22

EEBB

AA DD

1u 1v2v

1r 2r

t

1n

12 vtu

1n2n

PHYSICS CHAPTER 1

81

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

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

Surface DE ( (r r == +r +r22))

Assuming the lens is very thin thus very thin thus tt = 0= 0,

1

12

1

2

1

1 )(

r

nn

v

n

u

n

2

12

2

1

1

2 )(

r

nn

v

n

vt

n

(1)(1)

2

12

2

1

1

2

r

nn

v

n

v

n2

12

2

1

1

2 )(

r

nn

v

n

v

n

(2)(2)

PHYSICS CHAPTER 1

82

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

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

1

12

2

12

2

1

1

1 )(

r

nn

r

nn

v

n

u

n

2

12

1

12

2

1

1

1 )()(

r

nn

r

nn

v

n

u

n

(3)(3)

211

2

21

111

11

rrn

n

vu

211

2 111

1

rrn

n

fLens maker’s Lens maker’s equationequation

where length focal :fsurface refracting 1for curvature of radius : st

1r

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

surface refracting 2for curvature of radius : nd2r

PHYSICS CHAPTER 1

83

By equating eq. (3) and the lens maker’s equation, thus

therefore in general,

Note : If the medium is airair (n1= nair=1) thus the lens maker’s

equation can be written as

For thin lenses and lens maker’s equations, use the sign sign conventionconvention for refractionrefraction.

fvu

111

21

vuf

111 Thin lens formulaThin lens formula

where material lens theofindex refractive :n

21

111

1

rrn

f

PHYSICS CHAPTER 1

84

Linear magnification, Linear magnification, mm is defined as thethe ratio between image height, ratio between image height, hhii and object and object

height, height, hhoo.

Since the linear magnification equation can be

written as

u

v

h

hm

o

i

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

vuf

111

vvuf

111

1u

v

f

v1

f

vm

PHYSICS CHAPTER 1

85

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 curvature 7.69 mm. The lens made from the crown glass of refractive index 1.52.

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 maker’s equation in air, thus

Example 11 :

;52.1 m; 50.2 m; 1.75o nuhm 1069.7 3

21rr

21

111

1

rrn

f

PHYSICS CHAPTER 1

86


a.

b. The ray diagram for the case is

;52.1 m; 50.2 m; 1.75o nuhm 1069.7 3

21rr

33 1069.7

1

1069.7

1152.1

1

fm 1039.7 3f

FF11FF22 2F2F222F2F11

FrontFront backback

OOI

PHYSICS CHAPTER 1

87


c. The position of the image formed is

By using the linear magnification equation, thus

d. The characteristics of the image are realreal invertedinverted diminisheddiminished formed at the back of the lensformed at the back of the lens

vuf

111

m 1041.7 3vv

1

50.2

1

1039.7

13

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

u

v

h

hm

o

i

50.2

1041.7

75.1

3i

h

m 1019.5 3i

h OR mm 19.5

PHYSICS CHAPTER 1

88

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 ten times 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 12 :

10 cm; 10 1.66; mun

10u

vm uv 10

vuf

111

uuf 10

111

Virtual imageVirtual image

PHYSICS CHAPTER 1

89


a.

b. Since the thin lens is plano-convex thus

Therefore

10 cm; 10 1.66; mun

2r

21

111

1

rrn

f

11

166.11.11

1

1r

cm 33.71 r

1010

1

10

11

f

cm 1.11f

PHYSICS CHAPTER 1

90

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

a. in air,

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


a. By applying the lens maker’s equation in air,

Example 13 :

2/32 n

21

111

1

rrn

f

cm 201 r cm 102 r

2/32 nnand

PHYSICS CHAPTER 1

91


a.

b. Given

By using the general lens maker’s equation, therefore

3/41 n

211

2 111

1

rrn

n

f

cm 160f

cm 40f

2/32 n

10

1

20

11

2

31

f

10

1

20

11

1

34

23

f

PHYSICS CHAPTER 1

92

Many optical instruments, such as microscopes and microscopes and telescopestelescopes, use two converging lensestwo converging lenses together to produce an image.

In both instruments, the 1st lens (closest to the objectclosest to the object )is called the objectiveobjective and the 2nd lens (closest to the eyeclosest to the eye) is referred to as the eyepiece eyepiece or ocular ocular.

The image formed image formed by the 1 1stst lens lens is treatedtreated as the object for object for the 2the 2ndnd lens lens 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 be determined by applying the thin lens formula to each lens thin lens formula to each lens separatelyseparately.

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

1.4.5 Combination of lenses

21mmm where

ionmagnificat overall :mlens 1 the todueion magnificat : st

1mlens 2 the todueion magnificat : nd

2m

Picture 1.7

Picture 1.8

PHYSICS CHAPTER 1

93

The objective and eyepiece of the compound microscope are both converging lenses and have focal lengths of 15.0 mm and 25.5 mm respectively. A distance of 61.0 mm separates the lenses. The microscope is being used to examine a sample placed 24.1 mm in front 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 14 :

mm; 61.0 mm; 5.25 mm; 0.15 21 dffmm 24.11 u

d

1u

1f1f 2f2fFF11 FF11 FF22 FF22O

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

PHYSICS CHAPTER 1

94

d

1u



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

mm 7.391 v111

111

vuf

mm; 61.0 mm; 5.25 mm; 0.15 21 dffmm 24.11 u

1

1

1.24

1

0.15

1

v

(real)(real)

1I

1v 2u

12 vdu 7.390.612 u

mm 3.212 u

PHYSICS CHAPTER 1

95


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

(eyepiece) is

mm; 61.0 mm; 5.25 mm; 0.15 21 dffmm 24.11 u

222

111

vuf

2

1

3.21

1

5.25

1

v

mm 1292 v

(in front of the 2(in front of the 2ndnd lens) lens)

2I

mm 1292 v

d

1u


1I

1v 2u

PHYSICS CHAPTER 1

96


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

b. The characteristics of the final image are virtualvirtual invertedinverted magnifiedmagnified formed in front of the 1formed in front of the 1stst and 2 and 2ndnd lenses lenses.

mm; 61.0 mm; 5.25 mm; 0.15 21 dffmm 24.11 u

21mmm where1

11 u

vm

2

22 u

vm and

2

2

1

1

u

v

u

vm

3.21

129

1.24

7.39 m 98.9m

PHYSICS CHAPTER 1

97

Exercise 1.4 :1. a. A glass of refractive index 1.50 plano-concave lens

has a focal length of 21.5 cm. Calculate the radius of the concave surface.

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

calculate 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 lens forms 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 of the lens.

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

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

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

PHYSICS CHAPTER 1

98

3. When a small light bulb is placed on the left side of a converging lens, a sharp image is formed on a screen placed 30.0 cm on the right side of the lens. When the lens is moved 5.0 cm to the right, the screen has to be moved 5.0 cm to the left 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 cm4. A converging lens of focal length 8.00 cm is 20.0 cm to the left

of a converging lens of focal length 6.00 cm. A coin is placed 10.0 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; 0.9245. A converging lens with a focal length of 4.0 cm is to the left of

a second identical lens. When a feather is placed 12 cm to the left of the first lens, the final image is the same size and orientation as the feather itself. Calculate the separation between the lenses.

ANS. :ANS. : 12.0 cm12.0 cm

99

PHYSICS CHAPTER 1

Next Chapter…CHAPTER 2 :Physical optics

matriculation physics ( geometrical optics )

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light light

angle of reflection

refraction reflection

types of reflection

reflection of lightis

image distance

vertical plane mirror