1T. Norah Ali Almoneef : We represent light using rays, which are straight lines emanating from an...

T . Norah Ali Almoneef 1

Chapter 24Mirror,lenses,and imaging systems


24.1 Mirrors:

We represent light using rays, which are straight lines emanating from an object. This is an idealization, but is very useful for geometric optics.


Reflection and RefractionWhen a light ray travels from one medium to

another, part of the incident light is reflected and part of the light is transmitted at the boundary between the two media.

The transmitted part is said to be refracted in the second medium.

incident ray reflected ray

refracted ray


Law of reflection:

the angle of reflection (that the ray makes with the normal to a surface) equals the angle of incidence.

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


(parallel rays will all be reflected in the same directions)

(parallel rays will be reflected in a variety of directions)


True of false: Reflection of light by a rough surface does not obey the laws of reflection.

(T/F)

• If the angle of incidence of a ray of light is 42°, what is each of the following?

• The angle of reflectiona. The angle the incident ray makes with the mirrorb. The angle between the incident ray and the reflected

example:

example

42°48°

90°


• All the light rays from the source that reflect from the mirror SEEM to come from the same point behind the mirror


Formation of image in a plane mirrorFormation of image in a plane mirror

Stand in front of a looking glass and look at your image.

2. Is the image erect or inverted?

1. Can you receive your image on a screen ?

3. Is the image the same size or larger or smaller?

6. Where is the image formed ?

4. What happens when you tilt your head to the right?5. How does the image move when you step forward or

backward


Types of Images for Mirrors and Lenses• A real image is one in which light actually passes

through the image point– Real images can be displayed on screens

• A virtual image is one in which the light does not pass through the image point– The light appears to come (diverge) from that point– Virtual images cannot be displayed on screens


plane Mirror

• Simplest possible mirror• Properties of the image

can be determined by geometry

• One ray starts at P, follows path PQ and reflects back on itself

• A second ray follows path PR and reflects according to the Law of Reflection

p=q


Notations and plane Mirror

• The object distance is the distance from the object to the mirror or lens– Denoted by p

• The image distance is the distance from the image to the mirror or lens– Denoted by q

• The lateral magnification of the mirror or lens is the ratio of the image height (h ’) to the object height (h)– Denoted by (M =h’/h)


Properties of the Image Formed by a Flat Mirror• The image is as far behind the mirror as the object

is in front (q = p)• The image is unmagnified

– The image height is the same as the object height (h’ = h and M = 1)

• The image is virtual (cannot be picked up on a screen)

• The image is upright– It has the same orientation as the object

• There is an apparent left-right reversal in the image


LEFT- RIGHT REVERSAL

AMBULANCE


Now you look into a mirror and see the image of yourself.

In front of the mirror.

On the surface of the mirror.

Behind the mirror.


ExampleA girl can just see her feet at the bottom edge of the mirror.

Her eyes are 10 cm below the top of her head.

150 m

150 m

(a) What is the distance between the girl and her image in the mirror? Distance = 150 2 = 300 cm


example

A boy of height 1.5 m stands 5 m in front of a plane mirror.

His image is ______ m tall and ______ m from him.1.5 10


exampleWhich of the following descriptions about the image formed by a plane mirror is INCORRECT ?

A Light rays come from the image to our eyes.

B The image cannot be projected on a screen.

C It is as far away from mirror as the object is in front.

D It is virtual.


Light Traveling through Materials• All electromagnetic waves have the same speed c = 3.00×108 m/s

in a vacuum• In a material medium (glass, water, etc.), an EM wave travels at a

speed v < c• An EM wave generally travels at different speeds in different

materials• The change in speed as a light ray goes from one material to

another causes the ray to deviate from its incident direction• This change in direction is called refraction


•Refraction is when waves speed up or slow down due to travelling in a different medium ,and it can cause light rays to change their direction

•A medium is something that light waves will travel through

•Light rays are slowed down by the water

•Causes the ruler to look bent at the surface

•The mediums in this example are water and air

The degree that light bends when it enters a new medium is called the “index of refraction”


24.2 Thin Lenses

• Lenses are commonly used to form images by refraction• Lenses are used in optical instruments

– Cameras, telescopes, microscopes• Images from lenses

– Light passing through a lens experiences refraction at two surfaces– The image formed by one refracting surface serves as the object for the

second surface

A thin lens is one whose thickness t is small in comparison to distances of optical properties (radius of curvature, focal length, image and object distances. Light is reflected from a mirror. Light is refracted through a lens. For a thin lens, the thickness, t, of the lens can be neglected

t

Rfocal point


• Used in cameras, telescopes, human eye

1. Used in cameras & telescopes to correct spherical aberation, and also eyeglasses

• These are examples of converging lenses

• They have positive focal lengths

• They are thickest in the middle

• These are examples of diverging lenses

• They have negative focal lengths• They are thickest at the edges


Lenses• Formed by two curved boundaries between transparent

media.• Lenses may have spherical surfaces (lens-maker’s equation).

Most modern lenses have non-spherical curved surfaces to avoid spherical aberration.


• The distance of F from C is the focal length f of the lens.focal length

FC

the two focal lengths are equal


• The line through the optical centre and 2 foci is called the principal axis.

principal axis

FF'C

• Refracted rays appear to spread from a point called the principal focus F.

• Parallel rays are refracted outwards.

principal focus

optical centre

focal length


Focal Length of a Converging Lens

• The parallel rays pass through the lens and converge at the focal point

– Lens that converges (brings together) light rays. – Forms real images and virtual images depending on

position of the object


Focal Length of a Diverging Lens

• The parallel rays diverge after passing through the diverging lens

• The focal point is the point where the rays appear to have originated

– Diverges light rays

– All images are erect and reduced.


The equations can be used for both converging and diverging lensesA converging lens has a positive focal lengthA diverging lens has a negative focal length

Lens Equations

Q

M

Q’

M’

F F’

x x’f f’s s’

T

A

S

h

h’


Q’TS and F’TA are similar triangles, h’ + hs’

=hf’

QTS and FAS are similar triangles, h’ + h

sh’f

=


h’ + hs’

=hf’

h’ + hs

h’f

=

Adding the two equations:

h’ + hs’

+hf’

h’ + hs

h’f

= +

Since f = f’ for a thin lens,

h’ + hs’

+h’ + h

s=

h’ + hf

Multiplying through by 1h + h’

1s

1s’

1f

+ = Lens Formula

T . Norah Ali Almoneef 33T . Norah Ali Almoneef 33

Magnification of Images Through A Thin Lens• The lateral magnification of the image is

• When M is positive, the image is upright and on the same side of the lens as the object

• When M is negative, the image is inverted and on the side of the lens opposite the object

ss

hh

M

ss

hh

M


Images by Refraction

Image can be captured by a screen.Image can be captured by a screen.

screen

O IHence called ‘real’.Hence called ‘real’.

Image formation by a lensA Real images

O I

Light rays converge to a point.Light rays converge to a point.

Since only convex lenses converge light rays, real images can only be formed by convex lenses

Image formation by a lens

Light rays diverge from a point.Light rays diverge from a point.

b Virtual images

O

I

convex lens

No rays actually come from the image.No rays actually come from the image.

Hence called ‘virtual’.Hence called ‘virtual’.

Image formation by a lens

Light rays diverge from a point.Light rays diverge from a point.

b Virtual images

OI

concave lens

Image Construction:Ray 1: A ray parallel to lens axis passes through the far focus of a converging lens or appears to come from the near focus of a diverging lens.

Ray 1: A ray parallel to lens axis passes through the far focus of a converging lens or appears to come from the near focus of a diverging lens.

Converging Lens Diverging Lens

F

Ray 1

F

Ray 1

Image Construction:Ray 2: A ray passing through the near focal point of a converging lens or proceeding toward the far focal point of a diverging lens is refracted parallel to the lens axis.

Ray 2: A ray passing through the near focal point of a converging lens or proceeding toward the far focal point of a diverging lens is refracted parallel to the lens axis.


F

Ray 1

F

Ray 1

Ray 2

Ray 2

Image Construction:Ray 3: A ray passing through the center of any lens continues in a straight line. The refraction at the first surface is balanced by the refraction at the second surface.

Ray 3: A ray passing through the center of any lens continues in a straight line. The refraction at the first surface is balanced by the refraction at the second surface.


F

Ray 1

F

Ray 1

Ray 2

Ray 2

Ray 3

Ray 3


The Thin-Lens Equation

This gives us the thin-lens approximation, as well as the magnification:

s

sM

'

s

sM

'

magnification


C1 C2

C1 & C2 are the centres of the spheres of which the surfaces of the lens form a part

The line through C1 & C2 form the principal axis

Converging Lens

Principal axis


C1 C2

Converging lens

FF

A ray of light on entering the lens is refracted towards the normal and on leaving away from the normal. Since surfaces are inclined towards each other the ray is refracted towards the principal axis.

Rays parallel to the principal axis converge towards a point called the principal focus F.

Since light can travel equally well in both directions, there are two foci.

Normal


F = Focal point

To locate the position of an image in a convex lens we use two of the following rays of light

FF

3 through the focus emerging parallel to principal axis.

1 parallel to the principal axis emerging through focus

2 striking the centre of the lens passes straight through (if lens is thin)


F = Focal point

object

Images in Convex lens

ss

f

ImageReal, inverted & diminished

Image formed in convex lens when the object is placed outside twice the focal length

FF

ssf

1

1

1


Image formed in convex lens when the object is placed at twice focal length

F = Focal point

object

Image Real, inverted & same size as object


ss

f

FF

ssf

1

1

1


Image formed in convex lens when the object is placed at the focus

F = Focal point

object

Image at Infinity


s

f

FF

ssf

1

1

1


Image formed in convex lens when the object is placed inside the Focus

F = Focal point

object


s

f

Image Virtual, magnified & upright

FF

ssf

1

1

1


raybox

lens

screen

s s

Experiment to find the focal length of a convex lens


F = Focal point

1. A ray which strikes the lens travelling parallel to principal axis is refracted as if it came from focus

2. A ray striking the centre of the lens

passes straight through (if lens is thin)

3. A ray heading for the focus on striking the

lens is refracted parallel to principal

axis

To locate the position of an image in a concave lens we use two of the following rays of light

Convave lens

FF


Image formed in concave lens when the object is placedin front of lens

F = Focal point

object

Images in Concave lens

s

s

f

FF

ssf

1

1

1

Images Tracing PointsDraw an arrow to represent the location of an object, then draw any two of the rays from the tip of the arrow. The image is where lines cross.

Draw an arrow to represent the location of an object, then draw any two of the rays from the tip of the arrow. The image is where lines cross.

3. Is it enlarged, diminished, or same size?

2. Is the image real or virtual?

1. Is the image erect or inverted?

• Real images are always on the opposite side of the lens. Virtual images are on the same side.

Object Outside 2F

1. The image is inverted, i.e., opposite to the object orientation.

2. The image is real, i.e., formed by actual light on the opposite side of the lens.

3. The image is diminished in size, i.e., smaller than the object. Image is located between F

and 2F

Image is located between F and 2F

F

F

2F

2F

Real; inverted; diminished

Object at 2F

F

F

2F

2F

Real; inverted; same size


2. The image is real, i.e., formed by actual light on the opposite side of lens.

3. The image is the same size as the object.Image is located at 2F on other side

Image is located at 2F on other side

Object Between 2F and F

F

F

2F

2F

Real; inverted; enlarged


2. The image is real; formed by actual light rays on opposite side

3. The image is enlarged in size, i.e., larger than the object. Image is located beyond 2FImage is located beyond 2F

Object at Focal Length F

F

F

2F

2F

When the object is located at the focal length, the rays of light are parallel. The lines never cross, and no image is formed.

When the object is located at the focal length, the rays of light are parallel. The lines never cross, and no image is formed.

Parallel rays; no image formed

Object Inside F

F

F

2F

2F

Virtual; erect; enlarged

1. The image is erect, i.e., same orientation as the object.

2. The image is virtual, i.e., formed where light does NOT go.

3. The image is enlarged in size, i.e., larger than the object. Image is located on near side of

lens

Image is located on near side of lens

Diverging Lens Imaging

Diverging Lens

F

Diverging Lens

F

All images formed by diverging lenses are erect, virtual, and diminished. Images get larger as object approaches.



An object is placed 6.0 cm in front of a convex thin lens of focal length 4.0 cm. Where is the image formed and what is its magnification and power?

s = 6.0 cm f = 4.0 cm

P = 1

0.04 m= 25.0 D

1s

1s’

1f

+ =1s

1 1f

=

s’

_

16

1 14

=

s’

- s’ = 12 cm

Negative means real, inverted image

M = - 12 / 6 = -2


1s

1s’

1f

+ =


light from

distant object

falls short of retina

light from

distant object

falls on retina

with help of a diverging (concave) lens

Short-sight defect

Corrected


light from

near object

falls ‘behind’ retina

Long-sight defect

Corrected

with help of a converging (convex) lens

falls on retina

light from

near object

Convex and concave lenses

a Converging or Diverging?

convex lens

(converging lens)

concave lens

(diverging lens)

A light ray is incident on a…A light ray is incident on a convex lens.

Which one represents the path of the light ray?

A Path X.

B Path Y.

C Path Z.

X

Y

Z

Q2 A light ray is incident on a…A light ray is incident on a concave lens.

Which one represents the path of the light ray?

A Path X.

B Path Y.

C Path Z.

X

Y

Z


Image Summary • For a converging lens, when the object distance is greater than

the focal length (s > ƒ)– The image is real and inverted

• For a converging lens, when the object is between the focal point and the lens, (s < ƒ)– The image is virtual and upright

• For a diverging lens, the image is always virtual and upright– This is regardless of where the object is placed


Image Formed by a Lens

• The lens has an index of refraction n and two spherical surfaces with radii of R1 and R2

– R1 is the radius of curvature of the lens surface that the light of the object reaches first

– R2 is the radius of curvature of the other surface

• The object is placed at point O at a distance of s in front of the first surface


Signs for Lensmaker’s Equation

1. R1 and R2 are positive for convex outward surface and negative for concave surface.

2. Focal length f is positive for converging and negative for diverging lenses.



R1 R2

+

-

R1 and R2 are interchangeable

1 2

1 1 1( 1)n

f R R

R1, R2 = Radii

n= index of glass

f = focal length


Lensmaker’s Equation

R1 R2

Surfaces of different radius

The Lensmaker’s Equation:

1 2

1 1 1( 1)n

f R R

The focal length f for a lens.The focal length f for a lens.

Negative (Concave)

Positive (Convex)

Sign convention

R

Signs for Lensmaker’s Equation





R1 R2

+

-


1 2

1 1 1( 1)n

f R R

R1, R2 = Radii

n= index of glass

f = focal length

T . Norah Ali Almoneef 72T . Norah Ali Almoneef 72

The quantity, f, is the focal length of the lens. It’s the single most important parameter of a lens. It can be positive or negative.

f > 0 f < 0

R1 positiveR2 positive

R1 negativeR2 negative

n=1

R1 R2

n≠1n=1

R1 positiveR2 negative

The (concave) lens is considered to have R1 and R2 both negative, and so f must be negative

The NORMAL (convex) lens above is considered to have R1 and R2 both positive, and so f must be positive


The quantity, f, is the focal length of the lens. It’s the single most important parameter of a lens. It can be positive or negative.

f > 0 f < 0

R1 positiveR2 positive

R1 negativeR2 negative

n=1

R1 R2

n≠1n=1

R1 positiveR2 negative

The (concave) lens is considered to have R1 and R2 both negative, and so f must be negative

The NORMAL (convex) lens above is considered to have R1 and R2 both positive, and so f must be positive


Focal Length for a Lens(Lens Makers’ Equation)• The focal length of a lens is related to the

curvature of its front and back surfaces and the index of refraction of the material

• If the medium is air

• ( n for the lens)This is called the lens maker’s equation

21

11)1(

1RR

nf

f1

= }R1

R1

{ x 1}-nn

{21medium

lens

s1

s1

= }R1

R1

{ x 1}-nn

{21air

lens

Summary: Lensmaker’s Equation





R1 R2

+

-


1 2

1 1 1( 1)n

f R R

R1, R2 = Radii

n= index of glass

f = focal length

Example 1. A glass meniscus lens (n = 1.5) has a concave surface of radius –40 cm and a convex surface whose radius is +20 cm. What is

the focal length of the lens.

R1 = 20 cm, R2 = -40 cm

-40 cm

+20 cm

n = 1.51 2

1 1 1( 1)n

f R R

1 1 1 2 1(1.5 1)

20 cm ( 40 cm 40 cmf

f = 80.0 cmf = 80.0 cm Converging (+) lens.

Example: What must be the radius of the curved surface in a plano-convex lens in order that the focal length be 25 cm?

R1 = , f= 25 cm

2

1 1 1( 1)n

f R

R1= R2=?

f = ?

0

2 2

1 1 0.500(1.5 1)

25 cm R R

R2 = 12.5 cmR2 = 12.5 cm Convex (+) surface.

R2 = 0.5(25 cm)

Be careful with substitution of signed numbers!Be careful with substitution of signed numbers!

Alternative SolutionsIt might be useful to solve the lens equation algebraically for each of the parameters:

fss

1

'

11

fs

sfs

'fs

fss

ss

ssf

Example. A magnifying glass consists of a converging lens of focal length 25 cm. A bug is 8 mm long and placed 15 cm from the

lens. What are the nature, size, and location of image.

F

F

p = 15 cm; f = 25 cm

s = -37.5 cm

The fact that S is negative means that the image is virtual (on same side as object).

The fact that S is negative means that the image is virtual (on same side as object).

fss

1

'

11

cmcmcmcmcm

fssf

s 5.3725152515

Example 3 Cont.) A magnifying glass consists of a converging lens of focal length 25 cm. A bug is 8 mm long and placed 15 cm from

the lens. What are size of image.

F

F

s = 15 cm; s = -37.5 cm

h’ = +20 mm

The fact that h’ is positive means that the image is erect. It is also larger than object.

The fact that h’ is positive means that the image is erect. It is also larger than object.

y’ y

s

s

h

hM

''

cmcm

mmh

155.37

8

Example : What is the magnification of a diverging lens (f = -20 cm) the object is located 35 cm from the center of the lens?

F

First we find q . . . then M

s = +12.7 cm

M = +0.364

fss

1

'

11

cmcmcmcmcm

fssf

s 7.12)20(35

2035

s

s

h

hM

''

364.035

)7.12('

cmcm

ss

M

SummaryA Converging lens is one that refracts and converges parallel light to a real focus beyond the lens. It is thicker near the middle.

A Converging lens is one that refracts and converges parallel light to a real focus beyond the lens. It is thicker near the middle.

FF

A diverging lens is one that refracts and diverges parallel light which appears to come from a virtual focus in front of the lens.

A diverging lens is one that refracts and diverges parallel light which appears to come from a virtual focus in front of the lens.

The principal focus is denoted by the red F.FF

Summary of Math Approach

F

F

2F

2F

p

f

q

y

-y’

Lens Equation: Magnification:

s

s

h

hM

''fss

1'

11

Terms for Image Construction


• The near focal point is the focus F on the same side of the lens as the incident light.

• The far focal point is the focus F on the opposite side to the incident light.

F

Near focus

F

Near focus

F

Far focus

F

Far focus


Image PropertiesConvex vs. Concave Lenses


Image Properties Sign Conventions-

Diverging Lens Imaging

Diverging Lens

F

Diverging Lens

F



Analytical Approach to Imaging

F

F

2F

2F

s

f

h

-h’

Lens Equation: Magnification:

fss

1

'

11 s

s

h

hM

''

s’


Parallel ray

focus (f)focus (f) 2 f2 fx x x x

Focal ray

Image is:RealInvertedReducedAppears between f and 2f

Object beyond 2f



Focal ray

Parallel ray

Image is:RealInvertedSame sizeAppears between f and 2f

Object at 2f



Focal ray

Parallel ray

Image is:RealInvertedEnlargedAppears beyond 2f

Object betweenf and 2f


focus (f)focus (f) 2 f2 f

x x x x

Image is:VirtualErectEnlargedAppears on sameSide as Object

ApparentConvergence

Of rays

ObjectInsidefocus


f2 f f 2 fFocal ray

Parallel ray

Ray thru center

Image is:VirtualErectReducedAppears on sameSide as object


Thin Lens Equation: sign conventions

s s’

fss

1

'

11f

object image

s is positive for objects to the left of lens, negative for objects to the right of lens (virtual objects).

s’ is positive for images to the right of lens, negative for images to the left of lens (virtual images).

f is positive for converging lenses, negative for diverging lenses.

A It gets larger till it gets totally blurred at some distance.

B It gets larger, keeping erect all the way.C It gets smaller and becomes totally blurred at some

distance.

D It gets smaller, keeping erect all the way.

A boy holds a magnifying glass at arm’s lengthHe looks at a poster through the glass and sees a magnified erect image. What happens to the image if he moves the lens closer to his eyes?

If you can capture an image...If you can capture an image of a doll on a screen using a lens,

which of the following may NOT be correct?

A The lens you use is a convex lens.B The image is magnified.

C The image is real.

D The image is erect.

In the diagram, a ray parallel...In the diagram, a ray parallel to the principal axis of the lens is reflected backwards.

What is the focal length of the cylindrical convex lens?

A 5 cm

B 10 cm

C 20 cm

D 40 cm

10 cm

What happens to the image…What happens to the image when the plane mirror is moved backwards?

A The image becomes blurred.

B The image becomes smaller.

C The image does not change.

Both convex and concave...

Both convex and concave lenses can produce _______ images, which must be _________ than the object if convex lenses are used.

virtual larger


General Image Trends real images are always inverted virtual images are always upright real images are always in frontof the mirror virtual images are alwaysbehind the mirror negative image distance means virtual image positive image distance means real image

101T .Norah Ali Almoneef


1. The image will be as it was, but much dimmer.2. The image will be right-side-up and sharp.3. The image will be right-side-up and blurry.4. The image will be inverted and blurry.5. There will be no image at all.

A lens produces a sharply-focused, inverted image on a screen. What will you see on the screen if the lens is removed?


The focal length of a converging lens is

1. the distance at which an image is formed.2. the distance at which an object must be

placed to form an image.3. the distance at which parallel light rays

are focused.4. the distance from the front surface to the

back surface.


ExampleAn object is placed 20 cm in front of a converging lens of

focal length 10 cm. Where is the image? Is it upright or inverted? Real or virtual? What is the magnification of the image?

Real image, magnification = -1

cmscmcmcms

cmcmsfs

cmf

cms

2020

120

120

2120

110

1111

10

20


ExampleAn object is placed 8 cm in front of a diverging lens

of focal length 4 cm. Where is the image? Is it upright or inverted? Real or virtual? What is the magnification of the image?

05.0/

0241

41111

4

111

(concave) 4

ssm

cmscmcmsfs

cmsfss

cmf


24(b). Given a lens with a focal length f = 5 cm and object distance p = +10 cm, find the following: i and m. Is the image real or virtual? Upright or inverted? Draw 3 rays.

Image is real, inverted.

. .F1 F2s

Virtual side Real side

m 10

10 1

Example

fss111

101

101

511

s

cms 10

ss

hh

m


24(e). Given a lens with the properties (lengths in cm) R1 = +30, R2 = +30, s = +10, and n = 1.5, find the following: f, s and m. Is the image real or virtual? Upright or inverted? Draw 3 rays.

cmf 30

m 15

101.5

Image is virtual, upright.

Virtual side Real side

R1. .F1 F2

pR2

21

111

1RR

nf

301

301

301

15.11

f

ss

hh

m

sfs111

151

101

3011

s

cms 15


ExampleAn object is placed 5 cm in front of a converging lens of

focal length 10 cm. Where is the image? Is it upright or inverted? Real or virtual? What is the magnification of the image?

Virtual image, as viewed from the right, the light appears to be coming from the (virtual) image, and not the object.

Magnification = +2109

cmscmcmcms

cmcmsfs

cmf

cms

1010

110

210

115

110

1111

10

5

fss111


Examples

A diverging lens with f = -20 cmh = 2 cm, s = 30 cm

cms 12

The image is virtual and uprightcmh

hM

8.0

4.030

12

2

A converging lens with f = 10 cm

(a) S = 30 cm

cmss

151011

301

(b) S = 10 cm

(c) s = 5 cm

2510

ss

M

The image is real and inverted

The image is virtual and upright

The image is at infinity

fss

1

'

11

2011

301

s

ss

hh

M

5.03015

ss

M

s

cmss10

1011

51


The thin lens equation,, can be used to find the image distance:

5.03015

ss

M

fss

1

'

11

cmsscmcmcm

ssf

15

130

230

110

1

1

1 -

1


cmsscmcmcm

ssf

10

110

15

110

1

1

1 -

1


Repeat Example for a diverging lens of focal length 10.0 cm.Solution(A) We begin by constructing a ray diagram as in Figure36.31a taking the object distance to be 30.0 cm. The diagramshows that we should expect an image that is virtual,smaller than the object, and upright. Let us now apply thethin lens equation with p = 30.0 cm:

cmsscmcmcm

ssf

5.7

130

430

1101

1

1 -

1

25.030

)5.7(

cmss

M


QUICK QUIZAn object is placed to the left of a converging

lens. Which of the following statements are true and which are false? (a) The image is always to

the right of the lens. (b) The image can be upright or inverted. (c) The image is always

smaller or the same size as the object..

(a) False. A virtual image is formed on the left side of the lens if s < f.

(b) True. An upright, virtual image is formed when s < f, while an inverted, real image is formed when s

> f.

(c) False. A magnified, real image is formed if 2f > s > f, and a magnified, virtual image is formed if s

> f.


When a real object is placed just inside the focal point F of a diverging lens, the image is

A.virtual, erect, and diminished. B.real, inverted, and enlarged. C.real, inverted, and diminished. D.virtual, erect, and enlarged. E.virtual, inverted, and diminished.


An object is placed in front of a plano-concave lens at r1/2. The image produced by the lens is

A.inverted, real and reduced in size.B.inverted, virtual and enlarged in size.C.upright, virtual and reduced in size.D.upright, virtual and enlarged in size.E.upright, real and enlarged in size.

r1

119

Which of the following statements is false?

A.The image produced by a diverging lens is always virtual, upright and reduced in size.

B.The image produced by a converging lens can be virtual, upright and magnified in size.

C.The image produced by a converging lens cannot be virtual, upright and reduced in size.

D.The image produced by a converging lens cannot be real, inverted and reduced in size.

T .Norah Ali Almoneef


To project an image onto a screen using a lens,

A. the lens must be diverging and the object must be farther from the lens than the second focal point.

B. the lens must be converging and the object must be between the first focal point and the lens.

C. the lens must be diverging and the image must be farther from the lens than the second focal point.

D. the lens must be converging and the object must be farther from the lens than the first focal point.

E. the lens must be diverging and the object must be between the first focal point and the lens.


A real image is formed by a converging lens. If a weak diverging lens is placed between the

converging lens and the image, where is the new

image located?

A. farther from the converging lens than the original image

B. closer to the converging lens than the original image C. at the original image position


Example• An object is 32 cm to the left of a convex lens of +8.0 cm

focal length. – Where is the image located?

– Is the image• real or virtual• upright or inverted• magnified or reduced

• Image is located +11 cm from lens• Real reduced (m = -.33)• inverted (m is negative)


Example

• An object is located 4.0 cm to the left of a convex lens, the focal length is 6.0 cm. Is the object– real or virtual– magnified or reduced– upright or inverted

• Virtual (s = -12 cm)• Magnified (m = 3)• Upright (m is positive)

T .Norah Ali Almoneef 124

T .Norah Ali Almoneef 125

One ray is shown as it leaves an object placed before a positive lens. If this ray were continued to show its path through the lens, it would pass

through which point? (F marks the two focal points.)


A point object O is placed in front of a thin converging lens. F marks the two focal points. Observers are at 1, 2, and 3. The image of point O is seen by the

A.observer at 1 only. B.observer at 2 only. C.observer at 3 only. D.observers at 1, 2, and 3. E.observers at 1 and 2.


A ray of light leaves point O and passes through a thin positive lens. It crosses the principal axis at which point? (F marks the two focal points.)


When the ray in the diagram is continued through the diverging lens, it passes through which point? (F marks the two focal points.)


The image of the encircled point on the object formed in the positive lens is at which circle? (F

marks the two focal points.)


The image produced by the converging lens is at which point? (F marks the two focal points.)


The image of the object formed by the diverging lens is located at which point? (F marks the two

focal points.)


A concave (diverging) lens can produce an image that is

A.virtual, inverted, and magnified. B.real, erect, and magnified. C.diminished, erect, and virtual. D.magnified, erect, and virtual. E.diminished, real, and erect.


A converging lens and a screen are so arranged that an image of the sun falls on the screen. The

distance from the lens to the screen is

A.the focal length. B.the object distance. C.the magnifying power. D.one-half the radius of curvature of one

of the lens faces. E. the average radius of curvature of the

two lens faces. 134T .Norah Ali Almoneef


The "power," P, of a lens is equivalent to the focal length, f. One can be found from the other.

• Definition of P in terms of f is

• Meaning of P• P is a measure of the ray

bending power of the lens• Large P means the lens

bends rays more than if P were small

• Your eyeglass or contact lens prescription is usually given in diopters (P)

P (in diopters) 1

f (meters)

• The power of a converging lens is always positive because f is a positive number for a converging lens– The converging lens always bends

rays towards the axis behind the lens• The power of a diverging (concave)

lens is always negative because f is negative for a diverging lens– The diverging lens always bends rays

away from the axis behind the lens


24.4 the power of a Lens, aberations

• For lens in contact (separation is negligible)

power)each of sum is(power

111

:or 111

21

21

21

PPP

fff

fff


Lens and Mirror Aberrations

• One of the basic problems is the imperfect quality of the images– Largely the result of defects in shape and form

• Two common types of aberrations exist• (a) Spherical aberration: Rays passing through different

regions of a lens and do not come together in a common focal plane

• (B)Chromatic Aberration: Different dispersion of red and blue


Chromatic Aberration• Different wavelengths of light refracted by a lens

focus at different points• Violet rays are refracted more than red rays so the

focal length for red light is greater than the focal length for violet light

• Chromatic aberration can be minimized by the use of a combination of converging and diverging lenses


Chromatic aberration can be minimized using additional lenses

In an A chromat, the second lens cancels the dispersion of the first.A chromats use two different materials, and one has a negative focal length.

This adds to the expense and is one reason why good cameras are so expensive


Chromatic aberration can be minimized using additional lensesIn an A chromat, the second lens cancels the dispersion of the first.

A chromats use two different materials, and one has a negative focal length.


Spherical Aberration

Results from the focal points of light rays farfrom the principle axis are different from thefocal points of rays passing near the axis For a mirror, parabolic shapes can be used tocorrect for spherical aberration

Spherical aberration can be also minimized using additional lenses

The additional lenses cancel the spherical aberration of the first.


Multi-element lenses• Are used to reduce aberration.

1T. Norah Ali Almoneef : We represent light using rays, which are straight lines emanating from an...

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