mirrors and lenses optics 18/05/33 1 t.norah ali al-moneef king saud university 24-1 flat mirrors...

Mirrors and Lenses

Optics

18/05/331T.Norah Ali Al-moneef

king Saud university

24-1 Flat mirrors

24-2 Thin Lenses

Regular vs. Diffuse Reflection

Smooth, shiny surfaces have a regular reflection:

Rough, dull surfaces have a diffuse reflection.

Diffuse reflection is when light is scattered in different directions



ReflectionWe describe the path of light as straight-line raysReflection off a flat surface follows a simple rule:

angle in (incidence) equals angle out (reflection)angles measured from surface “normal” (perpendicular)

1 )The incident ray,

the reflected ray

and the normal all lie in the same plane.

Laws of reflection

normal

incident ray reflected ray

mirror

ˊ

2)The incident angel = the reflected angel



24-1 Flat mirrors



An object viewed using a flat mirror appears to be located behind the mirror, because to the observer the diverging rays from the source appear to come from behind the mirror

The image distance behind the mirror equals the object distance from the mirror The image height h’ equals the object height h so that the lateral magnification

The image has an apparent left-right reversal The image is virtual, not real!

S S

1- Virtual images - light rays do not meet and the image is always upright or right-side-up“ and also it cannot be projected Image only seems to be there

Real images - always upside down and are formed when light rays actually meet



example

• If the angle of incidence of a ray of light is 42owhat is each of the following?

A-The angle of reflection (42o)

B-The angle the incident ray makes with the mirror (48o)

C-The angle between the incident ray and the reflected

(90o)



7

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

a) In front of the mirror.

b) On the surface of the mirror.

C)Behind the mirror.



T.Norah Ali Al-moneef king Saud university8

Example

A 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

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Signs: Image size and magnification

images can be upright (positive image size h’) or inverted (negative image size h’)Define magnification m = h’/hPositive magnification: image orientation unchanged relative to objectNegative magnification: image inverted relative to objectl m l < 1 if image is smaller than objectl m l > 1 if image is bigger than objectl m l = 1 if image is same size as object



24-2 Thin Lenses



A lens is a transparent material made of glass or plastic that refracts light rays and focuses (or appear to focus) them at a point

A converging lens will bend incoming light that is parallel to the principal axis toward the principal axis.Any lens that is thicker at its center than at its edges is a converging lens with positive f.

A diverging lens will bend incoming light that is parallel to the principal axis away from the principal axis.Any lens that is thicker at its edges than at its center is a diverging lens with negative f 18/05/3311

T.Norah Ali Al-moneef king Saud university

Rules For Converging Lenses1) Any incident ray traveling parallel to the

principal axis of a converging lens will refract through the lens and travel through the focal point on the opposite side of the lens.

2) Any incident ray traveling through the focal point on the way to the lens will refract through the lens and travel parallel to the principal axis.

3) An incident ray which passes through the center of the lens will in effect continue in the same direction that it had when it entered the lens.



S

S

S-

S-

S-

s

ssf

111




king Saud university15

21

11)1(

1

RRn

n

f medium

lens

The lens maker’s formula )lens in a medium)

)n lens = index of refraction of the lens material( )nmedium= index of refraction of the medium(

1 2

1 1 1( 1)n

f R R

1 2

1 1 1( 1)n

f R R

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

)n = index of refraction of the lens material(

Ray Diagram for Converging Lens, Ray Diagram for Converging Lens, S > S > ff

The image is realrealThe image is invertedinvertedThe image is on the back sideon the back side of the lens

18/05/33 17T.Norah Ali Al-moneef



king Saud university18

1 1 1

p q f

1 1 1

p q f

qpf

q p

qpf

q p

qf

pq f

qfp

q f

pf

qp f

pfq

p f

Ray Diagram for Converging Lens, Ray Diagram for Converging Lens, SS < < ff

The image is virtualvirtualThe image is uprightuprightThe image is largerlarger than the objectThe image is on the front sidethe front side of the lens



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 2FImage is located between F and 2F

Real; inverted; diminished



Object at 2F

FF

FF

2F2F

2F2F

Real; inverted; same size

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 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

FF

FF

2F2F

2F2F

Real; inverted; enlarged

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

2 .The image is real; formed by actual light rays on opposite side3 .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

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

FF

FF

2F2F

2F2F

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 lensImage is located on near side of lens



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.

S = 15 cm; f = 25 cm

(15 cm)(25 cm)

15 cm - 25 cm

pfq

p f

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



. ExampleExample Where must an object be placed to have unit magnification )

M = 1.00) (a) for a converging lens of focal length 12.0 cm ? (b) for a diverging lens of focal length 12.0 cm ?

cmss

ss

ssf

24

2

12

1

11

12

1

111

ba

cmss

ss

ssf

24

2

12

1

11

12

1

111



exampleexample

A person uses a converging lens that has a focal length of 12.5 cm to inspect a gem. The lens forms a virtual image 30.0 cm away.

Determine the magnification. Is the image upright or inverted?

4.382.8

)30(

82.830

11

5.12

1

111

M

ss

ssf

solutionsolutionSince 0M ,the image is upright .



exampleA ray that starts from the top of an object and runs parallel to the axis of the lens, would then pass through the

a)principal focus of the lens

b)center of the lens

C)secondary focus of the lens



Example 5: Derive an expression for calculating the magnification of a lens when the object distance and focal length are given.

From last equation: = -s MSubstituting for q in second equation gives. . .

Thus . . . ,

fss

1

'

11 s

s

h

hM

fs

sfs

fs

fM

)( fss

sfsM

s



Diverging Thin Lens

Incoming parallel rays DIVERGE from a common point FOCALWe still call this the pointSame f on both sides of lensNegative focal lengthThinner in center



Ray Diagrams for Thin Lenses – Ray Diagrams for Thin Lenses –

DivergingDiverging

For a diverging lensdiverging lens, the following three rays are drawn:Ray 1Ray 1 is drawn parallel to the principal parallel to the principal

axisaxis and emerges directed away from the away from the focal point on the front sidefocal point on the front side of the lens

Ray 2Ray 2 is drawn through the centerthrough the center of the lens and continues in a straight linecontinues in a straight line

Ray 3Ray 3 is drawn in the direction toward the direction toward the focal point on the back sidefocal point on the back side of the lens and emerges from the lens parallel to the parallel to the principal axisprincipal axis 18/05/3331

T.Norah Ali Al-moneef king Saud university

Ray Diagram for Diverging LensRay Diagram for Diverging Lens

The image is virtualvirtualThe image is uprightuprightThe image is smallersmallerThe image is on the front sidethe front side of the lens



Sign Conventions for Thin Lenses

QuantityPositive When

Negative When

Object locatio (s)Object is in front of the lens

Object is in back of the lens

Image location (sˊ)Image is in back of the lens

Image is in front of the lens

Image height (h’)Image is uprightImage is inverted

R1 and R2Center of curvature is in back of the lens

Center of curvature is in front of the lens

Focal length (f)Converging lensDiverging lens



Example :

An object is placed 10 cm from a 15-cm-focal-length converging lens. Determine the image position and size (a) analytically, and (b) using a ray diagram.

. An object placed within the focal point of a converging lens produces a virtual image.Solution: a. The thin lens equation gives = -30 cm and M = 3.0. The image is virtual, enlarged, and upright.b. See the figure.

S ’



Example.

Where must a small insect be placed if a 25-cm-focal-length diverging lens is to form a virtual image 20 cm from the lens, on the same side as the object?

Since the lens is diverging, the focal length is negative. The lens equation gives = 100 cm.S



The power of lens

The reciprocal of the focal length = the power of lens

)(

1

mfP

If the focal length f is measured in meters then ;p measured in diopters

if two lenses with focal length f1 and f2 placed next to each other are equivalent to a single lens with a focal length f satisfying

21

21

111

PPP

fff



Spherical AberrationResults from the focal points of

light rays far from the principle axis are different from the focal points of rays passing near the axis



Spherical Aberration

41

With SA SA free



Chromatic AberrationDifferent wavelengths of light refracted

by a lens focus at different pointsViolet rays are refracted more than

red raysThe 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



Multiple lenses can be used to improve aberrationsSpherical Aberration

Chromatic Aberration



Lens Aberrations

Chromatic aberration can be improved by combining two or more lenses that tend to cancel each other’s aberrations. This only works perfectly for a single wavelength, however.




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

+ =1

s1 1f

=

s’

_

16

1 14

=

s’

- s’ =12 cm

Negative means real, inverted image

M = - 12 / 6 = -2

18/05/33


1s

1s’

1f

+ =

18/05/33

T.Norah Ali Al-moneef king Saud university47 18/05/33

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

--4040 cmcm

++2020 cmcm

n = 1.5n = 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 (+) lensConverging (+) 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 ? =

00

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)



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

FF

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




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 = cms

cmcmcms

cmcmsfs

cmf

cms

2020

1

20

1

20

21

20

1

10

1111

10

20

18/05/33


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/

02

4

1

4

1111

4

111

(concave) 4

ssm

cms

cmcmsfs

cms

fss

cmf

18/05/33


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.

m 10

10 1

Example

fss

111

10

1

10

1

5

11

s

cms 10

s

s

h

hm

18/05/33


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

f15

1

10

1

30

11

s

cms 1518/05/33

fss

111

s

s

h

hm


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 = +255

cmscmcmcms

cmcmsfs

cmf

cms

1010

1

10

2

10

11

5

1

10

1111

10

5

fss

111

18/05/33

T.Norah Ali Al-moneef king Saud university56 18/05/33

Summary: Lensmaker’s Equation

.1.1RR11 and and RR22 are positive for convex outward surface and negative are positive for convex outward surface and negative for concave surfacefor concave surface..

.2.2 Focal length Focal lengthff is positive for converging and negative for is positive for converging and negative fordiverging lensesdiverging lenses..

.1.1RR11 and and RR22 are positive for convex outward surface and negative are positive for convex outward surface and negative for concave surfacefor concave surface..

.2.2Focal length Focal length ff is positive for converging and negative for is positive for converging and negative for diverging lensesdiverging lenses..

R1 R2

+

-

R1 and R2 are interchangeable

1 2

1 1 1( 1)n

f R R

1 2

1 1 1( 1)n

f R R

R1, R2 = Radii

n= index of glass

f = focal length



Summary of Math Approach

FF

FF

2F2F

2F2F

f

y

-y’

Lens Equation: Magnification:

SSf

111

S

S

h

hM



SS

Summary of Sign Convention

1. Object p and image q distances are positive for real and images negative for virtual images.

2. Image height y’ and magnification M are positive for erect negative for inverted images

3. The focal length f and the radius of curvature R is positive for converging mirrors and negative for diverging mirrors.

SSf

111

S

S

h

hM



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:

SSf

111

fS

fSS

fS

SfS

SS

SSf

mirrors and lenses optics 18/05/33 1 t.norah ali al-moneef king saud university 24-1 flat mirrors...

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