Mirrors and Lenses

Optics

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 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 Almoneef8

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

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

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.

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

S

S

S-

S-

S-

s

ssf

111

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

11 . .The image is The image is inverted,inverted, i.e., i.e., opposite to the object opposite to the object orientationorientation..

22 . .The image is The image is real, real, i.e., formed i.e., formed by actual light on the opposite by actual light on the opposite

side of the lensside of the lens . .

33 . .The image is The image is diminished diminished in size, in size, i.e., smaller than the objecti.e., smaller than the object.. Image is located

between F and 2F

Image is located between F and 2F

Real; inverted; diminished

Example 3. 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 thatS- is negative means that the image isvirtual on) (same side as object.

fss

1

'

11

Object Between 2F and F

FF

FF

2F2F

2F2F

Real; inverted; enlarged

11 . .The image is The image is invertedinverted, i.e., , i.e., opposite to the object orientationopposite to the object orientation.. 22 . .The image is The image is realreal; formed by ; formed by

actual light rays on opposite actual light rays on opposite sideside33 . .The image is The image is enlarged enlarged in size, i.e., in size, i.e.,

larger than the objectlarger than the object.. Image is located beyond 2F

Image 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

. 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

fs

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

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

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 Aberration

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

For a mirror, parabolic shapes can be used to correct for spherical aberration

Spherical Aberration

37

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 aberrations

Spherical 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.

T . Norah Ali Almoneef41

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

T . Norah Ali Almoneef42

1s

1s’

1f

+ =

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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

T . Norah Ali Almoneef47

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 =

cmscmcmcms

cmcmsfs

cmf

cms

2020

120

120

2120

110

1111

10

20

T . Norah Ali Almoneef48

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

T . Norah Ali Almoneef49

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.

m10

10 1

Example

fss111

101

101

511

s

cms 10

ss

hh

m

T . Norah Ali Almoneef50

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

T . Norah Ali Almoneef51

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 = +251

cmscmcmcms

cmcmsfs

cmf

cms

1010

110

210

115

110

1111

10

5

fss111

T . Norah Ali Almoneef52