multiple lenses - the burns home page optics assignment.pdfmultiple lenses f= +8 = +12-40 cm -30 cm...

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Slide 2

An object 6 cm tall is placed 40 cm from a thin converging (double convex) lens

of f = 8 cm. A second converging lens of 12 cm focal length is placed 20 cm from

the first lens. Determine the position, size and character of the final image.

Multiple Lenses

f = +8 f = +12

-40 cm -10 cm-20 cm +10 cm +30 cm-30 cm

The image of the first lens can be used as the

object of the second lens

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Slide 3




Multiple Lenses

f = +8 f = +12

-40 cm -10 cm-20 cm +10 cm +30 cm-30 cm

The first lens

is a double

convex lens

1 1 1

s s f

1 1 1

s f s 40

?

8

s

s

f

Object is in front

of lens, so s is

positive

1 1 1

8 40

10

s

s

s„ is positive,

therefore image is

behind the lens

sM

s

10

40

0.25

Since the magnification is

negative, the image is inverted

at (0.25)(6cm)=1.5 cm tall

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Multiple Lenses

f = +8 f = +12

-40 cm -10 cm-20 cm +10 cm +30 cm-30 cm

The second

lens is a

double

convex lens

1 1 1

s s f

1 1 1

s f s 10

?

12

s

s

f

Object is 10 cm

in front of second

lens, so s is

positive

1 1 1

12 10

60

s

s

s„ is negative,,

therefore image is

60 cm in front of

the lens

sM

s

60

10

6


positive, the image is not inverted

and is (1.5cm)(6) = 9 cm tall

Therefore the final

image is inverted at 9

cm tall and is located

40 cm to the left of

the left most lens.

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Slide 5

An object is placed 20 cm from a thin diverging (double concave) lens of f = 10 cm.

A second lens (converging) of 20 cm focal length is placed 10 cm to the right of the

first lens. Determine the position, size and character of the final image.

Multiple Lenses

f = - 10 f = +20

-20 cm 30 cm 40 cm-10 cm

The image of the first lens can be used as the

object of the second lens

20 cm 50 cm 60 cm

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Slide 6




Multiple Lenses

f = - 10 f = +20

-20 cm 30 cm 40 cm-10 cm 20 cm 50 cm 60 cm

The first lens

is a double

concave lens 1 1 1

s f s 20

?

10

s

s

f

Object is in front

of lens, so s is

positive

1 1 1

10 20

6.7

s

s

s„ is negative,

therefore image is

in front of the lens

sM

s

6.67

20

1

3


positive, the image is upright

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Multiple Lenses

f = - 10 f = +20

-20 cm 30 cm 40 cm-10 cm 20 cm 50 cm 60 cm

The second

lens is a

double convex

lens 1 1 1

s f s 16.67

?

20

s

s

f

Object is in front

of lens, so s is

positive

1 1 1

20 16.67

100

s

s

s„ is negative,

therefore image is

in front of the lens

sM

s

100

16.67

6


positive, the image is upright

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Slide 8 Multiple Lenses We determine the effect of a system of lenses by considering the

image of one lens to be the object for the next lens.

For the first lens: s1 = +1.5, f1 = +1

For the second lens: s2 = +1, f2 = -4

\

\

f = +1 f = -4

-1 +3+10 +2 +6+5+4

'

1 3s '

11

1

2s

ms

'

1 1 1

1 1 1 1 11

1.5 3s f s

'

2 0.8s

'

22

2

4

5

sm

s

'

2 2 2

1 1 1 1 1 5

4 1 4s f s

1 2

8

5m m m

In front

behind

In front

In front

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Slide 9 Multiple Lenses Objects of the second lens can be virtual. Let‟s move the second

lens closer to the first lens (in fact, to its focus):

For the first lens: s1 = +1.5, f1 = +1

For the second lens: s2 = -2, f2 = -4

\

\

Note the negative object distance for the 2nd lens.

f = +1 f = -4

-1 +3+10 +2 +6+5+4

1 2 4m m m

'

11

1

2s

ms

'

1 3s

'

1 1 1

1 1 1 1 11

1.5 3s f s

'

22

2

2s

ms

'

2 4s

'

2 2 2

1 1 1 1 1 1

4 2 4s f s

Object opposite

side as light,

therefore negative.

The object for the second lens is VIRTUAL.

Therefore we will use the BST (Burns

Schlueter Theorem) for ray tracing. The

lens will pretend to have the negative of its

focal length and thus opposite properties.

The diverging lens will now pretend to be a

converging lens.

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Problem 1 Suppose we interchange the converging and diverging lenses in the

preceding case.

What is the relation of the new magnification m‟

to the original magnification m ?

• What is the nature of the final image?1B

(c) m’ > m(a) m’ < m (b) m’ = m

(a) real (b) virtual

1A

f = +1f = -4

-1 +3+10 +2 +6+5+4

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Slide

11

Problem 1

1A

-1 +3+10 +2 +6+5+4f = +1f = -4

• Since the formula for the magnification is equal to the product of the magnifications of each lens (m = m 1 m 2), you might think that interchanging the lenses does not change the overall magnification.• This argument misses the point that the magnification of a lens is not a property of the lens, but depends also on the object distance!• Consider the ray shown which illustrates that the magnification must be < 1!

Suppose we interchange the converging and diverging lenses in the preceding case.

What is the relation of the new magnification m‟

to the original magnification m ?

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Slide

12

Problem 1

-1 +3+10 +2 +6+5+4f = +1f = -4

1B

(a) real (b) virtual• What is the nature of the final image?

• The ray used in part A actually shows that the image is real and inverted.

• The equations:

'

2 2 2

1 1 1 11 121

23 23s f s

'

1 1 1

1 1 1 1 1 11

4 1.5 12s f s

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The Microscope

25

angular objective eyepiece

eyepiece objective

M m M

cm L

f f

L

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Slide

14 Microscope Question

A microscope has an objective of focal length 0.300 cm and a eyepiece focal

length of 2.00 cm.

a) Where must the image formed by the objective be for the eyepiece to

produce a virtual image 25.0 cm in front of the eyepiece?

b) If the lenses are 20.0 cm apart, what is the distance of the objective from the

object on the slide?

c) What is the total magnification of the microscope?

d) What distance would the object have to be from a single lens that gave the

same magnification? What would its focal length have to be?

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Slide



length of 2.00 cm.








1 1 1

es s f

1 1 1

25 2

1 1 1

2 25

1 25 2

2 25

50

23

2.17

e

e

e

e

s cm cm

s cm cm

cm cm

s cm cm

s cm

cm

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Microscope Question


length of 2.00 cm.





1 1 1

o os s f

1 1 1

17.8 0.3

1 1 1

0.3 17.8

1 17.8 0.3

0.3 17.8

0.305

o

o

s cm cm

s cm cm

cm cm

s cm cm

s cm

20

20

20 2.17

17.8

o e

o e

s s cm

s cm s

cm cm

cm

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Slide



length of 2.00 cm.








1 2M m m1 2

17.8 25

0.305 1.85

788.65

o e

o e

M m m

s s

s s

cm cm

cm cm

25

25 20

2.00 0.300

833.33

angular objective eyepiece

eyepiece objective

M m M

cm L

f f

cm cm

cm

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Slide



length of 2.00 cm.








sm

s

1 1 1

0.031 25

25 0.031 1

0.031 25

0.775

25.031

0.031

cm cm f

cm cm

cm cm f

f cm

cm

25

788

0.031

cms

cm

1 1 1

o os s f

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The Telescope

objective

angular

eyepiece

fM

f

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Slide

20 Telescope Question

An astronomical telescope has an objective of 50 cm focal length. The

eyepiece has a focal length of 3.5 cm. How far must these lenses be

separated when viewing and object 200 cm from the objective?

1 1 1

o is s f

1 1 1

200 50

1 1 1

50 200

200 50

50 200

66.67

i

i

i

cm s cm

s cm cm

cm cm

cm cm

s cm

Therefore the eyepiece must be placed

so that the principal focus is at to

location of the objective‟s image, to for a

virtual image at infinity. Thus the

separation of the two lenses will be:

66.67 cm+ 3.5 cm = 70.2 cm

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Slide

21 Amazing Eye

One of first organs to develop.

100 million Receptors

200,000 /mm2

Sensitive to single photons!

Candle from 12 miles

Ciliary Muscles

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The Physics of Focusing the Eye

Cornea n= 1.38

Lens n = 1.4

Vitreous n = 1.33

Which part of the eye does most of the light bending?

1) Lens 2) Cornea 3) Retina 4) Cones

Ciliary Muscles

Lens and cornea have similar shape, and index of refraction. Cornea has

air/cornea interface 1.38/1, 70% of bending. Lens has Lens/Vitreous interface

1.4/1.33. Lens is important because it can change shape.

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Slide

23 Eye (Relaxed)

25 mm

Determine the focal length of your eye when looking at an object far away.

1 1 1

25 mm f

os Object is far away:

25is mmImage at retina:

1 1 1

s s f

25 mmrelaxedf

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Slide

24 Eye (Tensed)

25 mm

Determine the focal length of your eye when looking at an object up close (25 cm).

25 cm

Object is up close: 25 250os cm mm

Want image at retina: 25is mm

1 1 1

s s f

1 1 1

250 mm 25 mm f

25 mmrelaxedf Recall:

22.7 mmtensef

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Near Point, Far Point

Eye‟s lens changes shape (changes f )

Object at any do can have image be at retina (di =

approx. 25 mm)

Can only change shape so much

“Near Point”

Closest do where image can be at retina

Normally, ~25 cm (if far-sighted then further)

“Far Point”

Furthest do where image can be at retina

Normally, infinity (if near-sighted then closer)

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Slide

26 If you are nearsighted...

Want to have (virtual) image of distant object, do = , at the far point, di = -dfar.

Too far for near-sighted eye to focus

dfar

Near-sighted eye can focus on this!

Contacts form virtual image at far point –

becomes object for eye.

do

(far point is too close)

flens = -dfar

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Slide

27 Refractive Power of Lens

Diopter = 1/f = POWER where f is focal length of lens in meters.

Person with far point of 5 meters, would need contacts with focal

length –5 meters.

Doctor‟s prescription reads:

1/(-5m) = –0.20 Diopters

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If you are farsighted...

When object is at do, lens must

create an (virtual) image at -dnear.

Want the near point to be at do.

Too close for far-sighted eye to focus

dnear=50 cm

Far-sighted eye can focus on this!

do =25cm

Contacts form virtual image at near

point – becomes object for eye.

(near point is too far)

0

1 1 1

near lensd d f

1 1 1

25 50

50

lenscm cm f

f cm

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Slide

29

The Eye The “Normal Eye”

Far Point distance that relaxed eye can focus onto retina =

Near Point closest distance that can be focused on to the retina

2.5cm

25cm

This is called “accommodation”

Diopter: 1/f where f is in metres

2.5 cmf

2.3 cmf

'

1 1 1 10

2.5 cmf s s

'

1 1 1 1 1

25 2.5 f s s

Therefore the normal eye acts as a lens with a focal length which can vary from 2.5 cm

(the eye diameter) to 2.3 cm which allows objects from 25 cm to infinity to be focused

on the retina

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Slide

30 An intuitive way to view eye corrections

Near-sighted eye is elongated, image of distant object forms in front of retina

Add diverging lens, image forms on retina

Far-sighted eye is short, image of close object forms behind retina

Add converging lens, image forms on retina

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www.jnvkannur.temple.at

DEFECTS OF VISION

Myopia – Short sightedness

Due to enlarged size of the eye ball, the images of

distant objects will be focused in front of the retina. This defect is

known as Myopia.

Remedy is the use of a concave lens as shown.

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Slide

32 Due to the reduced size of the eye ball, the images of nearby objects

will be focused behind the retina. This defect is known as

Hypermetropia.

Remedy is the use of a convex lens as shown.

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Slide

33 This little Piggy

In The Lord of the Flies,

Piggy‟s glasses are used to

focus the Sun‟s rays and

start a fire. What type of lens

do you need for this?

Later in the novel, Piggy‟s

glasses are broken, and

poor Piggy has a hard

time seeing because he is

nearsighted. What type of

lenses where in his

glasses?

Remember, do your research if you are going to be an author.

Convex lens

Concave lens

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Understanding

An arrow-shaped object is placed in front of a plane mirror as shown below.

The image would look like:

a)

b)

c)

d)

e)

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Slide

35 Understanding

An illuminated arrow is placed 2 cm in front of a diverging lens with focal

length of -6 cm. The image is:

a) real, inverted, smaller than the object

b) Virtual, inverted, larger than the object

c) Virtual, upright, larger than the object

d) Real, upright, larger than the object

e) Virtual, upright, smaller than the object

A diverging lens (has a negative focal length) will always create an

upright virtual image in front of the object. Since the image distance

is smaller than the object distance, the image will be smaller as well.

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Slide

36 Understanding

An object is placed in front of three different optical devices, two lenses and a

mirror, with focal points as shown in the figure. Which will produce real

images?

a) I only

b) II only

c) III only

d) I and II

e) II and III

A single concave lens produces only virtual images. An object

placed inside the focal length of a convex lens will result in a virtual

image. This eliminates I and II. An object outside the focal length of

a concave mirror will produce an inverted, real image

I II III

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Understanding

A concave mirror with a radius of curvature 1.5 m is used to collect light from a

distant source. The distance between the image formed and the mirror is

closest to:

a) 0.75 m

b) 1 m

c) 1.5 m

d) 2 m

e) 3 m

Since the object is distant, then the light rays that approach the

mirror are parallel. The focus is r/2 where r is the radius of

curvature. In this example, r=1.5 m, so f =0.75 m

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Slide

38 Understanding

A student sets up an optics experiment with a converging lens of focal length

10 cm. He places an illuminated arrow 2 cm high at 15 cm from the lens axis.

The size of the image:

a) 0.5 cm

b) 1 cm

c) 2 cm

d) 3.5 cm

e) 4 cm

Since the magnification is -2 times. The size of the image is 2 cm x 2

which is 4cm and the image is inverted.

1 1 1

15 10

1 1 1

10 15

1

3

1 1

0

o

i

i

i

i

cm s cm

s cm cm

s c

f

m

s s

0

30

15

2

i

c

cm

sm

m

s

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Slide

39 Understanding

An object is placed in front of a convex mirror. The location of the image is

closest to:

a) A

b) B

c) C

d) D

e) E

A convex (diverging) mirror will produce an upright, smaller, virtual

image.

object

A

B

CD

E

focus

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Understanding

For which of the cases will the image of the arrow be virtual and smaller than

the object:

a) I only

b) II only

c) III only

d) I and II

e) I and III

Diverging elements like I and III will always produce smaller virtual

images. II will produce a virtual image, but it will be larger. This is

basically a magnifying glass.

I II III

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Slide

41 Free Response Problem

The figure above shows an enlarged portion of the glass wall of a fish tank,

currently empty so that air (n=1) is on either side of the glass (n=1.5). The glass is

0.5 cm thick. A ray R is incident on the glass at a 300 angle with the normal as

shown.

a) On the left figure, continue the ray, showing qualitatively what happens at the

next interface.

b) At what distance above the normal line N will the transmitted ray emerge out of

the glass?

c) Determine the incident angle at the second interface that will ensure total

internal reflection. Could the initial ray R have its incident angle adjusted to

make this happen?

d) Suppose the tank is filled with water (n=1.33) as on the right figure. Show

qualitatively what happens at the glass water interface

0.5cm 0.5cm

glass glassair airair waterN N

300 300

RR

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Slide





shown.

a) On the left figure, continue the ray, showing qualitatively what happens at the

next interface.

0.5cm 0.5cm

glass

glass

air

air

air

waterN N

300 300

RR

The reflected light will also have an angle of reflection of 300

300

1

sin sin

sin

1sin sin 30

1.5

19.5

i i r r

ir i

r

r

n n

nsin

n

1

sin sin

sin

1.5sin sin 19.5

1

30

i i r r

ir i

r

r

n n

nsin

n

Thus the ray

exiting back into

air is parallel to

the original ray

300

19.50

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Free Response Problem

0.5cm 0.5cm

glass

glass

air

air

air

waterN N

300 300

RR

We can determine d from the geometry

300

0.5 tan 19.5

tan 19.50.5

0.18

d c

d

cm

m

cm

300

19.50




shown.

b) At what distance above the normal line N will the transmitted ray emerge out of

the glass?

d

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Slide


0.5cm 0.5cm

glass

glass

air

air

air

waterN N

300 300

RR

We require the critical angle

300

1

1 1sin

1.5

41

s

.

n

8

i rc

i

n

n

19.50




shown.

c) Determine the incident angle at the second interface that will ensure total

internal reflection. Could the initial ray R have its incident angle adjusted to

make this happen?

Since the ray that exits into the air has to

exit at 900 to be total reflected, and that we

have the incoming ray parallel to the

outgoing ray. We would need to incoming

ray have an incident angle of 900, thus

indicting that it would not enter the glass,

so we could not make it happen.

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Slide


0.5cm 0.5cm

glassglass

airair

airwater

N N

300 300

RR

300

19.50




shown.

d) Suppose the tank is filled with water (n=1.33) as on the right figure. Show

qualitatively what happens at the glass water interface

1

sin sin

sin

1.5sin sin 19.5

1.33

22.1

i i r r

ir i

r

r

n n

nsin

n

The ray exiting

into water is

NOT parallel to

the original ray,

with the angle of

refraction now

being:

19.50

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A converging lens with focal length 4 cm has an object placed 6 cm in front of

it. A diverging lens with focal length – 8 cm is placed 28 cm behind the first

lens.

a) Determine the position of the image formed by the first lens.

b) Draw a ray diagram needed to display the image from first lens.

c) What is the magnification of the image?

d) Determine the position of the image formed by the second lens.

e) Draw a ray diagram needed to display the image from the second lens.

f) Determine the overall magnification and image orientation of final image

4 16 28 480

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lens.


4 16 28 480

1 1 1

o is s f

2

1 1 1

6 4

1 1 1

4 6

6 4

24

12

i

i

i

cm s cm

s cm cm

cm cm

cm

s cm

Therefore the image is at the

12 cm + 8 cm = 20 cm position

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Slide




lens.

b) Draw a ray diagram needed to display the image from first lens.

4 16 28 480

Therefore the image is at the

12 cm + 8 cm = 20 cm position

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lens.

c) What is the magnification of the image?

i

o

sm

s

12

6

2

i

o

sm

s

cm

cm

Therefore the image is twice

as big and is inverted.

4 16 28 480

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Slide




lens.


Therefore the virtual image is located

at 36 cm - 5.33 cm = 30.67 cm

4 16 28 480

1 1 1

o is s f

2

1 1 1

16 8

1 1 1

8 16

16 8

128

5.33

i

i

i

cm s cm

s cm cm

cm cm

cm

s cm

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Slide




lens.

e) Draw a ray diagram needed to display the image from the second lens.

4 16 28 480

Therefore the final virtual

image is inverted.

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lens.

f) Determine the overall magnification and image orientation of final image

2 i

o

sm

s

2

5.332

16

0.667

i

o

sm

s

cm

cm

Therefore the image is 2/3 the

original size and is inverted.

4 16 28 480

Magnification

from first lens

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Slide


Two thin converging lenses of focal lengths 10 cm and 20 cm are separated

by 20 cm. An object is placed 15 cm in front of the first lens.


b) What is the magnification of the image?

c) Draw a ray diagram needed to display the image from first lens.


e) Determine the overall magnification and image orientation of final image.

f) Draw a ray diagram needed to display the image from the second lens.

5 10 30 500 20 25 4540 55 60

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Slide







5 10 30 500 20 25 4540 55 60

1 1 1

o is s f

1

1

1

1 1 1

15 10

1 1 1

10 15

30

i

i

i

cm s cm

s cm cm

s cm

1i

o

sM

s 1

30

15

2

cmM

cm

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5 10 30 500 20 25 4540 55 60

1 1 1

o is s f

2

2

2

1 1 1

10 20

1 1 1

20 10

26

3

i

i

i

cm s cm

s cm cm

s cm

2i

o

sM

s

2

26

3

10

2

3

cm

Mcm

1 2fM M M 2 4

23 3

fM

This line is back projected through the centre of lens 2 (which

would have come from lens 1 back to the object)

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Slide


A converging lens with focal length 25 cm has an object placed 150 cm in

front of it. A diverging lens with focal length – 15 cm is placed 20 cm behind

the first lens.







20 60 100 2400 140 22040 80 110

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Slide




the first lens.




20 60 100 2400 140 22040 80 110

1 1 1

o is s f

1

1

1

1 1 1

150 25

1 1 1

25 150

30

i

i

i

cm s cm

s cm cm

s cm

1i

o

sM

s 1

30

150

0.2

cmM

cm

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the first lens.




1 1 1

o is s f

2

2

2

1 1 1

10 15

1 1 1

15 10

30

i

i

i

cm s cm

s cm cm

s cm

2i

o

sM

s 2

30

10

3

cmM

cm

20 60 100 2400 140 22040 80 110

1 2fM M M 0.2 3 0.6fM

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multiple lenses - the burns home page optics assignment.pdfmultiple lenses f= +8 = +12-40 cm -30 cm...

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