an object 6 cm tall is placed 40 cm from a thin converging (double convex) lens of f = 8 cm. a...

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


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 4010

ss

s‘ is positive, therefore image is behind the lens

sM

s

10

400.25

Since the magnification is negative, the image is inverted at (0.25)(6cm)=1.5 cm tall


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 1060

ss

s‘ is negative,, therefore image is 60 cm in front of the lens

sM

s

60

106

Since the magnification is 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.

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


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


1 1 1

10 206.7

ss

s‘ is negative, therefore image is in front of the lens

sM

s

6.67

201

3

Since the magnification is positive, the image is upright


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


1 1 1

20 16.67100

ss

s‘ is negative, therefore image is in front of the lens

sM

s

100

16.676

Since the magnification is positive, the image is upright

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

'1

11

2s

ms

'1 1 1

1 1 1 1 11

1.5 3s f s

'2 0.8s

'2

22

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

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

'1

11

2s

ms

'1 3s

'1 1 1

1 1 1 1 11

1.5 3s f s

'2

22

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.

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

preceding case. 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

Problem 1

1A

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

Problem 1

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

The Microscope

25

angular objective eyepiece

eyepiece objective

M m M

cm L

f f

L

The object (O) to be magnified is placed just beyond the focus of the objective lens. A real image (I) is then formed between the two lenses as shown, which acts as an object for the eyepiece lens. The real image is virtual, inverted and magnified.

Microscope QuestionA 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?


1 1 1

es s f

1 1 1

25 2

1 1 1

2 25

1 25 2

2 25

50

271.85

e

e

e

e

s cm cm

s cm cm

cm cm

s cm cm

s cm

cm

Microscope QuestionA 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?

1 1 1

o os s f

1 1 1

18.15 0.3

1 1 1

0.3 18.15

1 18.15 0.3

0.3 18.15

0.305

o

o

o

o

s cm cm

s cm cm

cm cm

s cm cm

s cm

20

20

20 1.85

18.15

o e

o e

s s cm

s cm s

cm cm

cm


1 2M m m1 2

18.15 25

0.305 1.85

804.2

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


sm

s

1 1 1

0.032 25

25 0.032 1

0.032 25

0.032

cm cm f

cm cm

cm cm f

f cm

25

804

0.032

cms

cm

1 1 1

o os s f

The Telescope

objectiveangular

eyepiece

fM

f

Telescope QuestionAn 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

Amazing Eye One of first organs to develop.One of first organs to develop. 100 million Receptors 100 million Receptors

200,000 /mm200,000 /mm22

Sensitive to single photons!Sensitive to single photons! Candle from 12 milesCandle from 12 miles

Ciliary Muscles

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.

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

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

Near Point, Far Point Eye’s lens changes shape (changes Eye’s lens changes shape (changes ff ) )

Object at any do can have image be at retina (di = approx. 25 mm)

Can only change shape so muchCan only change shape so much ““Near Point” Near Point”

Closest do where image can be at retina Normally, ~25 cm (if far-sighted then further)

““Far Point”Far Point” Furthest dFurthest doo where image can be at retina where image can be at retina Normally, infinity (if near-sighted then closer)Normally, infinity (if near-sighted then closer)

If you are nearsighted (myopia)...

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

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 of –5 meters (negative because diverging lens).

Doctor’s prescription reads:

1/(-5m) = –0.20 Diopters

If you are farsighted (hypermetropia)...

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

50lenscm cm f

f cm

The Eye The “Normal Eye”The “Normal Eye”

Far Point Far Point distance that relaxed eye can focus onto retina distance that relaxed eye can focus onto retina == Near Point Near Point closest distance that can be focused on to the retina 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

An intuitive way to view eye correctionsNear-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

www.jnvkannur.temple.atwww.jnvkannur.temple.at

DEFECTS OF VISIONMyopia – 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.

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.

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 were in his glasses?

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

Convex lens

Concave lens

UnderstandingAn arrow-shaped object is placed in front of a plane mirror as shown below. The image would look like:

a)

b)

c)

d)

e)

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 objectb)Virtual, inverted, larger than the objectc)Virtual, upright, larger than the objectd)Real, upright, larger than the objecte)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.

UnderstandingAn 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 onlyb)II onlyc)III onlyd)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

UnderstandingA 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 mb)1 mc)1.5 md)2 me)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

UnderstandingA 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 cmb)1 cmc)2 cmd)3.5 cme)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

152

i

c

cm

sm

m

s

UnderstandingAn object is placed in front of a convex mirror. The location of the image is closest to:

a)Ab)Bc)Cd)De)E

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

object

A

B

CD

E

focus

UnderstandingFor which of the cases will the image of the arrow be virtual and smaller than the object:

a)I onlyb)II onlyc)III onlyd)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

Free Response ProblemThe 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

Free Response ProblemThe 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.

0.5cm 0.5cm

glassglass

airair

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

Free Response Problem

0.5cm 0.5cm

glassglass

airair

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

cmm

cm

300

19.50

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

d


0.5cm 0.5cm

glassglass

airair

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

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.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.


0.5cm 0.5cm

glassglass

airairair

water

N N300 300

RR

300

19.50

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.

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

Free Response ProblemA 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


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

2412

i

i

i

cm s cm

s cm cm

cm cm

cms cm

Therefore the image is at the 12 cm + 8 cm = 20 cm position


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


c) What is the magnification of the image?

i

o

sm

s

12

62

i

o

sm

s

cm

cm

Therefore the image is twice as big and is inverted.

4 16 28 480


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

1285.33

i

i

i

cm s cm

s cm cm

cm cm

cms cm


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.


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

Free Response ProblemTwo 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.

a)Determine the position of the image formed by the first lens.b)What is the magnification of the image?c)Draw a ray diagram needed to display the image from first lens.d)Determine the position of the image formed by the second 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


a)Determine the position of the image formed by the first lens.b)What is the magnification of the image?c)Draw a ray diagram needed to display the image from first lens.

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

152

cmM

cm


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

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

310

2

3

cmM

cm

1 2fM M M 2 42

3 3fM

This line is back projected through the centre of lens 2 (which would have come from lens 1 back to the object)


a)Determine the position of the image formed by the first lens.b)What is the magnification of the image?c)Draw a ray diagram needed to display the image from first lens.d)Determine the position of the image formed by the second 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.

20 60 100 2400 140 22040 80 110


a)Determine the position of the image formed by the first lens.b)What is the magnification of the image?c)Draw a ray diagram needed to display the image from 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

1500.2

cmM

cm


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

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

103

cmM

cm

20 60 100 2400 140 22040 80 110

1 2fM M M 0.2 3 0.6fM

an object 6 cm tall is placed 40 cm from a thin converging (double convex) lens of f = 8 cm. a...

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