Textbook sections 26-3 – 26-5, 26-8 Physics 1161: Lecture 17 Refraction & Lenses.

• Textbook sections 26-3 – 26-5, 26-8

Physics 1161: Lecture 17

Refraction & Lenses

Physics 1161: Lecture 17, Slide 2

Indices of Refraction

CheckpointRefraction

n1

n2

When light travels from one medium to another the speed changes v=c/n, but the frequency is constant. So the light bends:

q1

q21) n1 > n2

2) n1 = n2

3) n1 < n2

Compare n1 to n2.

n1

n2

Compare n1 to n2.

q2

q1

1) n1 > n2

2) n1 = n2

3) n1 < n2

q1 < q2

sinq1 < sinq2

n1 > n2

Which of the following is correct?

n1 sin(q1)= n2 sin(q2)

CheckpointRefraction

• A ray of light crossing the boundary from a fast medium to a slow medium bends toward the normal. (FST)

• A ray of light crossing the boundary from a slow medium to a fast medium bends away from the normal. (SFA)

FST & SFA

n1

n2

Snell’s Law Practice

n1sin1n2 sin2

norm

al

2

A ray of light traveling through the air (n=1) is incident on water (n=1.33). Part of the beam is reflected at an angle qr = 60. The other part of the beam is refracted. What is q2?

1 r

Usually, there is both reflection and refraction!

n1

n2

Snell’s Law Practice

n1sin1n2 sin2

norm

al

2

A ray of light traveling through the air (n=1) is incident on water (n=1.33). Part of the beam is reflected at an angle qr = 60. The other part of the beam is refracted. What is q2?

sin(60) = 1.33 sin(q2)

q2 = 40.6 degrees

q1 =qr =60

1 r

Usually, there is both reflection and refraction!

Refraction Applets

• Applet by Molecular Expressions -- Florida State University

• Applet by Fu-Kwung Hwang, National Taiwan Normal University

Parallel light rays cross interfaces from air into two different media, 1 and 2, as shown in the figures below. In which of the media is the light traveling faster?

1 2 3

0% 0%0%

1

air air

2

1. Medium 12. Medium 23. Both the same

Parallel light rays cross interfaces from air into two different media, 1 and 2, as shown in the figures below. In which of the media is the light traveling faster?

1 2 3

0% 0%0%

1

air air

2

1. Medium 12. Medium 23. Both the same

The greater the difference in the

speed of light between the two

media, the greater the bending of the

light rays.

Parallel light rays cross interfaces from medium 1 into medium 2 and then into medium 3. What can we say about the relative sizes of the indices of refraction of these media?

1 2 3 4 5

0% 0% 0%0%0%

1

3

21. n1 > n2 > n3

2. n3 > n2 > n1

3. n2 > n3 > n1

4. n1 > n3 > n2

5. none of the above

Parallel light rays cross interfaces from medium 1 into medium 2 and then into medium 3. What can we say about the relative sizes of the indices of refraction of these media?

1 2 3 4 5

0% 0% 0%0%0%

1

3

2

1. n1 > n2 > n3

2. n3 > n2 > n1

3. n2 > n3 > n1

4. n1 > n3 > n2

5. none of the aboveRays are bent toward the normal when crossing into #2, so n2 >

n1. But rays are bent away from the normal when going into

#3, so n3 < n2. How to find the relationship between #1 and #3?

Ignore medium #2! So the rays are bent away from the normal if they would pass from #1 directly into #3. Thus, we have: n2

> n1 > n3 .

Apparent Depth

• Light exits into medium (air) of lower index of refraction,  and turns left.

Spear-Fishing

• Spear-fishing is made more difficult by the bending of light.

• To spear the fish in the figure, one must aim at a spot in front of the apparent location of the fish.

n2

n1

d

d

d d

n2

n1

Apparent depth:

Apparent Depth

50

actual fish

apparent fish

To spear a fish, should you aim directly at the image, slightly above, or slightly below?

1 2 3

0% 0%0%

1. aim directly at the image

2. aim slightly above3. aim slightly below

To spear a fish, should you aim directly at the image, slightly above, or slightly below?

1 2 3

0% 0%0%

1. aim directly at the image

2. aim slightly above3. aim slightly below

Due to refraction, the image

will appear higher than the

actual fish, so you have to aim

lower to compensate.

To shoot a fish with a laser gun, should you aim directly at the image, slightly above, or slightly below?

1 2 3

0% 0%0%

1. aim directly at the image

2. aim slightly above3. aim slightly below

laser beam

light from fish

The light from the laser beam

will also bend when it hits the

air-water interface, so aim

directly at the fish.

Delayed Sunset

• The sun actually falls below below the horizon

• It "sets", a few seconds before we see it set.

Broken Pencil

Water on the Road Mirage

Palm Tree Mirage

Mirage Near Dana – Home of Ernie Pyle

Texas Mirage

Looming

Antarctic Looming

Looming

Looming

Types of Lenses

Lens Terms

Three Rays to Locate Image

• Ray parallel to axis bends through the focus.

• Ray through the focus bends parallel to axis.

• Ray through center of lens passes straight through.

Characterizing the Image

• Images are characterized in the following way1. Virtual or Real2. Upright or Inverted3. Reduced, Enlarged, Same Size

Object Beyond 2f

• Image is–Real–Inverted–Reduced

Object at 2f

• Image is–Real– Inverted–Same size

Object Between 2f and f

• Image is–Real– Inverted–Enlarged

Object at F

• No Image is Formed!

Object Closer than F

• Image is–Virtual–Upright–Enlarged

Converging Lens Images

Beacon Checkpoint

A beacon in a lighthouse is to produce a parallel beam of light. The beacon consists of a bulb and a converging lens. Where should the bulb be placed?

A. Outside the focal pointB. At the focal pointC. Inside the focal point

Lens in WaterCheckpoint

P.A.

F

Focal point determined by geometry and Snell’s Law:

n1 sin(q1) = n2 sin(q2)

Fat in middle = ConvergingThin in middle = DivergingLarger n2/n1 = more bending, shorter focal length.n1 = n2 => No Bending, f = infinity

Lens in water has _________ focal length!

n1<n2

Lens in WaterCheckpoint

P.A.

F

Focal point determined by geometry and Snell’s Law: n1 sin(q1) = n2 sin(q2)

Fat in middle = ConvergingThin in middle = DivergingLarger n2/n1 = more bending, shorter focal length.n1 = n2 => No Bending, f = infinity

Lens in water has larger focal length!

n1<n2

Half Lens Checkpoint

A converging lens is used to project a real image onto a screen. A piece of black tape is then placed over the upper half of the lens.

How much of the image appears on the screen?

1. Only the lower half will show on screen2. Only the upper half will show on screen3. The whole object will still show on screen

Half LensCheckpoint

A converging lens is used to project a real image onto a screen. A piece of black tape is then placed over the upper half of the lens.

Half LensCheckpoint

Still see entire image (but dimmer)!

Two very thin converging lenses each with a focal length of 20 cm are are placed in contact. What is the focal length of this compound lens?

1 2 3

0% 0%0%

1. 10 cm2. 20 cm3. 40 cm

Two very thin converging lenses each with a focal length of 20 cm are are placed in contact. What is the focal length of this compound lens?

1 2 3

0% 0%0%

1. 10 cm2. 20 cm3. 40 cm

Concave (Diverging) Lens

• Ray parallel to axis refracts as if it comes from the first focus.

• Ray which lines up with second focus refracts parallel to axis.

• Ray through center of lens doesn’t bend.

Image Formed by Concave Lens

• Image is always–Virtual–Upright–Reduced

Concave Lens Image Distance

• As object distance decreases– Image distance

decreases– Image size

increases

Image Characteristics

• CONVEX LENS – IMAGE DEPENDS ON OBJECT POSITION– Beyond F: Real; Inverted; Enlarged,

Reduced, or Same Size– Closer than F: Virtual, Upright, Enlarged– At F: NO IMAGE

• CONCAVE LENS – IMAGE ALWAYS SAME– Virtual– Upright– Reduced

Lens Equations

• convex: f > 0; concave: f < 0

• do > 0 if object on left of lens

• di > 0 if image on right of lens otherwise di < 0

• ho & hi are positive if above principal axis; negative below

1 1 1

o if d d

i i

o o

h dM

h d

do

di

Which way should you move object so image is real and diminished?

1 2 3

0% 0%0%

1. Closer to the lens2. Farther from the lens3. A converging lens can’t

create a real, diminished image.

F

F

Object

P.A.

Which way should you move object so image is real and diminished?

1 2 3

0% 0%0%

1. Closer to the lens2. Farther from the lens3. A converging lens can’t

create a real, diminished image.

F

F

Object

P.A.

Image Object

ImageObject

Object

Image

3 Cases for Converging Lenses

This could be used as a projector. Small slide on big screen

This is a magnifying glass

This could be used in a camera. Big object on small film

UprightEnlargedVirtual

InvertedEnlargedReal

InvertedReducedReal

Inside F

Past 2F

BetweenF & 2F

1) Rays parallel to principal axis pass through focal point.

2) Rays through center of lens are not refracted.

3) Rays toward F emerge parallel to principal axis.

Diverging Lens Principal Rays

F

F

Object

P.A.

Image is (always true): Real or Imaginary

Upright or Inverted

Reduced or Enlarged

1) Rays parallel to principal axis pass through focal point.

2) Rays through center of lens are not refracted.

3) Rays toward F emerge parallel to principal axis.

Diverging Lens Principal Rays

F

F

Object

P.A.

Image is virtual, upright and reduced.

Image

Which way should you move the object to cause the image to be real?

1 2 3

0% 0%0%

1. Closer to the lens2. Farther from the lens3. Diverging lenses can’t

form real images

F

F

Object

P.A.

Which way should you move the object to cause the image to be real?

1 2 3

0% 0%0%

1. Closer to the lens2. Farther from the lens3. Diverging lenses can’t

form real images

F

F

Object

P.A.

Multiple LensesImage from lens 1 becomes object for lens 2

1

f1 f2

2

Complete the Rays to locate the final image.

Multiple LensesImage from lens 1 becomes object for lens 2

1

f1 f2

2

Multiple Lenses: Magnification

f1 f2

do = 15 cm

f1 = 10 cm

di = 30 cmf2 = 5 cm

L = 42 cm

do=12 cm

di = 8.6 cm

1i

o

dm

d

2

8.6

12

cmm

cm

1 2 1.4netm mm

1 2

Net magnification: mnet = m1 m2

30

15

cm

cm 2

.717