Name:__________________________

Physics Chapter 14 Study Guide-----------------------------------------------------------------------------------------------------Useful Information:

!

c = 3 "108 m

s

!

1

i+1

o=1

f

!

M =h '

h=#i

o

!

n =c

v

!

ni sin$i = nr sin$r

!

sin$c =nr

ni-----------------------------------------------------------------------------------------------------A Basic Understanding:•Chapter 14 Odd Numbered Core Problems•Chapter 14 Review Problems:! 3, 10, 11, 17, 25, 29, 38, 41, 45, 57.•Ray Tracing for Lenses at least one problem.

Pushing to be Better:•Chapter 14 Your Try Problems•Ray Tracing for Lenses all 3 problems. •Chapter 14 Even Numbered Core Problems•Both supplemental assignment problems on the Refraction Lab.•Chapter 14 Review Problems: 1, 2, 4, 12, 14, 18, 20, 21, 24, 26, 27, 28, 30, 32, 34, 35, 36, 39, 43, 46, 47, 49, 50, 51, 55, 58, 62, 64.•Standardized Test Prep 1-10.

Digging Deeper:•Chapter 14 Review Problems: 53, 54, 59, 60, 65.•Individual Inquiry

Notes on Chapter 14 Equations:

!

1

i+1

o=1

f No change from Ch. 13. In the book they use p and q. That seems harder

than it has to be. i is the distance from the mirror to the image, o is the distance from the mirror to the object. f is the focal length of the mirror.

!

M =h '

h=#i

o Same substitutions as above. M is the magnification of the image. h’ is

the height of the image. h is the height of the object. Note, a positive magnification means that the image is upright and virtual. A negative magnification means the image is inverted and real. A magnification greater than one means the image is larger than the object. A magnification less than one means the image is smaller than the object.

!

n =c

v. The index of refraction (n) is a ratio between the speed of light in a vacuum (c)

and the speed of light in a material (v). Materials such as air, water, glass and plastic can Physics Ch.14 Study Guide page 1 of 2

slow down light, which causes the path of the light to bend or “refract.”

!

ni sin"i = nr sin"r This is Snell’s Law. When light travels from one material to another, you can use this equation to figure out what happens to that light. Remember to always measure your angles from the NORMAL!!! The index of refraction for one material, times the sine of the angle (as measured off normal) of the incident ray of light is equal to the index of refraction for the other material times the sine of the angle (again, measured off normal) of the refracted ray of light.

!

sin"c =nr

ni

. Sometimes, when a ray of light tries to go from a substance with a high

index of refraction, to a substance with a low index of refraction, Snell’s Law can’t return an answer. This is because the refracted ray would have to be greater that 90°, which can’t happen. Instead, what does happen is called “Total Internal Reflection.” The ray of light, instead of crossing the boundary, reflects off the boundary as if it were a perfect mirror. The angle where the switch occurs from Refraction to Total Internal Reflection is called the Critical Angle (

!

"c ). For angles less than

!

"c the light will refract.

For angles greater than

!

"c total internal reflection will occur.

!

ni is the index of

refraction of the material where the light is coming from.

!

nr is the index of refraction of the material the light is trying to go to.

Physics Ch.14 Study Guide page 2 of 2

Physics Chapter 14 Your Try Problems (Based on various worked examples.)

1) Video 14d: Watch the supplemental video “Snell's law of Refraction” and record at least four data points. Then use Snell’s Law to calculate the index of refraction of the material he used in the video.

2) Videos 14e-g: Show the path of the laser at at least two surface interactions. (See diagram below.)

Laser

n(air)=1.0

n(glass) = 1.7

Name:___________________

Refraction Lab Physics

Pin a piece of polar-graph paper to the cardboard. Place the water lens on the graph-paper. Place a pin on the flat side of the lens. Now, looking through the water (not over the water!!) place a pin on the round side of the lens so that optically it lines up the other pin and the center of the water lens.

Do this for 5 different angles.

Measure and record all angles on the paper and in a data table.

Use your data and Snell’s Law to calculate the index of refraction (n) for water.

Supplemental Assignment:Copy each diagram onto a separate sheet of graph paper. Calculate and trace the path of the light ray through the glass.

Radius = 4cmn = 1.35

Light Ray

2cm {

45°

45° 35°

Light Ray

n = 1.51) 2)

Physics Refraction Lab page 1 of 1

Name:___________________

Ray Tracing for Lenses Physics

Attached to the back of this are three example problems for ray-tracing. There are also good diagrams on pages 497, 498 and 500 of the text. But there is no substitute for doing a bunch of these yourself.

I want you to create ray tracing diagrams for the three problems listed below. I also want you to solve these problems mathematically using the thin lens equations.

Do them neatly on graph paper. Since you are using a graph to solve the problem, your grade will include a consideration of how neat your graphs are. You will want to do them once to learn, and then recopy them to make them neater.

The rules for Ray Tracing in lenses is almost identical as the rules for mirrors.

Ray Tracing Rules:Rule 1: A ray of light entering parallel to the optical axis, leaves through the focus.Rule 2: A ray of light entering along a line through the focus, leaves parallel to the optical axis. (opposite of Rule 1) Rule 3: A ray of light entering along a line through the center of the lens, leaves along that same line.

!

1

i+1

o=1

f In the book they use p and q. That seems harder than it has to be. i is the

distance from the lens to the image, o is the distance from the lens to the object. f is the focal length of the lens.

!

M =h '

h="i

o Same substitutions as above. M is the magnification of the image. h’ is

the height of the image. h is the height of the object. Note, a positive magnification means that the image is upright and virtual. A negative magnification means the image is inverted and real. A magnification greater than one means the image is larger than the object. A magnification less than one means the image is smaller than the object.

1) An object with a height of 5cm is placed 10cm from a convex lens with a focal length of 6cm. Find the location and magnification of the image.

2) An object with a height of 3cm is placed 6cm from a convex lens with a focal length of 10cm. Find the location and magnification of the image.

3) An object with a hegiht of 10cm is placed 10cm from a concave lens with a focal length of 10cm. Find the location and magnification of the image.

Physics Ray Tracing Lenses page 1 of 4

Ob

ject

h

= 5

cm

Imag

eh

= -

10cm

f =

10c

mO

pti

cal A

xis

Ru

le 1

Ru

le 2

Ru

le 3

o =

15c

m

i =

30

cm

Wo

rked

Exam

ple

#1:

Use

Ray

Tra

cin

g a

nd

th

e T

hin

Len

s E

qu

atio

n t

o d

eter

min

e th

e lo

cati

on

of

an i

mag

e fo

rmed

by

an

ob

ject

p

lace

d 1

5cm

fro

m a

co

nca

ve

mir

ror

wit

h a

fo

cal

len

gth

of

10cm

.

f =

10c

m

Physics Ray Tracing Lenses page 2 of 4

Imag

ei

= -

10cm

(vir

tual

)Wo

rke

d E

xa

mp

le #

2:

Use

Ray

Tra

cin

g a

nd

th

e T

hin

Len

s E

qu

atio

n t

o d

eter

min

e th

e lo

cati

on

o

f an

im

age

form

ed b

y a

n o

bje

ct p

lace

d 5

cm f

rom

a c

on

cav

e m

irro

r w

ith

a f

oca

l le

ng

th o

f 10

cm.

Ru

le 3

Ob

ject

f =

10c

mi

= -

10cm

Co

nv

erg

ing

L

en

s

Op

tica

l Axi

s

Ru

le 1

Ru

le 2

o =

5cm

Physics Ray Tracing Lenses page 3 of 4

f =

-10

cm

Div

erg

ing

L

en

s

Op

tica

l Axi

s

Ru

le 1

Ru

le 2

Ru

le 3

o =

6cm

i =

-3.

75 c

m

Wo

rke

d E

xa

mp

le #

3:

Use

Ray

Tra

cin

g t

o d

eter

min

e th

e lo

cati

on

of

an i

mag

e fo

rmed

by

an

ob

ject

p

lace

d 6

cm f

rom

a d

iver

gin

g l

ens

wit

h a

fo

cal

len

gth

of

-10c

m.

f =

-10

cm

Physics Ray Tracing Lenses page 4 of 4

Answers to Chapter 14 Your Try Problems

1) Video 14d: Watch the supplemental video “Snell's law of Refraction”...Acceptable answers for the index of refraction should be between 1.46 and 1.51.

2) Videos 14e-g: Show the path of the laser at at least two surface interactions. (See diagram below.)

Laser

n(air)=1.0

n(glass) = 1.7

Name:__________________________

Physics Chapter 14 Core Problems

1) (5pts) An object is placed 15cm from a converging lens with a focal length of 10cm. This image will be:a) Uprightb) In front of the lens.c) Virtuald) a, b, and c are correct.e) none of the other choices are correct.

2) (5pts) A diverging lens:a) always produces a blurry image.b) has a focal length that is negative.c) none of the other choices are correct.d) always produces a virtual image.e) both b and d.

3) (2pts) When light is refracted by droplets of water in the atmosphere to form a rainbow, the color of light that is bent the most is:a) redb) greenc) yellowd) blacke) violet

Physics Ch. 14 Core Problems page 1 of 2

4) (13pts) Trace the path of the laser through the piece of glass in the diagram below. Label all angles! The index of refraction is given on the diagram.

laser

n = 1.50

5) (13pts) An object is located 14cm from a converging lens with a focal length of 6cm. Use both ray tracing and the thin-lens equation to locate the image and give its magnification. (Do the ray tracing on graph paper.)

6) (13pts) An object is located 10cm from a diverging lens with a focal length of -15cm. Use both ray tracing and the thin-lens equation to locate the image and give its magnification. (Do the ray tracing on graph paper.)

7) (9pts) The index of refraction for water is 1.333. a) What is the speed of light in water?b) What is the critical angle for water?c) Describe what happens at the critical angle.

Physics Ch. 14 Core Problems page 2 of 2

A look at Convex Lenses (Converging)not a worked example

(1) (2)

Focus

Optical Axis

Rays of light that are coming in parallel to the optical axis (a line through the center of the lens and perpendicular to both surfaces) are bent TOWARDS the normal (1). This causes them to bend towards each other. Then, when they hit the second surface, they get bent AWAY FROM the normal. But now, the normal is in a different direction, so bending away from normal means that the two rays converge even more. Where they meet is the focus. For a good lens, any ray of light that enters parallel to the optical axis will also pass through the focus, making a sharp point of light.

A look at Concave Lenses (Diverging)not a worked example

(1)(2)

VirtualFocus

Optical Axis

Rays of light that are coming in parallel to the optical axis (a line through the center of the lens and perpendicular to both surfaces) are bent TOWARDS the normal (1). This causes them to bend away from each other. Then, when they hit the second surface, they get bent AWAY FROM the normal. But now, the normal is in a different direction, so bending away from normal means that the two rays keep moving away from each other. The two rays on the right never meet. But, if you drew imaginary line straight back through both of those lines they would meet at an imaginary point which is the ‘focus’ of this lens.