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Law of Reflection

Dispersion

Snells Law

Brewsters Angle

Preliminary topics before

mirrors and lenses

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Law of ReflectionDispersion

Snells Law

Brewsters Angle

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Geometrical Optics:Study of reflection and refractionof light from surfaces

The ray approximation states that light travels in straight lines

until it is reflected or refracted and then travels in straight lines again.

The wavelength of light must be small compared to the size of

the objects or else diffractive effects occur.

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Law of Reflection

1A

B

I = qR

i

r

Mirror

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Drawing Normals

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Fermats Principle

Using Fermats Principle you can prove theReflection law. It states that the path taken

by light when traveling from one point to

another is the path that takes the shortest

time compared to nearby paths.

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Two light rays 1 and 2 taking different paths

between points A and B and reflecting off a

vertical mirror

1

2

A

B Plane Mirror

Use calculus - method

of minimization

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t= 1

( 12 + 2 + 2

2 + ( )2 )

=

2

12 + 2

+2( )

22 + ( )2 )

= 0

y

h12 + 2=

( )

22 + ( )2 )

=

I=

qR

Write down time as a function of y

and set the derivative to 0.

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Law of Refraction: Snells Law

n1 sinq1 = n2 sinq2

n1n2

How do we prove it?

1

v1

sin1=

1

2

2

Air 1.0

Glass 1.33

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t= ( 1112 + 2 + 12

22 + ( )2 )

= 0

11

=12

n1 sin

= 2

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JAVA APPLET

Show Fermats principle simulator

http://www.aug.edu/~chmtmc/ntnujava/Fermat/Fermat.htmlhttp://www.aug.edu/~chmtmc/ntnujava/Fermat/Fermat.html

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What allows you to see various colors when

white light passes through a prism

Dispersion

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How does a Rainbow work?

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Why is light totally reflected inside a fiber

optics cable? Internal reflection

n1

sinq1

= n2

sinq2

(1.33) sinq1 = (1.00) sin 90 = 1.00

When 1 1 1

1.33 48.75

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Fiber Cable

Same here

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Corner Reflector?

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Show Total Internal reflection

simulator

Halliday, Resnick, Walker: Fundamentals of Physics, 7th Edition -

http://bcs.wiley.com/he-bcs/Books?action=chapter&bcsId=2037&itemId=0471216437&chapterId=12950http://bcs.wiley.com/he-bcs/Books?action=chapter&bcsId=2037&itemId=0471216437&chapterId=12950

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What causes a Mirage

1.06

1.09

1.08

1.07 1.07

1.08

1.09

sky eye

Hot road causes gradient in the index of refraction that increases

as you increase the distance from the road

Index of refraction

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Inverse Mirage Bend

Snells Law Example

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47. In the figure, a 2.00-m-long vertical pole extends from the bottom

of a swimming pool to a point 50.0 cm above the water. What is the

length of the shadow of the pole on the level bottom of the pool?

Consider a ray that grazes the top of the

pole, as shown in the diagram below. Here

1 = 35o, l1 = 0.50 m, and l2 = 1.50 m.

2

1

l2

l1

L x

air

water

shadow

x is given by

x = l1tan 1 = (0.50m)tan35o = 0.35 m.

The length of the shadow is x + L.

L is given by

L=l2tan 2

Use Snells Law to find 2

Snells Law Example

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According to the law of refraction, n2sin 2 =

n1sin 1. We take n1 = 1 and n2 = 1.33

oo

n55.25

33.1

35sinsin

sinsin 1

2

11

2 =

=

=

L is given by

.72.055.25tan)50.1(tan 22 mmlLo ===

The length of the shadow is L+x.

L+x = 0.35m + 0.72 m = 1.07 m.

2

1

l2

l1

L x

air

water

shadow

Calculation of L

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Polarization by Reflection Brewsters Law

(1.)sin = + = 90 100% = (90 ) = = 1

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Plane Mirrors Where is the image formed

Mirrors and Lenses

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Plane mirrors

Normal

Angle of

incidence

Angle of reflection

i = - p

Real side Virtual side

Virtual image

eye

Object distance = - image distance

Image size = Object size

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Problem: Two plane mirrors make an angle of 90o. How

many images are there for an object placed between

them?

object

eye

1

2

3

mirror

mirror

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Problem: Two plan mirrors make an angle of 60o. Find all

images for a point object on the bisector.

object

2

4

5,6

3

1

mirror

mirror

eye

U i th L f R fl ti t

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pocket

Assuming no spinAssuming an elastic collision

No cushion deformation

d

d

Using the Law of Reflection to

make a bank shot

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What happens if we bend the mirror?

i = - p magnification = 1

Concave mirror.

Image gets magnified.

Field of view is diminished

Convex mirror.

Image is reduced.

Field of view increased.

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Rules for drawing images for mirrors

Initial parallel ray reflects through focal point.Ray that passes in initially through focal point reflects parallel

from mirrorRay reflects from C the radius of curvature of mirror reflects along

itself. Ray that reflects from mirror at little point c is reflected

symmetrically

1

p +

1

=1

=

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Concept Simulator/Illustrations

Halliday, Resnick, Walker: Fundamentals of Physics, 7th Edition -

S h i l f i f

http://bcs.wiley.com/he-bcs/Books?action=chapter&bcsId=2037&itemId=0471216437&chapterId=12950http://bcs.wiley.com/he-bcs/Books?action=chapter&bcsId=2037&itemId=0471216437&chapterId=12950

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Spherical refracting surfaces

n1

p+2

=2 1

Using Snells Law and assuming small

Angles between the rays with the central

axis, we get the following formula:

A l thi ti t Thi L h th thi k i

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Apply this equation to Thin Lenses where the thickness is

small compared to object distance, image distance, and

radius of curvature. Neglect thickness.

Converging lens

Diverging lens

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Thin Lens Equation

1

f=

1

p+

1

i

Lensmaker Equation

1

f= (n - 1)(

1

r1-1

r2)

What is the sign convention?

Lateral Magnification for a Lensm =

Sign Convention

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Sign Convention

p

Virtual side - V Real side - R

i

Light

Real object - distance p is pos on V side (Incident rays are diverging)

Radius of curvature is pos on R side.

Real image - distance is pos on R side.

Virtual object - distance is neg on R side. Incident rays are converging)

Radius of curvature is neg on the V side.

Virtual image- distance is neg on the V side.

r2r1

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Rules for drawing rays to locate images from

a lens

A ray initially parallel to the central axis will pass through the focal point.

A ray that initially passes through the focal point will emerge from the lens

parallel to the central axis.

A ray that is directed towards the center of the lens will go straight

through the lens undeflected.

E l f d i i

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Example of drawing images

Example

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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.

pfi

111=

m =y

y= -

i

p

1

i=1

5-

1

10= +

1

10

Image is real,inverted.

. .F1 F2p

Virtual side Real side

m = -10

10= - 1

i = +10 cm

Example

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24(e). Given a lens with the properties (lengths in cm) r1 = +30, r2 =

-30, p = +10, and n = 1.5, find the following: f, i and m. Is the image

real or virtual? Upright or inverted? Draw 3 rays.

( )

=

21

111

1

rrn

f

( )30

1

30

1

30

115.1

1=

---=

f

cmf 30=

pfi

111=

15

1

10

1

30

11==

cmi 15=

m =y

y= -

i

p

m = -- 15

10= +1.5

Image is virtual,

upright.

Virtual sideReal side

r1. .F1 F2p

r2

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27. A converging lens with a focal length of +20 cm is located 10 cm to

the left of a diverging lens having a focal length of -15 cm. If an object is

located 40 cm to the left of the converging lens, locate and describe

completely the final image formed by the diverging lens. Treat each lensSeparately.

f1

f1

Lens 1 Lens 2

f2

f2

10

40

+20 -15

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f1

f1

Lens 1 Lens 2

f2

f2

10

40

+20 -15

Ignoring the diverging lens (lens 2), the image formed by the

converging lens (lens 1) is located at a distance1

i1

=1

f1

-1

p1

=1

20cm-

1

40cm. i1 = 40cm

40

This image now serves as a virtual object for lens 2, with p2 = - (40 cm - 10 cm) = - 30 cm.

30

Since m = -i1/p1= - 40/40= - 1 , the image is inverted

L 1 L 2

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1

i2

=1

f2

-1

p2

=1

- 15cm

-1

- 30cm

i2=-30cm.

Thus, the image formed by lens 2 is located 30 cm to the left of lens 2. It is

virtual (since i2 < 0).

f1

f1

Lens 1 Lens 2

f2

f2

10

40

+20 -15

40

30

The magnification is m = (-i1/p1) x (-i2/p2) = (-40/40)x(30/-30) =+1, so the image

has the same size orientation as the object.

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Optical Instruments

Magnifying lens

Compound microscope

Refracting telescope