23Physics_lecture_31 May 2014

5/26/2018 23Physics_lecture_31 May 2014

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MS101: Physics

Chapter 23: Reflection and Refraction of Light

Dr. Ahmed Amin Hussein

01007903935

[email protected]

2013-201429 April 2014 Prepared By: Dr. Ahmed Amin 1
mailto:[email protected]:[email protected]


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Chapter 23: Reflection and

Refraction of Light

Huygenss Principle

Reflection

Refraction

Total Internal Reflection

Polarization by Reflection

Formation of Images

Plane Mirrors

Spherical Mirrors

Thin Lenses

29 April 2014 Prepared By: Dr. Ahmed Amin 2


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23.1 Huygenss Principle

A set of points with equal phase is called a wavefront.



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A ray points in the direction of wave propagation and is

perpendicular to the wavefronts. Or a ray is a line in the

direction along which light energy is flowing.



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Huygenss principle: At some time t, consider every point on a

wavefront as a source of a new spherical wave. These wavelets move

outward at the same speed as the original wave. At a later time t+t,

each wavelet has a radius v

t, where v is the speed of propagation of thewave. The wavefront at t+t is a surface tangent to the wavelets.



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23.2 Reflection of Light

When light is reflected from a smooth surface the rays

incident at a given angle are reflected at the same angle.

This is specular reflection.



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Reflection from a rough surface is called diffuse reflection.

Smooth and rough are determined based on the

wavelength of the incident rays.



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The angle of incidence equals the angle of reflection. Theincident ray, reflected ray, and normal all lie in the same

plane. The incident ray and reflected ray are on opposite

sides of the normal.



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23.3 Refraction of Light

When light rays pass from one medium to another they

change direction. This is called refraction.



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Snells Law:

2211 sinsin nn

where the subscripts referto the two different media.

The angles are measured

from the normal.

When going from high nto low n,the ray will bend away

from the normal.



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The incident ray, transmitted ray, and normal all lie in the

same plane.

The incident and transmitted rays are on opposite sides of

the normal.



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Example (text problem 23.11): Sunlight strikes the surface of

a lake. A diver sees the Sun at an angle of 42.0with respect

to the vertical. What angle do the Suns rays in air make with

the vertical?

surfacen1 = 1.00; air

n2 = 1.33; water

Normal

42

Transmitted

wave

incident wave

1

1.63

8920.0sin42sin333.1sin00.1

sinsin

1

1

1

2211

nn



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23.4 Total Internal Reflection

The angle of incidence for when the angle of refraction is

90is called the critical angle c.

1

2

c

22c1

2211

sin

90sinsin

sinsin

n

n

nnn

nn



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If the angle of incidence is greater than or equal to thecritical angle, then no wave is transmitted into the other

medium. The wave is completely reflected from the

boundary.

Total internal reflection can only occur when the incident

medium has a larger index of refraction than the second

medium.



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Example (text problem 23.22): Calculate the critical angle

for sapphire surrounded by air.

4.34

565.0sin

90sin00.1sin77.1

sinsin

c

c

c

2211

nn

surface

n2 = 1.0; air

n1 = 1.77; sapphire

Normal

Transmitted wave

incident wave

2=90

1



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23.9 Thin Lenses

A diverging lens will bend light away from the principle axis.

A converging lens will bend light toward the principal axis.



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Magnification:

The thin lens equation:

p

q

h

hm

fqp

111



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Example (text problem 23.64): A diverging lens has a

focal length 8.00 cm.

(a) What are the image distances for objects placed at various distances

from the lens? Is the image real or virtual? Upright or inverted? Enlarged

or diminished?

Object

distance

Image

distance

Real/

virtual?

Upright/

inverted?

Enlarged/

diminished

5 cm 3.08 cm Virtual upright Diminished

8 cm 4.00 cm Virtual Upright Diminished

14 cm

5.09 cmVirtual Upright Diminished





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(b) If the object is 4.00 cm high, what is the height of theimage?

Object

distance

Image

distance

Magnification Image height

5 cm 3.08 cm 0.616 2.46 cm

8 cm 4.00 cm 0.500 2.00 cm

14 cm 5.09 cm

0.364 1.45 cm

16 cm 5.33 cm 0.333 1.33 cm

20 cm 5.71 cm 0.285 1.14 cm

Example continued:



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Summary

The Laws of Reflection

The Laws of Refraction

Condition for Total Internal Reflection

Condition for Total Polarization of Reflected Light

Real/virtual Images

Mirrors (plane & spherical)

Thin Lenses



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QUESTIONS ?



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X

Geometric optics is an approximation to the behavior of

light that applies when interference and diffraction arenegligible. In order for diffraction to be negligible, the sizes

of objects must be large compared to the wavelength of

light.


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24

X23.5 Polarization by Reflection

Brewsters angleis the angle of incidence for which the

reflected light is completely polarized.

Light is totally polarized when the reflected ray and the

transmitted ray are perpendicular.

i

t

tBti

ttii

n

n

nnn

nn

B

BB

tan

cos90sinsin

sinsin


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25

Example (text problem 23.32): (a) Sunlight reflected from

the still surface of a lake is totally polarized when the

incident light is at what angle with respect to the horizontal?

1.53

33.100.1

33.1tan

air

water

B

Bn

n

The angle is measured from the normal, so 9053.1

= 36.9is the angle from the horizontal.


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26

(b) In what direction is the reflected light polarized?

Example continued:

It is polarizedperpendicular

to the plane of

incidence.


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27

Example continued:

(c) Is any light incident at this angle transmitted into the

water? If so, at what angle below the horizontal does the

transmitted light travel?

9.36

6000.0sin

sin333.11.53sin00.1

sinsin

2

2

2

2211

nnFrom Snells Law:

The angle is measured from the normal, so 9036.9

= 53.1is the angle from the horizontal.


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28

X23.6 Formation of Images

An image is realif light rays from a point on the object

converge to a corresponding point on the image.

A camera

lens forms a

real image.


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29

Your eye focuses

the diverging raysreflected by the

mirror.

The light rays

appear to come

from behind themirror.

An image is virtualif the light

rays from a point on the object

are directed as if they divergedfrom a point on the image, even

though the rays do not actually

pass through the image point.


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30

Example (text problem 23.35): A defect in a diamond

appears to be 2.00 mm below the surface when viewed from

directly above that surface. How far beneath the surface is

the defect.

Surface

Actual location of

defect

Air

n2=1.00

Diamond

n1= 2.4191

2

1

2

y

y


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31

The angles 1and 2are related by Snells Law:

2211 sinsin nn

The actual depth of the defect is y and it appears to

be at a depth of y. These quantities are related by:

12 tantan yy

Example continued:


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32

Dividing the previous two expressions gives:

2211 coscos ynyn

As long as you are directly above the defect and its

image, the angles 1 and 2are nearly 0. Rays fromonly a narrow range of angles will enter your eye. The

above expression simplifies to:

1

2

21

n

n

y

y

ynyn

(general result)

Example continued:


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33

The actual depth of the defect in the diamond is then

mm.84.4mm00.200.1

419.2

2

1

y

n

ny

Example continued:


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34

X23.7 Plane Mirrors

A point source and its image are at the same distance from

the mirror, but on opposite sides of the mirror.

Treat an extended

object as a set of

point sources.


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35

Example (text problem 23.41): Entering a darkened room,

Gustav strikes a match in an attempt to see his

surroundings. At once he sees what looks like another

match about 4 m away from him. As it turns out, a mirror

hangs on one of the walls. How far is Gustav from the wall

with the mirror?

The image seems 4 m away, but the mirror is only

2 m away since the rays will appear to come from

a point 2 m behind the mirror.


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36

X23.8 Spherical Mirrors

A convex (or diverging) mirror curves

away from the observer.

Principal

axis

vertexCenter of

curvature

The focal point


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A ray parallel to

the principle axis is

reflected, and it

appears to have

come from point F,

the focal point of

the mirror.

For a convex mirror, the focal point is on the axis and is

located a distance 0.5R behind the mirror, where R is the

radius of curvature.


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38

Drawn in green, red, and blue are the principal rays.

1. A ray parallel to the principal axis is reflected as if it came

from the focal point. (green)

2. A ray along a radius is reflected back upon itself. (red)

3. A ray directed toward the focal point is reflected parallel to

the principal axis. (blue)


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39

For the pencil in the previous figure, the image is upright,

virtual, smaller than the object, and closer to the mirror

than the object.


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40

Principal

axis

vertexCenter of

curvature

The focal point

A concave (or converging) mirror

curves toward the observer.


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41

1. A ray parallel to the principal axis is reflected through the

focal point. (green)

2. A ray along a radius is reflected back upon itself. (red)

3. A ray along the direction from the focal point to the mirror is

reflected parallel to the principal axis. (blue)

Drawn in green, red, and blue are the principal rays.


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42

The magnification is defined as .sizeobject

sizeimage

h

hm

An inverted image has m < 0 and an upright image

has m > 0.

The expression for magnification can also be written as

p

qm where p is the object distance and

q is the image distance.


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43

The mirror equation:

fqp111 where f is the focal length of the mirror.f < 0 when the focal point is behind the

mirror.


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Example (text problem 23.46): An object 2.00 cm high is

placed 12.0 cm in front of a convex mirror with a radius of

curvature of 8.00 cm. Where is the image formed?

fqp

111

where p = 12.0 cm, f = 0.5R = 4.00 cm, and q is the

unknown image distance. Solving gives q = 3.00 cm.

The image is behind the mirror.


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45

Example: A concave mirror has a radius of curvature of

10 cm. (a) Describe the image formed if the object is

placed between the mirror and the focal point.

p (cm) q (cm) m

1 -1.25 1.25

2 -3.33 1.67

3 -7.5 2.5

4 -20 5

fqp

111

Using the mirror equation for a

range of p values, the values of

q and m can be determined.

The image is

virtual, upright,

and magnified.


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46

(b) Describe the image formed if the object is placed

between the focal point and a distance of twice the focal

length.

Example continued:

p (cm) q (cm) m

6 30.00 -5.00

7 17.50 -2.50

8 13.33 -1.67

9 11.25 -1.25

10 10.00 -1.00

The image is real,inverted, and magnified.

Note that at p = 2f the

image is the same size

as the object.


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47

(c) Describe the image formed if the object is placed at a

distance greater than twice the focal length.

Example continued:

The image is

real, inverted,

and diminished.

p (cm) q (cm) m

11 9.17 -0.83

12 8.57 -0.71

13 8.13 -0.63

14 7.78 -0.56

15 7.50 -0.50


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(d) Describe the image formed if the object is placed at

the focal point.

Example continued:

When p = f, q = .fqp111

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23Physics_lecture_31 May 2014