• Law of Reflection• Law of Refraction or Snell’s Law• Chromatic Dispersion• Brewsters Angle

Chapter 33 ContinuedProperties of Light

Dispersion: Different wavelengths have different velocities andtherefore different indices of refraction. This leads to differentrefractive angles for different wavelengths. Thus the light is dispersed.The frequency does not change when n changes.

v = f!

! changes when medium changes

f does not change when medium changes

Snells LawRed

Snells Lawblue

Fiber Cable

Total Internal Reflection SimulatorHalliday, Resnick, Walker: Fundamentals of Physics,7th Edition - Student Companion Site

Why is light totally reflected inside a fiberoptics cable? Internal reflection

�

n1sin!

1= n

2sin!

2

(1.509)sin!1 = (1.00)sin90 = 1.00

!1 " sin#1 1

1.509" 41.505 deg

Total Internal Reflection

What causes a Mirage

1.06

1.091.08

1.07 1.071.08

1.09

sky eye

Hot road causes gradient in the index of refraction that increasesas you increase the distance from the road

Index of refraction

Inverse Mirage Bend

Chromatic Dispersion

How does a Rainbow get made?

Primary rainbowSecondary rainbowSupernumeraries

Polarization by Reflection: Brewsters Law

47. In the figure, a 2.00-m-long vertical pole extends from the bottomof a swimming pool to a point 50.0 cm above the water. What is thelength of the shadow of the pole on the level bottom of the pool?

Consider a ray that grazes the top of thepole, as shown in the diagram below. Hereθ1 = 35o, l1 = 0.50 m, and l2 = 1.50 m.

θ2

θ1

l2

l1

L x

airwater

shadow

x is given by

x = l1tanθ1 = (0.50m)tan35o = 0.35 m.

The length of the shadow is x + L.

L is given by

L=l2tan θ2

Use Snells Law to find θ2

Snells Law Example

According to the law of refraction, n2sinθ2 =n1sinθ1. We take n1 = 1 and n2 = 1.33

o

o

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.

θ2

θ1

l2

l1

L x

airwater

shadow

Calculation of L

Lecture 14 Images Chapter 34

•Geometrical Optics•Fermats Principle

-Law of reflection-Law of Refraction

•Plane Mirrors and Spherical Mirrors•Spherical refracting Surfaces•Thin Lenses•Optical Instruments

-Magnifying Glass, Microscope, Refracting telescope• Polling Questions

Geometrical Optics:Study of reflection and refractionof light from surfaces

The ray approximation states that light travels in straight linesuntil it is reflected or refracted and then travels in straight lines again.The wavelength of light must be small compared to the size ofthe objects or else diffractive effects occur.

Law of Reflection

1A

B

!I= !

R

!i

!r

Mirror

Fermat’s PrincipleUsing Fermat’s Principle you can prove theReflection law. It states that the path taken by light when traveling from one point toanother is the path that takes the shortesttime compared to nearby paths.

JAVA APPLET

Show Fermat’s principle simulator

Two light rays 1 and 2 taking different pathsbetween points A and B and reflecting off a

vertical mirror

1

2

A

B Plane Mirror

Use calculus - method of minimization

t = 1

C( h

1

2+ y

2+ h

2

2+ (w ! y)

2)

dt

dy=

2y

h1

2+ y

2+

!2(w ! y)

h2

2+ (w ! y)

2)= 0

y

h1

2+ y

2=

(w ! y)

h2

2+ (w ! y)

2)

sin" I = sin"R

!I= !

R

Write down time as a function of yand set the derivative to 0.

t = (1

v1

h1

2+ y

2+

1

v2

h2

2+ (w ! y)

2)

dt

dy= 0

1

v1

sin" I = 1

v2

sin"R

n1sin!

I= n

2 sin!

R

Plane Mirrors Where is the image formed

Mirrors and Lenses

Plane mirrors

Normal

Angle ofincidence

Angle of reflectioni = - p

Real sideVirtual side

Virtual image

eye

Object distance = - image distanceImage size = Object size

Problem: Two plane mirrors make an angle of 90o. Howmany images are there for an object placed betweenthem?

object

eye

1

2

3

mirror

mirror

pocket

Assuming no spinAssuming an elastic collisionNo cushion deformation

d

d

Using the Law of Reflection tomake a bank shot

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.

Rules for drawing images for mirrors

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

i=1

f

m =!i

p

Spherical refracting surfaces

n1

p+n2

i=n2! n

1

r

Using Snell’s Law and assuming small Angles between the rays with the central axis, we get the following formula:

Thin Lenses

n1

p+n2

i=n2! n

1

r

Apply this equation to Thin Lenses where the thickness issmall compared to object distance, image distance, and

radius of curvature. Neglect thickness.

Converging lens

Diverging lens

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 Lens m = !i

p

m = !image height

object height

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

Rules for drawing rays to locate images froma lens

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

2) A ray that initially passes through the focal point will emerge from the lens parallel to the central axis.

3) A ray that is directed towards the center of the lens will go straightthrough the lens undeflected.

Real image ray diagram for aconverging lens

Object Distance=20 cmand lens focal length is +10 cm

Find: Image distanceMagnificationHeight of imageIs image erect or invertedIs it real or virtualDraw the 3 rays

Object Distance =5 cm and focal length is + 10 cm

Find: Image distanceMagnificationHeight of imageIs image erect or invertedIs it real or virtualDraw the 3 rays

p=Object Distance =10 cm

Virtual image ray diagram forconverging lens

Virtual image ray diagram for adiverging lens

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 the 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

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 imagereal 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!=!=

i

cmi 15!=

�

m =! y

y= "

i

p

�

m = !!15

10= +1.5

Image is virtual,upright.

Virtual side Real side

r1. .F1 F2pr2

27. A converging lens with a focal length of +20 cm is located 10 cm tothe left of a diverging lens having a focal length of -15 cm. If an object islocated 40 cm to the left of the converging lens, locate and describecompletely the final image formed by the diverging lens. Treat each lensSeparately.

f1

f1

Lens 1 Lens 2

f2

f2

10

40

+20 -15

f1

f1

Lens 1 Lens 2

f2

f2

10

40

+20 -15

Ignoring the diverging lens (lens 2), the image formed by theconverging lens (lens 1) is located at a distance

�

1

i1

=1

f1

!1

p1

=1

20cm!

1

40cm. i

1= 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

�

1

i2

=1

f2

!1

p2

=1

!15cm!

1

!30cm i

2= !30cm.

Thus, the image formed by lens 2 is located 30 cm to the left of lens 2. It isvirtual (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 imagehas the same size orientation as the object.

Optical Instruments

Magnifying lensCompound microscopeRefracting telescope

In Figure 34-35, A beam of parallel light rays from a laser is incident on asolid transparent sphere of index of refraction n.

Fig. 34-35(a) If a point image is produced at the back of the sphere, what is the

index of refraction of the sphere?

(b) What index of refraction, if any, will produce a point image at thecenter of the sphere? Enter 'none' if necessary.

Chapter 34 Problem 32

Chapter 34 Problem 91 Figure 34-43a shows the basic structure of a human eye. Lightrefracts into the eye through the cornea and is then further redirectedby a lens whose shape (and thus ability to focus the light) iscontrolled by muscles. We can treat the cornea and eye lens as asingle effective thin lens (Figure 34-43b). A "normal" eye can focusparallel light rays from a distant object O to a point on the retina atthe back of the eye, where processing of the visual informationbegins. As an object is brought close to the eye, however, themuscles must change the shape of the lens so that rays form aninverted real image on the retina (Figure 34-43c).

(a) Suppose that for the parallel rays of Figure 34-43a and Figure34-43b, the focal length f of the effective thin lens of the eye is2.52 cm. For an object at distance p = 48.0 cm, what focal lengthf' of the effective lens is required for the object to be seen clearly?

(b) Must the eye muscles increase or decrease the radii ofcurvature of the eye lens to give focal length f'?