Lecture 14 Images Chapter 34 Geometrical Optics Fermats Principle -Law of reflection -Law of...

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

Left over from Lecture 13 and Chapter 33

1) Show electric and magnetic fields are out of phase in demo.2) Reflection and Refraction

1) Red or green laser through smoky block2) Show maximum bending angle

3) Total internal reflection1) Show it with block

1) Dispersion through prism2) Brewsters Law

1) Show using polaroid paper

Reflection and Refraction of Light

Dispersion: Different wavelengths have different velocities and therefore 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 = fλλ changes when medium changesf does not change when medium changes

Snells Law

n2=1.509

Red

€

v1 sinθ1 = v2 sinθ2

n1=1.000 θ1

θ2

€

n1 sinθ1 = n2 sinθ2

€

sinθ2 =1.00sinθ1

1.509

sinθ2 =1.00sin90

1.509θ2 = 41.505deg

Snell’s Law at Work

Fiber Cable

Why is light totally reflected inside a fiber optics cable? Internal reflection

€

n1 sinθ1 = n2 sinθ2

(1.509)sinθ1 =(1.00)sin90 =1.00θ1 ≥ sin−1 1

1.509 ≥ 41.505 deg

9

nλ increases with frequencyBlue light bends more than red

Equilateral prismdispersing sunlight late afternoon.

Sin θrefr= 0.866nλ

How much light is polarized when reflected from a surface?Polarization by Reflection: Brewsters Law

Fresnel EquationsThe parallel and perpendicular components of the electric vector of the reflected and transmitted light wave is plotted below when light strikes a piece of glass. Now the intensity of polarized light is proportional to the square of the amplitude of the oscillations of the electric field. So, we can express the intensity of the incoming light as I0 = (constant) E 2.

E

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

Geometrical Optics:Study of reflection and refraction of 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

θ I = θ R

θi

θr

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 to another is the path that takes the shortesttime compared to nearby paths.

JAVA APPLET

Show Fermat’s principle simulatorhttp://www.phys.hawaii.edu/~teb/java/ntnujava/index.html

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

t = 1C ( h1

2 + y2 + h22 + (w−y)2 )

dtdy

=2y

h12 + y2

+−2(w−y)

h22 + (w−y)2 )

=0

y

h12 + y2

=(w−y)

h22 + (w−y)2 )

sinθ I = sinθR

θ I = θR

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

t =(1v1

h12 + y2 +

1v2

h22 + (w−y)2 )

dtdy

=0

1v1sinθ I =

1v2sinθR

n1 sinθ I =n2 sinθR

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. How many images are there for an object placed between them?

object

eye

1

2

3

mirror

mirror

pocket

Assuming no spinAssuming an elastic collisionNo cushion deformation

d

d

Using the Law of Reflection to make 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+1i=1f

m=−ip €

f =r

2

Lateral magnification = ratio of image height/object height

One simple way to measure focal length of a concave mirror?

Spherical refracting surfaces

n1

p+n2

i=n2 −n1

r

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

In Figure 34-35, A beam of parallel light rays from a laser is incident on a solid 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 the center of the sphere? Enter 'none' if necessary.

Chapter 34 Problem 32

n1

p+n2

i=n2 −n1

r

n1

p+n2

i=n2 −n1

r

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

Simple Lens Model

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 =−

ip

m =−image heightobject 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 from a 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 straight through the lens undeflected.

Real image ray diagram for a converging lens

Virtual image ray diagram for converging lens

Virtual image ray diagram for a diverging lens

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.

. .F1 F2p

Virtual side Real side

Example

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

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.

Virtual side Real side

r1. .F1 F2

pr2

Example

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−=−=

i

cmi 15−=

€

m =′ y

y= −

i

p

€

m = −−15

10= +1.5

Image is virtual, upright.

Virtual side Real side

r1. .F1 F2

pr2

Example

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

Example

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. 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

€

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.

Optical Instruments

Magnifying lensCompound microscopeRefracting telescope

Galileo - converging + diverging lensKeplerian - converging + converging lens

Reflecting Telescope

Chapter 34 Problem 91 Figure 34-43a shows the basic structure of a human eye. Light refracts into the eye through the cornea and is then further redirected by a lens whose shape (and thus ability to focus the light) is controlled by muscles. We can treat the cornea and eye lens as a single effective thin lens (Figure 34-43b). A "normal" eye can focus parallel light rays from a distant object O to a point on the retina at the back of the eye, where processing of the visual information begins. As an object is brought close to the eye, however, the muscles must change the shape of the lens so that rays form an inverted real image on the retina (Figure 34-43c).

(a) Suppose that for the parallel rays of Figure 34-43a and Figure 34-43b, the focal length f of the effective thin lens of the eye is 2.52 cm. For an object at distance p = 48.0 cm, what focal length f' of the effective lens is required for the object to be seen clearly?

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