Lecture 14 Images Chapter 34

• Law of Reflection• Dispersion• Snell’s Law• Brewsters Angle

Preliminary topics before mirrors and lenses

•Law of Reflection•Dispersion•Snell’s Law•Brewsters Angle

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

Drawing Normals

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.

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.

Law of Refraction: Snells Law

€

n1 sinθ1 = n2 sinθ2

n1

n2

How do we prove it?

1

v1

sinθ1 =1v2sinθ2

Air 1.0Glass 1.33

t =(1v1

h12 + y2 +

1v2

h22 + (w−y)2 )

dtdy

=0

1v1sinθ I =

1v2sinθR

n1 sinθ I =n2 sinθR

JAVA APPLET

Show Fermat’s principle simulator

What allows you to see various colors when white light passes through a prism

Dispersion

How does a Rainbow work?

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 dose not change when n changes.

v = fλλ changes when medium changesf does not change when medium changes

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

€

n1 sinθ1 = n2 sinθ2

€

(1.33)sinθ1 = (1.00)sin90 =1.00

When θ1 ≥sin−1 1

1.33 ≥48.75 deglight won't get out of the material

Fiber Cable

Same here

Corner Reflector?

Show Total Internal reflection simulator

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

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

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 θ1 = 35o, l1 = 0.50 m, and l2 = 1.50 m.

θ2

θ1

l2

l1

L x

air

water

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

oo

n55.25

33.1

35sinsin

sinsin 1

2

112 =⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛= −− θθ

L is given by

.72.055.25tan)50.1(tan 22 mmlL o === θ

The length of the shadow is L+x.

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

θ2

θ1

l2

l1

L x

air

water

shadow

Calculation of L

Polarization by Reflection Brewsters Law

(1.)sinθB =nsinθr

θB +θr =90 we get 100% polarized reflected wavesin θB =nsin(90 −θB) =ncosθB

θB =tan−1n Brewsters Law

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

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

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

Concept Simulator/Illustrations

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

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:

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

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

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

•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.

Example of drawing images

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

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−=−=

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

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

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