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Lecture 14 Images Chapter 34PowerPoint Presentation

Lecture 14 Images Chapter 34

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Lecture 14 Images Chapter 34

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Lecture 14 Images Chapter 34

- Geometrical Optics
- Fermats Principle
- Law of reflection
- Law of Refraction

- Magnifying Glass, Microscope, Refracting telescope

Left over from Lecture 13 and Chapter 33

- Show electric and magnetic fields are out of phase in demo.
- Reflection and Refraction
- Red or green laser through smoky block
- Show maximum bending angle

- Total internal reflection
- 1) Show it with block
- Dispersion through prism
- Brewsters Law
- 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.

n2=1.509

Red

θ1

n1=1.000

θ2

Equilateral prism

dispersing sunlight

late afternoon.

Sin θrefr= 0.866nλ

How much light is polarized when reflected from a surface?

Polarization by Reflection: Brewsters Law

Fresnel Equations

The 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

eye

sky

1.09

1.09

1.08

1.08

1.07

1.07

Index of refraction

1.06

Hot road causes gradient in the index of refraction that increases

as you increase the distance from the road

Geometrical Optics:Study of reflection and refraction of 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.

Mirror

B

A

1

Using Fermat’s Principle you can prove the

Reflection 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.

JAVA APPLET

Show Fermat’s principle simulator

http://www.phys.hawaii.edu/~teb/java/ntnujava/index.html

B

2

A

1

Plane Mirror

Use calculus - method

of minimization

Write down time as a function of y

and set the derivative to 0.

Mirrors and Lenses

Angle of

incidence

Virtual side

Real side

Normal

Virtual image

Angle of reflection

i = - p

eye

Object distance = - image distance

Image size = Object size

eye

object

2

1

3

Problem: Two plane mirrors make an angle of 90o. How many images are there for an object placed between them?

mirror

mirror

Using the Law of Reflection to make a bank shot

Assuming no spin

Assuming an elastic collision

No cushion deformation

d

d

i = - p magnification = 1

What happens if we bend the mirror?

Concave mirror.

Image gets magnified.

Field of view is diminished

Convex mirror.

Image is reduced.

Field of view increased.

- 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

Lateral magnification = ratio of image height/object height

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

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

Lensmaker Equation

Lateral Magnification for a Lens

What is the sign convention?

p

Real side - R

Virtual side - V

Light

r1

r2

i

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.

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.

Example

Virtual side

Real side

.

.

F1

p

F2

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.

Example

Virtual side

Real side

.

.

F1

p

F2

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.

Image is real,

inverted.

.

.

r1

r2

F1

p

F2

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.

Real side

Virtual side

.

.

r1

r2

F1

p

F2

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.

Real side

Virtual side

Image is virtual, upright.

Lens 1

Lens 2

+20

-15

f1

f2

f1

f2

40

10

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 lens

Separately.

Lens 1

Lens 2

30

40

+20

-15

f1

f2

f1

f2

40

10

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

converging lens (lens 1) is located at a distance

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

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

Lens 1

Lens 2

30

40

+20

-15

f1

f2

f1

f2

40

10

Thus, the image formed by lens 2 is located 30 cm to the left of lens 2. It is virtual (since i2 < 0).

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.

Magnifying lens

Compound microscope

Refracting telescope

Galileo - converging + diverging lens

Keplerian - converging + converging lens

Reflecting Telescope

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'?