+ All Categories
Home > Documents > Electromagnetic Waves

Electromagnetic Waves

Date post: 22-Feb-2016
Category:
Upload: mohawk
View: 31 times
Download: 3 times
Share this document with a friend
Description:
Electromagnetic Waves. CHARITY I. MULIG. Def’n : EM Wave. Energy-carrying wave emitted by vibrating charges (often electrons) that is composed of oscillating electric and magnetic fields that regenerate one another. . The EM Spectrum. - PowerPoint PPT Presentation
Popular Tags:
34
Electromagne tic Waves CHARITY I. MULIG
Transcript
Page 1: Electromagnetic Waves

ElectromagneticWaves

CHARITY I. MULIG

Page 2: Electromagnetic Waves

Def’n: EM Wave

• Energy-carrying wave emitted by vibrating charges (often electrons) that is composed of oscillating electric and magnetic fields that regenerate one another.

Page 3: Electromagnetic Waves

The EM Spectrum

Range of frequencies over which electromagnetic radiation can be propagated.

Page 4: Electromagnetic Waves

Change in frequency of a wave of sound or light due to the motion of the source or the receiver.

os

oL f

vvvvf

Where•fl is the apparent frequency•f0 is the original frequency•v is the speed of the wave in the medium•v0 is the speed observer relative to the medium; positive if the observer is moving towards the source•vs is the speed of the source relative to the medium; positive if the source is moving away from the observer.

Page 5: Electromagnetic Waves

Doppler Effect for EM Waves

Observed Frequency

• vs,r = vs – vr is the velocity of the source relative to the receiver; it is positive when the source and the receiver are moving further apart.

• λo is the wavelength of the transmitted wave in the reference frame of the source.

Change in Frequency

Page 6: Electromagnetic Waves
Page 7: Electromagnetic Waves

Def’n: Polarization• Aligning of

vibrations in a transverse wave, usually by filtering out waves of other directions.

Page 8: Electromagnetic Waves

Wavefronts vs. RaysHuygen’s Principle

“The wave fronts of light waves spreading out from a point source can be regarded as the overlapped crests of tiny secondary waves – wave fronts are made up of tinier wave fronts”

Page 9: Electromagnetic Waves

Properties of EM Waves1. Reflection2. Refraction3. Diffraction4. Dispersion5. Scattering6. Interference7. Polarization

Page 10: Electromagnetic Waves

Geometric Optics

Page 11: Electromagnetic Waves

Reflection

Page 12: Electromagnetic Waves

Types of ReflectionSpecular/Regular Diffused/Irregular

Page 13: Electromagnetic Waves

The open-mesh parabolic dish is a diffuse reflector for short-wavelength light but a polished reflector for long-wavelength radio waves.

Page 14: Electromagnetic Waves

Law of Reflection1. The incident,

reflected and normal ray all lie in the same plane.

2. The angle of incidence is equal to the angle of reflection.

Page 15: Electromagnetic Waves

Reflection at a Plane Surface

Page 16: Electromagnetic Waves

Locating Plane Mirror Image

Page 17: Electromagnetic Waves

Guidelines for Ray Diagrams

Page 18: Electromagnetic Waves

Ray Diagram For Concave Mirrors

Page 19: Electromagnetic Waves

Ray Diagram for Convex Mirrors

Page 20: Electromagnetic Waves

Mirror Equation and Lateral Magnification

0

0

1211

dd

hhm

fRdd

i

o

i

i

Page 21: Electromagnetic Waves

Mirror Equation Sign Convention

Quantity Positive Negatived0 Real object Virtual Object

di Real image Virtual Image

f Concave Mirror Convex Mirror

m Upright/Erect Inverted

Page 22: Electromagnetic Waves

Sample Problems•A concave mirror forms an image, on a wall 3 m from the mirror, of the filament of a headlight lamp 10 cm in front of the mirror. (a) What are the radius of curvature and focal length of the mirror? (b) What is the height of the image if the height of the object is 5 mm?•Suppose that in the previous example, the left half of a mirror’s reflecting surface is covered with non-reflective soot. What effect will this have on the image of the filament?

The image of a tree just covers the length of a plane mirror 4 cm tall when the mirror is held 35 cm from the eye. The tree is 28 m from the mirror. What is its height?

Page 23: Electromagnetic Waves

RefractionDefinition:

“The bending of light as it passes obliquely from one medium to another.”

Cause of Refraction - The change in the average speed of light as enters a different medium”

The direction of the light waves changes when one part of each wave slows down before the other part.

Page 24: Electromagnetic Waves

Fermat’s Principle of Least Time

• Pierre Fermat • Out of all possible paths that light might travel

to get from one point to another, it travels the path that requires the shortest time.

Phet

Page 25: Electromagnetic Waves

Index of Refraction

• Describes how much light the speed of light in a material differs from its speed in a vacuum.

• The index of refraction of vacuum is 1.

vcn

Page 26: Electromagnetic Waves

Snell’s Law

• aka Snell-Descartes law • aka the law of

refraction• Follow’s from Fermat’s

principle of least time

1

2

2

1

2

1

sinsin

nn

vv

When light slows down in

going from one medium to another such as going from air to water, it refracts toward the normal. When it speeds up in traveling from one medium to another, such as going from water to air it refracts away from the normal.

Page 27: Electromagnetic Waves

Few Phenomena

Due to Refraction

Because of refraction, a submerged object appears to be nearer to the surface than it actually is.Because of atmospheric refraction, when the sun is near the

horizon, it appears to be higher in the sky.

The apparent wetness of the road is not reflection of the sky by water but, rather, refraction of sky light through the warmer and less-dense air near the road surface.

Because of refraction, the full root-beer mug appears to hold more root beer than it actually does.

Page 28: Electromagnetic Waves

Lenses

A prism

A curved prismA converging lens

Page 29: Electromagnetic Waves

Types of Lenses

Page 30: Electromagnetic Waves

Ray Diagrams

Finding the image produced by a THIN CONVERGING LENS. To emphasize that the mirror is thin the ray QAQ’ is shown as bent at the midplane of the lens rather than at the two surfaces and ray QOQ’ is shown as a straight line.

Page 31: Electromagnetic Waves

Graphical Method for Thin Lenses

• A ray through (or proceeding toward) the first focal point F1 emerges parallel to the axis.

•A ray parallel to the axis emerges from the lens in a direction that passes through the second focal point F2 of a converging lens, or appears to come from the second focal point of a diverging lens.•A ray through the center of the lens is not appreciably deviated; at the center of the lens the two surfaces are parallel, so this ray emerges at essentially the same angle at which it enters and along the same line.

Page 32: Electromagnetic Waves

Important Equations and Conventions

fdd io

111

Thin Lens Equation

21

1111RR

nf

Lensmaker’s Equation for Thin Lenses

Quantity Positive Negative

d0 Real object Virtual Object

di Real image Virtual Image

f Converging Lens Diverging Lens

R Converging Lens Diverging lens

m Upright Image Inverted Image

o

i

o

i

dd

hhm

Lateral Magnification for Thin Lenses

Page 33: Electromagnetic Waves

Sample Problemsa. Suppose the absolute values of the radii of

curvature of the lens surfaces in a double convex lens are both equal to 10 cm and the index of refraction is n = 1.52. What is the focal length of the lens?

b.Suppose a double concave lens also has n = 1.52, and the absolute values of the radii of curvature of its lens surfaces are also both equal to 10 cm. What is the focal length of this lens?

A converging lens has a focal length of 20 cm. Describe the image when an object is placed the following distances from the lens: (a) 50 cm; (b) 20 cm; (c) 15 cm; (d) -40 cm. Determine the magnification in each case.

You are given a thin diverging lens. You find that a beam of parallel ray spreads out after passing through the lens, as though all the rays came from a point 20 cm from the center of the lens. You want to use this lens to form an erect virtual image that is 1/3 the height of the object. (a) Where should the object be placed? (b) Draw a principal-ray diagram.

An object 8 cm high is placed 12 cm to the left of a converging lens of focal length 8 cm. A second converging lens of focal length 6 cm is placed 36 cm to the right of the first lens. Both lenses have the same optic axis. Find the position, size, and orientation of the image produced by the two lenses in combination. (Combinations of converging lenses are used in telescopes and microscopes.)

Page 34: Electromagnetic Waves

Sample Problems

An insect 3.75 mm tall is placed 22.5 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the right surface has a radius of curvature of magnitude 13 cm, and the index of refraction of the lens material is 1.70 cm. (a) Calculate the location and size of the image this lens forms of the insect. Is it real or virtual? Erect or inverted? (b) Repeat part (a) if the lens is reversed.


Recommended