THE WHY’S & HOW’S OF THE PHENOMENON OF ELECTROMAGNETIC RADIATION.

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THE WHY’S & HOW’S OF THE PHENOMENON OF ELECTROMAGNETIC RADIATION

Transcript of THE WHY’S & HOW’S OF THE PHENOMENON OF ELECTROMAGNETIC RADIATION.

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THE WHY’S & HOW’S OF THEPHENOMENON OF ELECTROMAGNETICRADIATION

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ORIGIN OF LIGHT Where does light come from? How is it produced? The least complex answer is that light comes from the atom itself - the motion of the electrons around the nucleus. When an electron drops an energy level - a packet of light (photon) is produced.

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http://library.advanced.org/10796/ch10/ch10.htm

ELECTROMAGNETIC SPECTRUMIs a “collection” of the total range of waves–

All are forms of “light”

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VISIBLE LIGHT Of all the electromagnetic waves, visible light is the only portion of electromagnetic waves that can be detected by the human eye. It is a very small section of the spectrum and visible wavelengths run from 7.5 x 10-7 m (red) to 3.5 x 10 -7 m (purple).

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Luminous and illuminated bodies

 Light is produced by a luminous body. A light bulb is a luminous body that emits light in almost every direction. Light travels in straight lines at 299,792,458 m/sec in a vacuum. When light hits an object OR another medium, it is reflected or refracted . An illuminated body reflects light. When a ray of light reaches our eyes, the receptors in our eyes will produce a different color sensation depending on the wavelength of the light wave.  

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COLORS

Red, green and blue are known as primary colors, because when they are added together white light is formed. By mixing primary colors in pairs we obtain secondary colors. Blue and red produce magenta, and blue and green produce cyan.  

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http://library.advanced.org/10796/ch10/ch10.htm

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Moving from one medium to another

Once a wave (incident wave) has reached a change in media, part of the energy is transferred to the medium that is immediately next to it (transmitted wave) and part is reflected backward (reflected wave).The energy transferred depends on the difference between the mediums. If there is a significant difference, almost all the energy will be reflected.  

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If the mediums are similar, most of the energy will be transferred. However, the reflected waves will be inverted if the medium that comes next is more dense or it won't be inverted if the medium is less dense. 

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Reflection of light  If we draw a line perpendicular to a surface, this line is the normal of the surface. When a ray of light hits the surface of an object, part of the light is reflected. If the ray of light is at an angle with the surface, then the angle between the incident ray and the normal [incident angle]will be the same angle between the normal and the reflected ray [reflected angle]. This is called the law of reflection.

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Most surfaces are not completely flat. When millions of rays of light hit the rough surface of an object, they are reflected in all directions. This is how we can see illuminated objects.  

http://library.advanced.org/10796/ch10/ch10.htm

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Incident Ray

Incident Angle

Reflected Angle

Reflected RayNormal Line

Boundary / Mirror

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White Light:White light is not a distinct color. Instead, white light is the combination of all the other colors.

White light

Another discovery by Isaac Newton

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Selective Reflection:When we perceive an object to be a particular color, we actually are receiving only one particular color of

light in our eyes.

Ex: A banana looks yellow because it reflects only yellow light. It absorbs all the other colors.

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Black:

When an object appears black, it means that all colors ( frequencies) of light are being absorbed by

that object. None are being reflected.

Often, black objects are hotter because they

are absorbing more light/energy.

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If a rose were illuminated with a red light, you would see the red rose, but the stem and leaves

would look nearly black. Since they are green, that means they reflect only green light ( absorb red).

reflected

absorbed

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Refraction of light  When a ray of light passes from one medium to another, it bends. Depending of the new medium the light will travel faster or slower. If the light travels faster in the second medium, then this medium is called the rarer medium. On the other hand, the medium in which the light travels slower, in this case the first one, is called the denser medium. When a ray of light enters a denser medium, it is bent towards the normal. When a ray of light enters a rarer medium, it is bent away from the normal.  

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http://theory.uwinnipeg.ca/physics/light/node5.html

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There is an index of refraction (n) between the two mediums.To get a value, we have to divide the sine of the angle in vacuum or air by the sine of the angle in the denser medium. The index of refraction is given by the formula:

n = sin a / sin b In addition, if the new medium is more dense, the light bends because it slows down. How much do you ask? This decrease in speed is given by the formula: v = c / n where v - is the new speed of light  

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Law of Refraction: Snell’s Law

• There is a formula to predict how much a wave will “bend” as it travels into a new medium:

n1sin1 = n2sin2

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Angle of Incidence

Angle of Refraction

1 = 40

2 = 32

AIR - n1 = 1.00

? - n2 =

“New” Angle of Incidence

“New” Angle of Refraction

n1 sin 1 = n2 sin 2

(1.00) (0.643) = (x) (0.53)

X = 1.21

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Collision of wavesWhen two waves traveling in opposite directions through the same medium collide, the amplitude of the resulting wave will be the sum of the two initial waves. This is called interference and there are of two types: Constructive interference is when the amplitudes of the initial waves are in the same direction. The resulting wave will be larger than the original waves. The highest point of a constructive interference is called an antinode.  

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Destructive interference is when the amplitudes of the initial waves are opposite. The amplitude of the resulting wave will be zero. The point in the middle of a destructive interference is called a node and it never moves (in light - it would be a dark spot)

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#3: Diffraction When a wave travels through a small hole in a barrier, it bends around the edges. This is called diffraction.

• The bending of a wave around an obstacle

http://micro.magnet.fsu.edu/primer/java/diffraction/basicdiffraction/index.html

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#4: Interferenceof

Waves

The Addition of waves can add (Constructive) or subtract(Destructive).

Standing waves are a result of waves combining in phasehttp://micro.magnet.fsu.edu/primer/java/interference/doubleslit/index.html

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Apparent Depth

• Light exits into medium (air) of lower index of refraction,  and turns left.

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Spear-Fishing

• Spear-fishing is made more difficult by the bending of light.

• To spear the fish in the figure, one must aim at a spot in front of the apparent location of the fish.

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Delayed Sunset

• The sun actually falls below below the horizon

• It "sets", a few seconds before we see it set.

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Green Flash

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Broken Pencil

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Water on the Road Mirage

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In general, females have better color vision than males.

In fact, a high % of males have some degree of color blindness.

This doesn’t mean that they see the world in B&W, but instead their color vision isn’t as vivid.

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You should see a “25” in the dot pattern. The following tests will be more difficult if you are

color blind…

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Normal color vision: 45

Red/Green color blindness: no pattern

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Normal color vision: 6

Red/Green color blindness: no pattern

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Optical Illusions:Here are some illusions to illustrate how our

eye/brain makes judgements. The light our eyes receive may be objective, but we often interpret

this data subjectively. Our vision perception is not always very “scientific”.

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Are the horizontal lines bent?

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What do you see?

A dalmation?

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What do you see?

A man’s face, or the word liar written in

cursive.

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Do you see the spiral?Actually they are concentric circles.

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The square inscribed in the circle is NOT kinked.

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How many black dots do you see? None

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Two faces?

Wine goblet?

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Can you say the color of each word without reading the actual word?

For example, for the first word you would say: “green” since the letters are green.

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LIGHT ORIGIN REVIEW QUIZ1. WHERE DOES LIGHT COME FROM? - EXPLAIN THE MECHANISM.

2. WHY DO WE SEE COLORS? - HOW IS IT WE SEE COLORS?

3. WHAT ARE THE COLORS OF THE VISIBLE SPECTRUM?

4. NAME AT LEAST 4 OTHER FORMS OF ELECTROMAGNETIC RADIATION.

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Some of the light is always reflected. However, when a ray of light goes from a denser medium to a rarer medium, all the light will be reflected if the angle of incidence is greater than the critical angle. The critical angle is the angle of incidence for which the refracted ray is at 90 degrees with the normal.  

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LIGHT WAVES VISUAL DEVICE CRITERIA

1. MUST USE AT LEAST 1 REGULAR SIZE PIECE OF CONSTRUCTION PAPER2. AESTICHALLY PLEASING3. USE MIND-MAP DESIGN OR FLOW CHART DESIGN4. MUST INCLUDE THE FOLLOWING ITEMS: - ORIGIN OF LIGHT

- BOHR ATOM- EMISSION & ABSOPTION SPECTRA ORIGINS- TRANSFER OF ENERGY- TYPE OF WAVE MOTION- ELECTROMAGNETIC SPECTRUM

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CONVERGING (CONVEX)LENS - GENERAL IMAGEFORMATION

DIVERGING (CONCAVE)LENS - GENERAL IMAGEFORMATION

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CONVERGING LENSES

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CONVERGING LENSES

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The diagrams above shows that in each case, the image is •located behind the lens •a virtual image •an upright image •reduced in size (i.e., smaller than the object)

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LENS EQUATION

Sample Problem #1A 4.0-cm tall light bulb is placed a distance of 45.7 cm from a double convex lens having a focal length of 15.2 cm. Determine the image distance and the image size.

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Sample Problem #1A 4.0-cm tall light bulb is placed a distance of 45.7 cm from a double convex lens having a focal length of 15.2 cm. Determine the image distance and the image size.

ho = 4.0 cm do = 45.7 cm f = 15.2 cm

1/f = 1/do + 1/di

1/(15.2 cm) = 1/(45.7 cm) + 1/di

0.0658 cm-1 = 0.0219 cm-1 + 1/di

0.0439 cm-1 = 1/di

di = 22.8 cm

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Sample Problem #1A 4.0-cm tall light bulb is placed a distance of 45.7 cm from a double convex lens having a focal length of 15.2 cm. Determine the image distance and the image size.

ho = 4.0 cm do = 45.7 cm f = 15.2 cm

hi/ho = - di/do

hi /(4.0 cm) = - (22.8 cm)/(45.7 cm)

hi = - (4.0 cm) * (22.8 cm)/(45.7 cm)

hi = -1.99 cm

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Sample Problem #2A 4.0-cm tall light bulb is placed a distance of 8.3 cm from a double convex lens having a focal length of 15.2 cm. (NOTE: this is the same object and the same lens, only this time the object is placed closer to the lens.) Determine the image distance and the image size

ho = 4.0 cm do = 8.3 cm f = 15.2 cm 1/f = 1/do + 1/di

1/(15.2 cm) = 1/(8.3 cm) + 1/di

0.0658 cm-1 = 0.120 cm-1 + 1/di

-0.0547 cm-1 = 1/di

di = -18.3 cm

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Sample Problem #2A 4.0-cm tall light bulb is placed a distance of 8.3 cm from a double convex lens having a focal length of 15.2 cm. (NOTE: this is the same object and the same lens, only this time the object is placed closer to the lens.) Determine the image distance and the image size

ho = 4.0 cm do = 8.3 cm f = 15.2 cm hi/ho = - di/do

hi /(4.0 cm) = - (-18.2 cm)/(8.3 cm)

hi = - (4.0 cm) * (-18.2 cm)/(8.3 cm)

hi = 8.8 cm

http://www.glenbrook.k12.il.us/gbssci/phys/Class/refrn/u14l5f.html