all about light. light is a part of the electromagnetic spectrum
TRANSCRIPT
All About Light
Light is a part of the electromagnetic spectrum.
Light can travel through empty space as well as air
and water.
Light is a transverse wave.
Light comes in discreet packages called photons.
Light has properties of a particle and wave, known as
the dual nature of light.
Similarities of Light to Sound• Source determines the frequency.• Speed depends on what it is
traveling through.• Light travels fastest in a vacuum,
slowest through water.• v = f λ still applies.• The speed of light in a vacuum is
constant and represented by c.• c = f λ c = 3.00 ˣ 108 m/s
The primary colors of light are red, green, and blue.
The secondary colors of light are yellow, cyan, and magenta.
All of the colors combined make white light.
The mixing of all of these colors makes every color we
see.
Polarization of Light
Incident light is unpolarized and is at all angles.
The first filter removes the horizontal light.
The second filter removes the vertical light.
No light gets past the second filter!
This is how sunglasses work!
Glare is horizontal light reflected off of the ground.
Vertical Polarizers eliminates the horizontal light leaving only
vertical.
More on Reflection and Refraction.
The Law of Reflection always applies.
A Smooth surface gives a regular reflection where
light comes back in parallel beams.
A rough surface gives a diffuse reflection because light is
scattered.
Keep in mind we see nothing unless it reflects light!!
Refraction goes along with reflection!!
Notice the refracted ray is smaller than the incident ray.
This is because light is slower in water than in air!
Facts about RefractionWhen light goes from a faster to
slower medium, it will bend towards the normal.
• The Angle of Refraction is always dependent on the Angle of Incidence!!!
• The higher the Index of Refraction the slower light travels through the medium!!
The Light travels towards water at 60°
Normal
So the Angle of Incidence is 60°. (Θi = 60°)
NormalΘi
So the Angle of Refraction is 40.6°. (Θr = 40.6°)
NormalΘi
Θr
Since light travels through glass even slower it will bend more towards the normal!!
Normal
So the Angle of Incidence is still 60°. (Θi = 60°)
NormalΘi
Since light moves slower through glass than water it
bends more as it crosses the boundary!!
NormalΘi
So the Angle of Refraction is 34.5°. (Θr = 34.5°)
NormalΘi
Θr
Using this information we can calculate the Index of
Refraction (n).
n =sin Θr
sin Θi
Finding the Index of Refraction for water…
n =sin 40.6°
sin 60°
n = 1.33
The Index of Refraction can also be done in terms of the
speed of light!
n =v
c
n = 1.33
=3.00 × 108 m/s
2.25 × 108 m/s
Use the speed of light in a vacuum over the speed of
light in water.
n =v
c
n = 1.33
=3.00 × 108 m/s
2.25 × 108 m/s
Finding the Index of Refraction for glass…
n =sin 34.5°
sin 60°
n = 1.53
Remember Index of Refraction can never be less than 1
(vacuum) !!
n =sin 34.5°
sin 60°
n = 1.53
Using Snell’s Law we can predict how light will bend if
we know the indexes of refraction!!
Θi
Θi
ni sin Θi = nr sin Θr
If Θi = 30° and the light is traveling from water to air…
Θi
ni sin Θi = nr sin Θr
1.33 sin 30° = 1.0 sin Θr
Θi
1.33 sin 30° =
1.0 sin Θr
Θi
1.33 0.5 =
1.0 sin Θr
Θi
= 0.665 sin Θr
Θi
= 0.665 Θr sin-1
Θi
= 41.7° Θr
Θi
Θi
Θi
When light went from air to water it bent towards the normal, but water to air it
bends away from the normal!!
Θi
Θi
Why???
Θi
Θi
Because light travels faster in air than water!!!
Θi
Θi
When light travels from a slower to faster medium it
bends away from the normal
Common Indexes of Refraction
• Vacuum• Air• Water• Ethanol• Crown Glass• Quartz• Flint Glass• Diamond
• 1.00• 1.0003• 1.33• 1.36• 1.52• 1.54• 1.61• 2.42
Mirrors and Lenses
First we will look at curved mirrors, which are just a
section of a circle.
The focal point of a concave mirror is the point halfway between the center of the
circle and the surface of the mirror.
Following three rules we can find the location of any
image.
FC
#1 Draw a line from the top of the object reflected
through the focal point.
FC
#2 Draw a ray through the focal point and reflected
parallel.
FC
#3 draw a line through the center and straight back to
the object.
FC
We now know the location of the image. Any two of our
lines will tell us.
FC
It is an inverted image because it is upside down.
FC
It is a real image because the beams of light actually
pass through.
FC
We can locate this image with the same rules.
FC
What do we do?
FC
Extend the lines behind the mirror!!!
FC
Is this image real?
FC
No, the light does not actually travel there!
FC
This image is said to be virtual !
FC
Is the image inverted or erect?
FC
Erect, because it is not upside down!
FC
Convex mirrors produce small images. The focal
point is behind the mirror.
The images are never real.
These mirrors are valued for their wide angle views!!
We apply the rules we learned before.
F C
We treat the reflected ray as if it were coming from the
focus.
F C
The second line is to the focus and straight outward.
F C
The image is smaller giving us our “wide angle view”
F C
Now on to Lenses!!
Locating an image with a lens
FF
#1 Draw a line from the top of the object and then through a focal point.
FF
#2 Draw a line to the top of the object straight through
the center.
FF
The image is inverted and real.
FF
If the object is in front of the focal point….
FF
What do we do????
FF
Extend the lines backwards……
FF
The image is erect and virtual……
FF
Convex Lens Image Zones
FF2F 2F
Zone 1: Image is on the same side, erect, magnified, and virtual.
Convex Lens Image Zones
FF2F 2F
Zone 2: Image is on the opposite side, inverted, magnified, and real.
Convex Lens Image Zones
FF2F 2F
Zone 3: Image is on the opposite side, inverted, diminished, and real.
Zone I
At F image at infinity and does not exist.
Zone II
At 2F image at 2F
Zone III
A convex lens focuses light on a point…..
A concave lens spreads light out….
A concave lens is used to help people with myopia.
OpticsThe Math of Mirrors and Lenses
I See You!!
The Mirror Equation
1 1 1+ =so si fso is the object’s distance from the mirror.si is the image’s distance from the mirror.f is the focal length of the mirror.
The Mirror Equation
1 1 1+ =so si fIf the answer is positive the image is real !!If the answer is negative the image is virtual !!
The Magnification Equation
= ˗ sosi
If the answer is positive the image is upright !!If the answer is negative the image is inverted !!
mm in this equation means magnification!!
More on Magnification
× ho hi∣m ∣Use the absolute value for magnification!!=
ho is the height of the object!!
hi is the height of the image!!
20 cmfocal pointso30 cm
Where is the image?130 + 1si = 120
0.033 + 1si = 0.05si = 59 cm
59 cm
The positive result means a real image.
20 cmfocal pointso30 cm
What is the size of the image?? si=59so= 20
-m59 cm
-m
si
m = - 2.95The negative result means an inverted image.
20 cmfocal pointho30 cm
What is the size of the image?? 59 cmhi
2.95∣m ∣ ×× ho5 == hihi
hi = 14.75 cm
Light can be used to send digital signals through Total
Internal Reflection…
This is used in fiber optic cables…
Total Internal Reflection• This occurs because an incident beam enters a
medium with a lower index of refraction, such as from water to air. (Light is refracted away from the normal)
• The critical angle (θc) is reached when the light is refracted at 90° and travels along the surface.
• At angles greater than the critical angle (θc) total internal reflection occurs.
Critical Angle (θc)
Notice Total Internal Reflection only occurs when the angle of the Incident beam is greater than the
critical angle. θi ˃ θc
Summary of Requirements for Total Internal Reflection
• Index of Refraction must be lower for medium incident beam in entering.
• ni n˂ r or n1 n˂ 2 • The angle of incidence must be greater
than the critical angle!
• θi ˃ θc or θ1 ˃ θc
Calculating Critical Angle
n1n2sin θc =
n1 is the index of refraction for the medium where the
incident beam originated
n2 is the index of refraction for the medium the
incident beam enters
From Water to Air
n1n2sin θc = = 1.001.33sin θc = 0.7518sin-1 θc
= (0.7518) θc = 48.8°
If a light source is 5 m underwater, how far away on the surface will total internal
reflection occur??
Air n = 1.00
Water n = 1.33
y = 5 m
x = ? m
Air n = 1.00
Water n = 1.33
y = 5 m
x = ? m
tan 48.8° = yx = 5 m
xx = 5 m tan 48.8°
x = 5.71 m
Dispersion of Light
• Different colors of light have different indexes of refraction. This causes the “rainbow effect” under the proper conditions.
• For some media, such as air, this difference is miniscule and rarely noticed.
• For glass the difference in indexes of refraction can be profound for each color of light.
Indexes of Refraction for a piece of glass.
• Red 1.502• Orange 1.506• Yellow 1.511• Green 1.517• Blue 1.523• Violet 1.530
θ1 = 60°
We make use that the sum of the angles in a triangle are always 180°
θ2 θ3
θ4
θ1 = 60°
(90° - θ2) + 60° + (90° - θ3) = 180°
θ2 θ3
θ4
θ1 = 60°
(90° - θ2) + (90° - θ3) = 120°
θ2
θ3
θ4
60°
60° 60°
θ1 = 60°
θ2 + θ3 = 60°
θ2
θ3
θ4
60°
60° 60°
θ1 = 60°
θ3 = 60° - θ2
θ2 θ3
θ4
θ1 = 60°
Now we can use Snell’s Law to solve for the rest!!!
θ2 θ3
θ4
θ1 = 60°
1 sin 60° = n sin θ2
n sin θ3 = 1 sin θ4
θ2 θ3
θ4
θ1 = 60°
We will do this for red n = 1.502
θ2 θ3
θ4
θ1 = 60°
1 sin 60° = 1.502 sin θ2
θ2 θ3
θ4
θ1 = 60°
sin-1 0.866/ 1.502 = θ2
θ2 θ3
θ4
θ1 = 60°
35.21° = θ2
θ2 θ3
θ4
θ1 = 60°
θ3 = 60° - θ2
θ3 = 60° - 35.21°
θ2 θ3
θ4
θ1 = 60°
θ3 = 24.79°
θ2 θ3
θ4
θ1 = 60°
n sin θ3 = 1 sin θ4
θ2 θ3
θ4
θ1 = 60°
1.502 sin 24.79° = 1 sin θ4
θ2 θ3
θ4
θ1 = 60°
1.502 sin 24.79° = 1 sin θ4
θ2 θ3
θ4
θ1 = 60°
39.05° = θ4
θ2 θ3
θ4
Now calculate the rest!!
• Color n θ2 θ3 θ4 • Red 1.502 35.21° 24.79° 39.05°• Orange 1.506• Yellow 1.511• Green 1.517• Blue 1.523• Violet 1.530
Indexes of Refraction for a piece of glass.
• Color n θ2 θ3 θ4 • Red 1.502 35.21° 24.79° 39.05°• Orange 1.506 35.10° 24.90° 39.35°• Yellow 1.511 34.97° 25.03° 39.74°• Green 1.517 34.81° 25.19° 40.22°• Blue 1.523 34.65° 25.35° 40.70°• Violet 1.530 34.47° 25.53° 41.25°
The End
