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 Refraction
When 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°)
Normal
Θi
So the Angle of Refraction is
40.6°. (Θr = 40.6°)
Θr
Normal
Θi
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°)
Normal
Θi
Since light moves slower
through glass than water it
bends more as it crosses
the boundary!!
Normal
Θi
So the Angle of Refraction is
34.5°. (Θr = 34.5°)
Θr
Normal
Θi
Using this information we
can calculate the Index of
Refraction (n).
n=
sin Θi
sin Θr
Finding the Index of
Refraction for water…
n=
sin 60°
sin 40.6°
n = 1.33
The Index of Refraction can
also be done in terms of the
speed of light!
n=
c
v
=
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=
c
v
=
n = 1.33
3.00 × 108 m/s
2.25 × 108 m/s
Finding the Index of
Refraction for glass…
n=
sin 60°
sin 34.5°
n = 1.53
Remember Index of
Refraction can never be less
than 1 (vacuum) !!
n=
sin 60°
sin 34.5°
n = 1.53
Using Snell’s Law we can
predict how light will
bend if we know the
indexes of refraction!!
Θi
If Θi = 30° and the light is
traveling from water to air…
ni sin Θi = nr sin Θr
Θi
ni sin Θi = nr sin Θr
1.33 sin 30° = 1.0 sin Θr
Θi
sin Θr =
1.33
1.0
Θi
sin 30°
sin Θr =
1.33
1.0
Θi
0.5
sin Θr = 0.665
Θi
Θr = sin-1 0.665
Θi
Θr = 41.7°
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
Θi
Θi
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.
C
F
#1 Draw a line from the top
of the object reflected
through the focal point.
C
F
#2 Draw a ray through the
focal point and reflected
parallel.
C
F
#3 draw a line through
the center and straight
back to the object.
C
F
We now know the location
of the image. Any two of
our lines will tell us.
C
F
It is an inverted image
because it is upside
down.
C
F
It is a real image because
the beams of light
actually pass through.
C
F
We can locate this image
with the same rules.
C
F
What do we do?
C
F
Extend the lines behind
the mirror!!!
C
F
Is this image real?
C
F
No, the light does not
actually travel there!
C
F
This image is said to be
virtual !
C
F
Is the image inverted or
erect?
C
F
Erect, because it is not
upside down!
C
F
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
F
F
#1 Draw a line from the top
of the object and then
through a focal point.
F
F
#2 Draw a line to the top
of the object straight
through the center.
F
F
The image is inverted and
real.
F
F
If the object is in front of
the focal point….
F
F
What do we do????
F
F
Extend the lines
backwards……
F
F
The image is erect and
virtual……
F
F
Convex Lens Image Zones
2F
F
F
2F
Zone 1: Image is on the same
side, erect, magnified, and
virtual.
Convex Lens Image Zones
2F
F
F
2F
Zone 2: Image is on the opposite
side, inverted, magnified, and
real.
Convex Lens Image Zones
2F
F
F
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.
Optics
The Math of Mirrors and Lenses
I See You!!
The Mirror Equation
1
so
+
1
si
=
1
f
so 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
so
+
1
si
=
1
f
If the answer is positive the image is
real !!
If the answer is negative the image is
virtual !!
The Magnification Equation
m =˗
si
so
m in this equation means
magnification!!
If the answer is positive the image is
upright !!
If the answer is negative the image is
inverted !!
More on Magnification
∣m ∣ × ho = hi
Use the absolute value for
magnification!!
ho is the height of the object!!
hi is the height of the image!!
Where is the
image?
so
focal point
59 cm
1
1
+
30
si
1
0.033 +
si
=
20 cm
30 cm
1
20
= 0.05
si = 59 cm
The positive
result means
a real image.
What is the
size of the
image??
si
so
focal point
59 cm
m =-
20 cm
30 cm
si
so
59
m =20
m = - 2.95
The negative
result means
an inverted
image.
What is the
size of the
image??
hi
ho
focal point
59 cm
∣m ∣ × ho = hi
2.95 × 5
= hi
hi = 14.75 cm
20 cm
30 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 ˂ nr or n1 ˂ n2
• The angle of incidence must be greater
than the critical angle!
• θi ˃ θc or θ1 ˃ θc
Calculating Critical Angle
sin θc =
n2
n1
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
sin θc =
n2
n1
=
1.00
1.33
sin θc = 0.7518
θc = sin-1 (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
x=?m
y=5m
tan 48.8°=
x
y
x
=
5m
x = 5 m tan 48.8°
x = 5.71 m
Air n = 1.00
Water n = 1.33
x=?m
y=5m
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
Orange
Yellow
Green
Blue
Violet
1.502
1.506
1.511
1.517
1.523
1.530
We make use that the sum of the
angles in a triangle are always 180°
θ1 = 60°
θ2
θ3
θ4
(90° - θ2) + 60° + (90° - θ3) = 180°
θ1 = 60°
θ2
θ3
θ4
(90° - θ2) + (90° - θ3) = 120°
60°
θ1 = 60°
60°
θ2
θ3
θ4
60°
θ2 + θ3 = 60°
60°
θ1 = 60°
60°
θ2
θ3
θ4
60°
θ3 = 60° - θ2
θ1 = 60°
θ2
θ3
θ4
Now we can use Snell’s Law to
solve for the rest!!!
θ1 = 60°
θ2
θ3
θ4
1 sin 60° = n sin θ2
n sin θ3 = 1 sin θ4
θ1 = 60°
θ2
θ3
θ4
We will do this for red n = 1.502
θ1 = 60°
θ2
θ3
θ4
1 sin 60° = 1.502 sin θ2
θ1 = 60°
θ2
θ3
θ4
sin-1 0.866/ 1.502 = θ2
θ1 = 60°
θ2
θ3
θ4
35.21° = θ2
θ1 = 60°
θ2
θ3
θ4
θ3 = 60° - θ2
θ3 = 60° - 35.21°
θ1 = 60°
θ2
θ3
θ4
θ3 = 24.79°
θ1 = 60°
θ2
θ3
θ4
n sin θ3 = 1 sin θ4
θ1 = 60°
θ2
θ3
θ4
1.502 sin 24.79° = 1 sin θ4
θ1 = 60°
θ2
θ3
θ4
1.502 sin 24.79° = 1 sin θ4
θ1 = 60°
θ2
θ3
θ4
39.05° = θ4
θ1 = 60°
θ2
θ3
θ4
Now calculate the rest!!
•
•
•
•
•
•
•
Color
Red
Orange
Yellow
Green
Blue
Violet
n
1.502
1.506
1.511
1.517
1.523
1.530
θ2
θ3
θ4
35.21° 24.79° 39.05°
Indexes of Refraction for a piece of
glass.
•
•
•
•
•
•
•
Color
Red
Orange
Yellow
Green
Blue
Violet
n
1.502
1.506
1.511
1.517
1.523
1.530
θ2
35.21°
35.10°
34.97°
34.81°
34.65°
34.47°
θ3
24.79°
24.90°
25.03°
25.19°
25.35°
25.53°
θ4
39.05°
39.35°
39.74°
40.22°
40.70°
41.25°
The End