Download Refraction

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Doctor Light (Kimiyo Hoshi) wikipedia , lookup

Bioluminescence wikipedia , lookup

Doctor Light (Arthur Light) wikipedia , lookup

Transcript
Refraction
Behavior of waves at boundaries
When the energy in a wave hits a
fixed boundary, the disturbance in
the medium through which the
energy is traveling (the wave) is
inverted.
When the wave hits a free
boundary, the pulse is reflected
but not inverted.
Behavior of waves at boundaries
When the energy in a wave
travels through media with
different density (), part of
the wave is reflected, part is
transmitted.
low  to high 
high  to low 
Think about it…
Make a rule summarizing the
reflection and transmission
characteristics of waves traveling
through media with different
densities.
Include:
• Height of wave
• Width of wave
• Speed of wave
Possible explanation?
low  to high 
high  to low 
Transmission of energy
From less dense to more dense:
•
•
Speed decreases
Wavelength decreases
From more dense to less dense
•
•
Speed increases
Wavelength increases
low to high   v, 
high to low  v, 
Waves (and light) at boundaries
When light passes from
air (less dense) to glass
(more dense), the speed
and wavelength of light
decrease.
Change in speed
 change in direction
Refraction
Refraction: bending the path
of light
The broken pencil
Years of experience has trained us
into thinking that light travels in
straight lines from objects to our
eyes.
When the image forms somewhere
unexpected, we have to fight that
training to make sense of what we
see.
The broken pencil, explained
Light is reflected from pencil
Light is refracted when it passes
from water to glass and glass to air.
Light travels straight to our eyes.
In this example, the image appears
to the left of the object’s location
and appears broken.
Other examples
• Magnifying water
•
Using water to start a fire
• Spearfishing
• Uncovering root beer deceit
• Best profile pics ever!
Marching soldiers
1. Stand shoulder to shoulder
forming a straight line,
connected with meter sticks.
2. A line of tape separates the
room into two ‘media’
3. When students approach the
line, they use baby steps. When
they pass the line, they abruptly
change pace
4. Observe what happens to the
direction of travel.
Conditions of refraction
1. Change in speed
i.e., density change in media
2. Approach the boundary
at an angle
Otherwise, speed would
change at the same time
Passing through…
When light passes through material, it
is actually absorbed and re-emitted
and absorbed and re-emitted and
absorbed and ….
The speed at which light is
transmitted is called optical density
Index of refraction
One indicator of optical density is
called index of refraction,
abbreviated n
𝑛𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙
3.00 𝑥 108 𝑚 𝑠
=
𝑣𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙
Material
Index of Refraction
Vacuum
1.0000
Air
1.0003
Ice
1.31
Water
1.333
Ethyl Alcohol
1.36
Plexiglas
1.51
Crown Glass
1.52
Flint Glass
1.66
Zircon
1.923
Diamond
2.417
Quantified
What is the speed of light in a diamond?
8
n
Trydiamond
it first = 2.42; c = 3.00 x 10 m/s
vTry=it ?first
𝑐
𝑛=
Try
it first
𝑣
=> 𝑣 =
𝑐
𝑛
3.00 𝑥 108 𝑚/𝑠
Try
it first
𝑣=
2.42
𝑣
x 108 m/s
Try =1.23
it first
Predicting the direction of bending
Traveling from fast (less
dense) to slow (more dense),
bent TOWARDS the normal
fast
𝜃𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡 > 𝜃𝑟𝑒𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛
Predicting the direction of bending
Traveling from slow (more
dense) to fast (less dense),
bent AWAY the normal
𝜃𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡 < 𝜃𝑟𝑒𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛
Tractor analogy
True if …
• sides are parallel
• medium on either
side of transparent
medium is the
same.
Think about it…
Copy this diagram onto a sheet of paper. What is the
path of the tractor and light?
In real life…
Think about it…
Draw the missing ray in each situation
Think about it…
The fish is shown in its actual
position.
Where will the image of the
fish appear to the person
above the water? If the
person wants to hit the fish,
where should s/he throw the
spear?
Justify your answer.
Quantifying refraction, 1 of 3
The optical density
makes a difference
𝜃𝑟𝑒𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝛼 𝑛
Collect some (virtual) data
Go to PHeT Bending Light
simulator.
Set medium 1 to air.
Set medium 2 to water.
Turn on the ray.
Measure and record the angle of
incidence and angle of refraction.
Repeat 10 or more times for
angles between 0 and 90
Mathematical model
𝑦 = 𝑚𝑥 + 𝑏
sin 𝜃𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑐𝑒
= 1.33 sin 𝜃𝑟𝑒𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 + 0
1.33 is the index of
refraction of water
Snell’s Law
𝑛1 sin 𝜃1 = 𝑛2 𝑠𝑖𝑛𝜃2
Where
𝑛1 is index of refraction of medium 1
𝜃1 is the angle of incidence
𝑛2 is index of refraction of medium 2
𝜃2 is the angle of refraction
More? You want more?
In real life…
Examples
Examples
Quantified
A jeweler wants to determine the index of refraction of a gem stone.
She shines a laser through it so that the incident beam strikes the face
at 45 from the normal. If the light travels through the gem at an angle
of 17 from the normal, what is the index of refraction?
Try
first45;
𝜃1it=
𝜃2 = 17; nair = 1.00
Try2 it=first
n
?
Try
sinit first
𝜃 𝑛
1 1
= sin 𝜃2 𝑛2 => 𝑛2 =
(sin 45°) 1.00
Try
first
𝑛2it=
sin 17°
𝑛
=
2.42
2
Try it first
sin 𝜃1 𝑛1
sin 𝜃2
Quantified
If your friend shone a laser from underwater to the air at an angle of 40
from the normal, at what angle would you expect it to travel?
n
Trywater
it first= 1.33; water = 40; nair = 1.00

Trywater
it first= ?
sinit first
𝜃1 𝑛1
Try
= sin 𝜃2 𝑛2 => 𝜃2 =
(sin 40°)(1.33)
−1
𝜃
= 𝑠𝑖𝑛
Try2it first
1.00
𝜃
Try2it=
first59°
sin 𝜃1 𝑛1
−1
𝑠𝑖𝑛 (
)
𝑛2
= 𝑠𝑖𝑛−1 (0.85)
Demo: triangular prism
1. Place a triangular prism upright
on the table.
2. Place a pin (or friend’s eye or
whatever) so that the light
passes parallel to the normal
into one face of the prism.
3. Look through for the object
through the other face of the
prism.
Where is it? Why can’t you see it?
Critical Angle
When light travels from an optically
dense material (e.g., water) to less
optically dense material (e.g., air), it
bends away from the normal.
At some incident angle, it will refract
parallel to boundary of the substance,
i.e., no light passes the boundary.
This angle is called the critical
angle.
Total Internal Reflection
Beyond the critical angle, light is
reflected back into the original
substance. This is called total
internal reflection
Total Internal Reflection
Total internal reflection
From fun swimming…
to fascinating demonstrations…
to cool party tricks…
to cool dorm room decorations…
to optics…
to revolutions in communications
Think about it…
Is it possible for light to undergo
total internal reflection as it travels
from air into water?
Explain your answer.
Collect some more (virtual) data
Go to PHeT Bending Light simulator.
Set medium 1 to water. Set medium 2 to
air.
At what angle does light totally internally
reflect at a water / air boundary?
Repeat for glass / water boundary.
Repeat for a glass / air boundary.
What patterns do you notice?
Critical angle
For values n1 < n2, (i.e., traveling from more dense to less materials)
@ 2 = 90, sin 𝜃𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 = sin 90°
𝑛2
𝑛1
sin 𝜃𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 =
𝑛2
𝑛1
NOTE:
𝑛
𝑛1
If n2 > n1, 𝜃𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 = sin−1 ( 2 ) is undefined
Think about it
Go to Physics Classroom Refraction simulator.
http://www.physicsclassroom.com/PhysicsClassroom/media/interactive/Refraction/i
ndex.html
Turn on protractor.
Randomize the top or bottom medium.
Using the Snell’s, calculate its index of refraction.
Predict its critical angle.
Summary of mathematical relationships
𝑛𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 =
𝑐
𝑣𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙
𝑛1 sin 𝜃1 = 𝑛2 𝑠𝑖𝑛𝜃2
sin 𝜃𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 =
𝑛2
𝑛1
Deriving Snell’s Law
Law of refraction
A wave or line of soldiers or whatever
hits the surface at an angle 1.
At time 0, A2 reaches boundary.
At some time t, A2 has traveled some
distance into new medium (l2)and A1 has
reached boundary after traveling some
distance (l1).
The wave or soldiers or whatever has
been deflected by angle 2.
i
r
Notice that the angle of incidence is
the same as the angle formed by
the wave front with the boundary.
sin 𝜃𝑖 =
𝑙1
𝑎
where l is the distance the wave
travels in a certain amount
of time, so l = v1t
𝑣1 𝑡
sin 𝜃𝑖 =
𝑎
i
l1
i
a
Notice that the angle formed by the
angle of refraction is the same as
the angle between the boundary
and the wave front.
a
sin 𝜃𝑟 =
𝑙2
𝑎
r
l2
where l is the distance the wave
travels in a certain amount
of time, so l = v2t
𝑣2 𝑡
sin 𝜃𝑟 =
𝑎
r
Law of refraction
Divide to obtain law of refraction:
𝑣1 𝑡
sin 𝜃1
= 𝑣𝑎 𝑡
sin 𝜃2
2
𝑎
𝑐
sin 𝜃1 𝑣1
𝑛1
= =𝑐
sin 𝜃2 𝑣2
𝑛2
Clever!
𝑛 sin 𝜃 = 𝑛 𝑠𝑖𝑛𝜃
1
1
2
2