Transcript Refraction

By
Mark Jordan
©
The Professional Development Service for Teachers is funded by the
Department of Education and Skills under the National Development Plan
OUTLINE OF THE DAY
To Be Completed by Class Teacher
TheThe
Professional
Development
Service
for Teachers
is funded
by the
Professional
Development
Service
for Teachers
is funded
by Department
the
of
Education
and
Science
under
the
National
Development
Plan
Department of Education and Skills under the National Development Plan
A bowl has a coin placed on the bottom just out of view. When water is
poured into the bowl suddenly the coin can be seen – the water seems to
‘lift’ the coin. Why do you think this happens?
The pencil on the left seems broken where it enters the
water. Why?
These are both examples of light bending or
Refraction. Can you explain how this is caused?
Hint next Slide.
3
• Light is stated to travel at , approx, 3.00 x 108 m/s. Does light always
travel at that speed, or can it travel faster or slower?
• How does the line of a marching band make a turn without getting out
of step? How could this be related to the bending of light through
glass
4
Normal
Refraction
Air
Glass
r
i
Angle of
incidence
Ray of light travelling from less
dense medium (e.g. air) to more
dense medium (e.g. glass) changes
direction or bends – called
Refraction.
A normal (90o) to point where the
light enters dense medium (glass)
shows ray bending into the normal.
Snell, a Dutch mathematician, found
that :-
sin i  sin r
5
Normal
Refraction
r
Air
Glass
Angle of
refraction Light ray travelling from a more
dense medium (glass) to a less
dense medium (air) bends away
from the normal - Snell’s Law
again applies i.e. sin i α sin r
Refraction is the bending of
a wave at the boundary when
it is going from one medium
to another
i
We can verify
Snell’s Law with
an Experiment
6
Using a ray box and a block of glass record the values for
the angle of incidence & angle of refraction as shown.
7
Find the sine of angles of incidence and refraction and record
i/ o
r/o
sin i
sin r
35o
23o
0.57
0.39
40o
26o
0.64
0.44
45o
29o
0.7
0.49
50o
32o
0.76
0.53
55o
34o
0.81
0.56
60o
36o
0.86
0.59
65o
38o
0.90
0.62
Draw graph of sin i (y- axis) against sin r (x-axis)
8
Draw graph of sin i (y-axis) against sin r (x- axis)
Refractive Index
(0.68, 1.0)
Straight line graph through the origin proves Snell’s
Law
1
0.9
0.8
i.e. Sin i  Sin r
Sin i
0.7
0.6
0.5
0.4
(0.14, 0.2)
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Sin r
Choose coordinates on
Refractive index (n) =
line
9
sin i
sin r
= 1.48
Waves going from air
to glass at angle other
than 90o
 velocity decreases
 frequency remains
constant
 Wavelength
decreases (from c = f λ)
c
n
c
In slowing and changing
wavelength, the angle at
which they proceed
MUST change i.e. they
REFRACT
1
2
10
Waves going from air
to glass at 90o
 velocity decreases
 frequency remains
constant
 Wavelength
decreases (from c = f λ)
If the waves arrive
perpendicular to the
boundary they do not
REFRACT
11
r
r= 90°
Air
Air
Glass
Glass
i
 light ray
travelling from
more dense to less
dense medium
 refraction
occurs
 refracted ray
bends away from
the normal.
Air
Glass
ic
As the angle of
incidence gets
bigger, angle of
refraction gets
bigger &
eventually
becomes 90o. This
angle of incidence
is called the
critical angle
12
i r
Angle of
incidence becomes
bigger than the
critical angle
then..
Total internal
reflection occurs
Refraction
r= 90°
Air
Glass
It is possible to calculate the
refractive index using the
critical angle
c
sin c
n 
sin 90
sin 90
 n 
sin c
1
 n 
sin c
glass
As the angle of
incidence gets bigger,
the angle of refraction
gets bigger &
eventually becomes 90o.
This angle of incidence
is called the critical
angle
air
13
air
glass
14
Critical angle of glass in
prism 41.9o (approx)
 Light from air to glass at 90o
No Refraction
450
450
450
450
 Light attempting to go from
glass to air but angle of
incidence greater than critical
angle.
Total Internal Reflection
 Light from glass to air at 90o
No Refraction
15
Critical angle of glass in
prism 41.9o (approx)
 Light from air to glass at 90o
No Refraction
 Light attempting to go from
glass to air but angle of
incidence greater than critical
angle.
450
Total Internal Reflection
450
 Light from glass to air at 90o
No Refraction
16
Mirage ----
Total Internal Reflection of light from sky
Cool air
High density
Warm air
Hot air
Low density
Hot ground
17
Refraction
n = Real depth
Apparent depth
Glass of
water
18
Refraction
Cork
Pin
To find refractive
index of a liquid
(water) by measuring
Real depth over
Apparent depth of an
object in the liquid.
Apparent depth
Mirror
Real depth
Water
Image
Pin
19
An optical fiber is a thin, flexible, transparent fiber of glass that acts
as a "light pipe”. See image. Using your knowledge of TIR can you
explain the following.
• How can a beam of laser light travel
through a clear optical fibre without
any significant amount of light
escaping when the beam strikes the
side of the fibre?
• Do you think the laser light could
stay in a fiber that is bent? Why or….
why not?
Click the links below to view a video (must have Quicktime installed)
and for more information/images.
http://www.teachersdomain.org/asset/lsps07_vid_laserfall/
http://en.wikipedia.org/wiki/File:Fiber_optic_illuminated.jpg
20
Optic Fibre
Glass cladding of
low refractive index
21
Glass core of high
refractive index
Glass of high refractive index
Glass of low refractive index
N
N
Normal
22
N
The Professional Development Service for Teachers is funded by the
Department of Education and Skills under the National Development Plan