Download Refraction

Refraction 1
Copyright © Mark Jordan
Davitt College,
Castlebar
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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
The Laws of Refraction
1)The incident ray, the normal at the point of incidence and the
refracted ray all lie ón the same plane.
2)The ratio of the sin of the angle of incidence to the sine of the
angle of refraction is a constant.
Refractive index of a medium
The refractive index is the ratio of the sine of the angle of
incidence to the sine of the angle of refraction when light travels
from a vacuum into that medium.
Normal
Refraction
Air
Glass
i
r
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
We can verify
Snell’s Law with
an Experiment
Refractive Index
The refractive index between two media – ratio of sin i to sin r
when light travels from one media into the other.
The order the light passes through the media is
important.
e.g light passing from glass into water - refractive index =
gnw
But when light passes from water into glass – refractive
index = wng
1
The relation between two is gnw =
w ng
Refractive Index
Example 1
A ray of light enters water from air. If the angle of incidence is 40o
find the angle of refraction given n for water is 1.33
Sin i
Sin 400
= 1.33
Gnw = Sin r =
Sin r
Sin 400
Sin r =
= 0.4833 r = Sin-1 (0.4833)
1.33
0
=
28.9
Example 2
The refractive index between glass and water is 0.91. What is the
refractive index between water and glass?
1
1
=
= 1.1
Wng =
n
0.91
g w
Experiment : To prove Snell's Law i.e sin i / sin r = constant
Using a ray box and a block of glass record the values for
the angle of incidence & angle of refraction as shown.
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)
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
sin i
sin r
= 1.48
Refractive index in terms of relative speed. Waves going from air
HONOURS ONLY
to glass at angle other
than 90o
 velocity decreases
 frequency remains
constant
 Wavelength
decreases (from c = f λ)
c
n
c
1
2
In slowing and changing
wavelength, the angle at
which they proceed
MUST change i.e. they
REFRACT
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
Total Internal Reflection
When a light ray travels from a denser medium into a rarer medium
it is refracted away from the normal.
As the angle of incidence increases so does the angle of refraction.
Eventually the angle i is reached that give an angle r of 90o.
The angle i for which this occurs is called the critical angle. C
If the angle of incidence is
increased beyond the critical angle
then all the light is reflected back
into the medium and none goes
into the 2nd medium. This is called
Total internal reflection.
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.
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
Air
Glass
i r
Angle of
incidence becomes
bigger than the
critical angle
then..
Total internal
reflection occurs
Refractive index and critical angle
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
air
glass
Total Internal Reflection and the Critical Angle
Example
The refractive index of glass is 1.5. Find the critical angle.
1
1
1
n
=

Sin
C
=
=
= 0.667
a g
Sin C
1.5
ang
Critical angle of glass in
prism 41.9o (approx)
Turning light through 180o using a
prism
 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
Critical angle of glass in
prism 41.9o (approx)
Turning light through 90o using a
prism
 Light from air to glass at 90o
No Refraction
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
Snell's Window
Since light can retrace it's path it is found that if light is to enter
water from air and arrive at a point below the surface only light
within a radius r can reach that point below the surface.
This is because any light from the point below the surface that has
an angle of incidence greater that the critical angle is reflected back
down i.e. Total internal reflection.
Mirage ----
Total Internal Reflection of light from sky
A ray of light is continually refracted away from the
normal as goes from denser to a rarer medium until the
angle of incidence is greater than the cricital angle causeing
total internal reflection
Cool air
High density
Warm air
Hot air
Low density
Hot ground
Refractive index in terms of real and apparent depth
An object viewed through a glass block appears to be nearer to you
than it really is.
The bottom of a swimming pool appears to be less deep that it
actually is.
The next slide shows why because of refraction an object appears
to be less deep that is actually is.
Refractive index in terms of real and apparent depth
Apparent
depth
Real
depth
n = Real depth
Apparent depth
Glass of
water
Refractive index in terms of real and apparent depth
It can be proved that the:
Refractive index =
Real depth
Apparent depth
Example
A block of glass of thickness 4 cm is placed ón a mark ón the
bench. When the mark is viewed perpendicularly a virtual image
appears 2.67 cm from the top of the glass block. Find the refractive
index of the glass.
Refractive index =
Real depth
Apparent depth =
4
2.67
= 1.5
Finding the refractive index of a liquid using real and apparent
Cork
depth
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
Optic Fibre another use of total internal reflection
Glass cladding of
low refractive index
Glass core of high
refractive index
Total internal reflection and optical fibres
Light enters the fibre at an angle greater than the critical angle. Total
internal reflection occurs and the ray is reflected in such a way as to
ensure continued total internal reflection.
Light can in escape if:
a) If the fibre is bend at too large an angle.
b) If two parts of the optical fibre come in contact with each other.
This prevented by coating the fibre with a layer of glass of lower
refractive index i.e a rarer or less dense medium.
Uses of optical fibres.
Telecommunications -electrical telephone signals are converted to
light and travel along optical fibres to their destination.
ADVANTAGES
Less energy loss than electrical cables
Much smaller than electrical wires.
Less interference with optical fibres
Medicine – optical fibres are used in endoscopes to look at
inaccessible parts of the body e.g. stomach.
Dentist's drill to carry light near the drill bit and so make it easier to
see the work area.