Refraction of Light-Notes.pdf

Chapter 35
Chapter 35 -
- Refraction
Refraction
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
© 2007

Objectives:
Objectives: After completing this
After completing this
module, you should be able to:
module, you should be able to:
•
• Define and apply the concept of the
Define and apply the concept of the index
index
of refraction
of refraction and discuss its effect on the
and discuss its effect on the
velocity and wavelength of light.
velocity and wavelength of light.
•
• Determine the changes in
Determine the changes in velocity
velocity and/or
and/or
wavelength
wavelength of light after refraction.
of light after refraction.
•
• Apply
Apply Snell
Snell’
’s law
s law to the solution of problems
to the solution of problems
involving the refraction of light.
involving the refraction of light.
•
• Define and apply the concepts of
Define and apply the concepts of total internal
total internal
reflection
reflection and the
and the critical angle
critical angle of incidence.
of incidence.

Refraction
Refraction
Water
Air
Refraction is the
bending of light as it
passes from one
medium into another.
Refraction is the
bending of light as it
passes from one
medium into another.
refraction
N
w
A
Note: the angle of
Note: the angle of
incidence
incidence 
A
A in air
in air
and the angle of
and the angle of
refraction
refraction 
A
A in water
in water
are each measured
are each measured
with the normal
with the normal N
N.
.
The incident and refracted
rays lie in the same plane
and are reversible.
and are reversible.
and are reversible.

Refraction Distorts Vision
Refraction Distorts Vision
Water
Air
Water
Air
The eye, believing that light travels in straight
lines, sees objects closer to the surface due to
refraction. Such distortions are common.

The Index of Refraction
The Index of Refraction
The index of refraction for a material is the ratio
of the velocity of light in a vacuum (3 x 108 m/s)
to the velocity through the material.
The
The index of refraction
index of refraction for a material is the ratio
for a material is the ratio
of the velocity of light in a vacuum (3 x 10
of the velocity of light in a vacuum (3 x 108
8 m/s)
m/s)
c
v
c
n
v

Index of refraction
c
n
v

Examples: Air n= 1; glass n = 1.5; Water n = 1.33
Examples: Air n= 1; glass n = 1.5; Water n = 1.33

Example 1.
Example 1. Light travels from air (
Light travels from air (n =
n = 1
1) into glass,
) into glass,
where its velocity reduces to only
where its velocity reduces to only 2 x 10
2 x 108
8 m/s
m/s.
.
What is the index of refraction for glass?
What is the index of refraction for glass?
8
8
3 x 10 m/s
2 x 10 m/s
c
n
v
 
vair = c
vG = 2 x 108 m/s
Glass
Air
For glass: n = 1.50
If the medium were
If the medium were water
water:
: n
nW
W =
= 1.33
1.33. Then
. Then
you should show that the velocity in water
you should show that the velocity in water
would be reduced from
would be reduced from c
c to
to 2.26 x 10
2.26 x 108
8 m/s
m/s.
.

Analogy for Refraction
Analogy for Refraction
Sand
Pavement
Air
Glass
Light bends into glass then returns along
original path much as a rolling axle would
when encountering a strip of mud.
3 x 108 m/s
3 x 108 m/s
2 x 108
m/s
vs < vp

Deriving Snell
Deriving Snell’
’s Law
s Law
Medium 1
Medium 2


 

Consider two light
Consider two light
rays. Velocities are
rays. Velocities are
v
v1
1 in medium 1
in medium 1
and
and v
v2
2 in med. 2.
in med. 2.
Segment
Segment R
R is common
is common
hypotenuse to two
hypotenuse to two rgt
rgt.
.
triangles. Verify shown
triangles. Verify shown
angles from geometry.
angles from geometry.
v1
v1t
v2
v2t 
1
1
R
R
1 2
1 2
sin ; sin
v t v t
R R
 
 
1
1 1
2
2 2
sin
sin
v t
v
R
v t v
R


 

Snell
Snell’
’s Law
s Law
1
2
Medium 1
Medium 2
The ratio of the sine of the
angle of incidence 1 to the
sine of the angle of refraction
2 is equal to the ratio of the
incident velocity v1 to the
refracted velocity v2 .
The ratio of the
The ratio of the sine
sine of the
of the
angle of incidence
angle of incidence 
1
1 to the
to the
sine
sine of the angle of refraction
of the angle of refraction

2
2 is equal to the ratio of the
is equal to the ratio of the
incident
incident velocity
velocity v
v1
1 to the
to the
refracted
refracted velocity
velocity v
v2
2 .
.
Snell’s
Law:
1 1
2 2
sin
sin
v
v



v1
v2

Example 2:
Example 2: A laser beam in a darkened room
A laser beam in a darkened room
strikes the surface of water at an angle of
strikes the surface of water at an angle of
30
300
0. The velocity in water is 2.26 x 10
. The velocity in water is 2.26 x 108
8 m/s.
m/s.
What is the angle of refraction?
What is the angle of refraction?
The incident angle is:
The incident angle is:

A
A = 90
= 900
0 –
– 30
300
0 = 60
= 600
0
sin
sin
A A
W W
v
v



8 0
8
sin (2 x 10 m/s)sin 60
sin
3 x 10 m/s
W A
W
A
v
v

   W = 35.30
W = 35.30
Air
Air
H2O
30
300
0
W

A
A

Snell
Snell’
’s Law and Refractive Index
s Law and Refractive Index
Another form of Snell
Another form of Snell’
’s law can be derived from
s law can be derived from
the definition of the index of refraction:
the definition of the index of refraction:
from which
c c
n v
v n
 
1 1 2
1
2 2 1
2
;
c
v v n
n
c
v v n
n
 
1 1 2
2 2 1
sin
sin
v n
v n


 
Snell
Snell’
’s law for
s law for
velocities and indices:
velocities and indices:
Medium 1
1
2
Medium 2

A Simplified Form of the Law
A Simplified Form of the Law
1 1 2
2 2 1
sin
sin
v n
v n


 
Since the indices of refraction for many common
Since the indices of refraction for many common
substances are usually available, Snell
substances are usually available, Snell’
’s law is
s law is
often written in the following manner:
often written in the following manner:
1 1 2 2
sin sin
n n
 

The product of the index of refraction and the
sine of the angle is the same in the refracted
medium as for the incident medium.

Example 3.
Example 3. Light travels through a block of glass,
Light travels through a block of glass,
then remerges into air. Find angle of emergence
then remerges into air. Find angle of emergence
for given information.
for given information.
Glass
Air
Air
Air
Air
n=1.5
First find
First find 
G
G inside glass:
inside glass:
sin sin
A A G G
n n
 

50
500
0
G

 0
sin (1.0)sin50
sin
1.50
A A
G
G
n
n

  
G = 30.70
G = 30.70
From geometry, note
From geometry, note
angle
angle 
G
G same for
same for
next interface.
next interface.
G
sin sin sin
A G G A
A A
n n n

 
 
Apply to each interface:
Apply to each interface:
e = 500
e = 500
Same as entrance angle!
Same as entrance angle!

Wavelength and Refraction
Wavelength and Refraction
The energy of light is determined by the frequency
The energy of light is determined by the frequency
of the EM waves, which remains constant as light
of the EM waves, which remains constant as light
passes into and out of a medium. (Recall
passes into and out of a medium. (Recall v = f
v = f
.
.)
)
Glass
Air n=1
n=1.5
A
G
fA= fG
G A
;
A A A G G G
v f v f
 
 
; ;
A A A A
G G G G
v f v
v f v
 
 
 
1 1 1
2 2 2
sin
sin
v
v
 
 
 

The Many Forms of Snell
The Many Forms of Snell’
’s Law:
s Law:
Refraction is affected by the index of refraction,
the velocity, and the wavelength. In general:
1 2 1 1
2 1 2 2
sin
sin
n v
n v
 
 
  
All the ratios are equal. It is helpful to recognize
that only the index n differs in the ratio order.
All the ratios are equal. It is helpful to recognize
that only the index n differs in the ratio order.
Snell’s
Law:

Example 4:
Example 4: A helium neon laser emits a beam of
A helium neon laser emits a beam of
wavelength
wavelength 632 nm
632 nm in air (
in air (n
nA
A = 1
= 1). What is the
). What is the
wavelength inside a slab of glass (
wavelength inside a slab of glass (n
nG
G = 1.5
= 1.5)?
)?
n
nG
G = 1.5;
= 1.5; 
A
A = 632 nm
= 632 nm
;
G
A A A
G
G A G
n n
n n
 


 
(1.0)(632 nm)
1.5
421 nm
G
  
Note that the light, if seen
Note that the light, if seen inside
inside the glass, would
the glass, would
be
be blue
blue. Of course it still appears red because it
. Of course it still appears red because it
returns to air before striking the eye.
returns to air before striking the eye.
Glass
Air
Air
Air
Air
n=1.5


G


G

Dispersion by a Prism
Dispersion by a Prism
Red
Orange
Yellow
Green
Blue
Indigo
Violet
Dispersion is the separation of white light into
its various spectral components. The colors
are refracted at different angles due to the
different indexes of refraction.
Dispersion
Dispersion is the separation of white light into
is the separation of white light into

Total Internal Reflection
Total Internal Reflection
Water
Air
light
The critical angle c is the
limiting angle of incidence
in a denser medium that
results in an angle of
refraction equal to 900.
The
The critical angle
critical angle 
c
c is the
is the
refraction equal to 90
0.
.
When light passes at an angle from a medium of
When light passes at an angle from a medium of
higher index to one of lower index, the emerging
higher index to one of lower index, the emerging
ray bends away from the normal.
ray bends away from the normal.
When the angle reaches a
When the angle reaches a
certain maximum, it will be
certain maximum, it will be
reflected internally.
reflected internally.
i = r
Critical
angle

c
c
90
900
0

Example 5.
Example 5. Find the critical angle of incidence
Find the critical angle of incidence
from water to air.
from water to air.
For critical angle,
For critical angle, 
A
A = 90
= 900
0
n
nA
A = 1.0;
= 1.0; n
nW
W = 1.33
= 1.33
sin sin
W C A A
n n
 

0
sin90 (1)(1)
sin
1.33
A
C
w
n
n
  
Critical angle: c = 48.80 Water
Air

c
c
90
900
0
Critical angle
In general, for media where
n
n1
1 >
> n
n2
2 we find that:
we find that:
1
2
sin C
n
n
 

Summary
Summary
1 2 1 1
2 1 2 2
sin
sin
n v
n v
 
 
  
Snell’s
Law:
c = 3 x 108 m/s
v
Index of refraction
c
n
v

Medium
n

Summary (Cont.)
Summary (Cont.)
The critical angle c is the
refraction equal to 900.
The
The critical angle
critical angle 
c
c is the
is the
0.
.
n1 > n2 we find that:
1
2
sin C
n
n
 
n1 > n2

c
c
90
900
0
Critical angle
n1
n2

CONCLUSION: Chapter 35
CONCLUSION: Chapter 35
Refraction
Refraction

Refraction of Light-Notes.pdf

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