Chapter 35 - Refraction
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
©
2007
Objectives: After completing this
module, you should be able to:
• Define and apply the concept of the index
of refraction and discuss its effect on the
velocity and wavelength of light.
• Apply Snell’s law to the solution of problems
involving the refraction of light.
• Determine the changes in velocity and/or
wavelength of light after refraction.
• Define and apply the concepts of total internal
reflection and the critical angle of incidence.
Refraction
Refraction is the
bending of light as it
passes from one
medium into another.
Note: the angle of
incidence qA in air
and the angle of
refraction qA in water
are each measured
with the normal N.
Air
qA
qw
N
Water
refraction
The incident and refracted
rays lie in the same plane
and are reversible.
Refraction Distorts Vision
Air
Air
Water
Water
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 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.
Index of refraction
c
n
v
c
c
n
v
v
Examples: Air n= 1; glass n = 1.5; Water n = 1.33
Example 1. Light travels from air (n = 1) into glass,
where its velocity reduces to only 2 x 108 m/s.
What is the index of refraction for glass?
Air
vair = c
8
c 3 x 10 m/s
n 
8
v 2 x 10 m/s
Glass
vG = 2 x
108
m/s
For glass: n = 1.50
If the medium were water: nW = 1.33. Then
you should show that the velocity in water
would be reduced from c to 2.26 x 108 m/s.
Analogy for Refraction
3 x 108 m/s
Pavement
Air
Glass
2 x 108
m/s
Sand
vs < vp
3 x 108 m/s
Light bends into glass then returns along
original path much as a rolling axle would
when encountering a strip of mud.
Deriving Snell’s Law
Consider two light
rays. Velocities are
v1 in medium 1
and v2 in med. 2.
Segment R is common
hypotenuse to two rgt.
triangles. Verify shown
angles from geometry.
v1t
v2t
sin q1  ; sinq 2 
R
R
Medium 1
v1
q1
v1t
q1 R q2
v t q1
2
q2
v2
Medium 2
v1t
sin q1
v1
R


sin q 2 v2t
v2
R
q2
Snell’s Law
q1
Medium 1
v1
q2
v2
Medium 2
Snell’s
Law:
The ratio of the sine of the
angle of incidence q1 to the
sine of the angle of refraction
q2 is equal to the ratio of the
incident velocity v1 to the
refracted velocity v2 .
sin q1 v1

sin q 2 v2
Example 2: A laser beam in a darkened room
strikes the surface of water at an angle of
300. The velocity in water is 2.26 x 108 m/s.
What is the angle of refraction?
300
Air
qA
H2O
The incident angle is:
qA = 900 – 300 = 600
sin q A vA

sin qW vW
qW
vW sin q A (2 x 10 m/s)sin 60
sin qW 

8
vA
3 x 10 m/s
8
0
qW = 35.30
Snell’s Law and Refractive Index
Another form of Snell’s law can be derived from
the definition of the index of refraction:
q1
Medium 1
q2
Medium 2
c
c
n  from which v 
v
n
c
v1
v1 n2
n1

;

c
v2
v2 n 1
n2
Snell’s law for
velocities and indices:
sin q1 v1 n2
 
sin q 2 v2 n1
A Simplified Form of the Law
Since the indices of refraction for many common
substances are usually available, Snell’s law is
often written in the following manner:
sin q1 v1 n2
 
sin q 2 v2 n1
n1 sin q1  n2 sin q 2
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. Light travels through a block of glass,
then remerges into air. Find angle of emergence
for given information.
First find qG inside glass:
Air Glass
qG
500
q
qG
n=1.5 Air
From geometry, note
angle qG same for
next interface.
nA sin q A  nG sin qG
nA sin q A (1.0)sin 500
sin q G 

nG
1.50
qG = 30.70
Apply to
qe each
= 500interface:
nSame
nG sin q G angle!
 nA sin q A
A sin qas
A entrance
Wavelength and Refraction
The energy of light is determined by the frequency
of the EM waves, which remains constant as light
passes into and out of a medium. (Recall v = fl.)
Air
lA
fA= fG
Glass n=1
n=1.5
lG
lG < lA
v A  f Al A ;
vG  fG lG
vA f lA vA lA

;

;
vG f lG vG lG
sin q1 v1 l1
 
sin q 2 v2 l2
The Many Forms of Snell’s Law:
Refraction is affected by the index of refraction,
the velocity, and the wavelength. In general:
Snell’s
Law:
sin q1 n2 v1 l1
  
sin q 2 n1 v2 l2
All the ratios are equal. It is helpful to recognize
that only the index n differs in the ratio order.
Example 4: A helium neon laser emits a beam of
wavelength 632 nm in air (nA = 1). What is the
wavelength inside a slab of glass (nG = 1.5)?
Air Glass
qG
q
q
qG
n=1.5 Air
nG = 1.5;
lA = 632 nm
l A nG

;
lG nA
nAl A
 lG
nG
(1.0)(632 nm)
lG 
 421 nm
1.5
Note that the light, if seen inside the glass, would
be blue. Of course it still appears red because it
returns to air before striking the eye.
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.
Total Internal Reflection
When light passes at an angle from a medium of
higher index to one of lower index, the emerging
ray bends away from the normal.
Air
900
qc
i=r
Critical
light angle Water
When the angle reaches a
certain maximum, it will be
reflected internally.
The critical angle qc is the
limiting angle of incidence
in a denser medium that
results in an angle of
refraction equal to 900.
Example 5. Find the critical angle of incidence
from water to air.
For critical angle, qA = 900
nA = 1.0; nW = 1.33
nW sin q C  nA sin q A
nA sin 900 (1)(1)
sin q C 

nw
1.33
Critical angle: qc = 48.80
In general, for media where
n1 > n2 we find that:
Critical angle
Air
900
qc
Water
n1
sin q C 
n2
Summary
c = 3 x 108 m/s
Medium
n
v
Index of refraction
c
n
v
Refraction is affected by the index of refraction,
the velocity, and the wavelength. In general:
Snell’s
Law:
sin q1 n2 v1 l1
  
sin q 2 n1 v2 l2
Summary (Cont.)
The critical angle qc is the
limiting angle of incidence
in a denser medium that
results in an angle of
refraction equal to 900.
In general, for media where
n1 > n2 we find that:
Critical angle
n2
900
qc
n1
n1 > n2
n1
sin q C 
n2
CONCLUSION: Chapter 35
Refraction