Transcript Phys132 Lecture 5 - University of Connecticut

Physics 1502: Lecture 28
Today’s Agenda
• Announcements:
– Midterm 2: Monday Nov. 16 …
– Homework 08: due next Friday
• Optics
– Waves, Wavefronts, and Rays
– Reflection
– Index of Refraction
Waves, Wavefronts, and Rays
• Consider a light wave (not necessarily visible) whose E
field is described by,
• This wave travels in the +x direction and has no
dependence on y or z, i.e. it is a plane wave.
3-D Representation
RAYS
Wave Fronts
EM wave at an interface
• What happens when light hits a surface of a material?
• Three Possibilities
– Reflected
– Refracted (transmitted)
– Absorbed
incident
ray
reflected
ray
MATERIAL 1
MATERIAL 2
refracted
ray
Geometric Optics
• What happens to EM waves (usually light) in
different materials?
– index of refraction, n.
• Restriction: waves whose wavelength is much
shorter than the objects with which it interacts.
• Assume that light propagates in straight lines,
called rays.
• Our primary focus will be on the REFLECTION and
REFRACTION of these rays at the interface of two
materials.
incident
ray
reflected
ray
MATERIAL 1
MATERIAL 2
refracted
ray
Reflection
• The angle of incidence equals the angle of reflection
 qi = qr , where both angles are measured from the normal:
· Note also, that all rays lie in the “plane of incidence”
qi qr
· Why?
» This law is quite general; we supply a limited
justification when surface is a good conductor,
• Electric field lines are perpendicular to the
conducting surface.
• The components of E parallel to the surface of the
incident and reflected wave must cancel!!
Ei
Er
qi
qr
qi qr
x
Index of Refraction
• The wave incident on an interface can not only
reflect, but it can also propagate into the second
material.
• Claim the speed of an electromagnetic wave is
different in matter than it is in vacuum.
– Recall, from Maxwell’s eqns in vacuum:
– How are Maxwell’s eqns in matter different?
 e0 e , m0 m
· Therefore, the speed of light in matter is related to the
speed of light in vacuum by:
where n = index of refraction of the material:
· The index of refraction is frequency dependent: For example
nblue > nred
Refraction
•
How is the angle of refraction related to the angle of
incidence?
– Unlike reflection, q1 cannot equal q2 !!
q1
n1
n2
» n1  n2  v1  v2
but, the frequencies (f1,f2) must be the same  the
wavelengths must be different!
Therefore, q2 must be different from q1 !!
q2
Snell’s Law
• From the last slide:
q2
q1 L
q2
q1
q1
q2
q2
The two triangles above each have hypotenuse L
\
But,
n1
n2
1
Lecture 28, ACT 1
• Which of the following ray diagrams could represent the passage of
light from air through glass and back to air?
(a)
(b)
(c)
air
air
air
glass
glass
glass
air
air
air
Lecture 28, ACT 1
• Which of the following ray diagrams could represent the passage of
light from air through glass and back to air?
(a)
(b)
(c)
air
air
air
glass
glass
glass
air
air
air
Lecture 28, ACT 2
• Which of the following ray diagrams could represent the passage of
light from air through glass and back to air?
(a)
(b)
(c)
(d)
A prism does two
things,
1. Bends light the
same way at both
entrance and exit
interfaces.
2. Splits colours due to
dispersion.
Index of refraction
Prisms
1.54
ultraviolet
absorption
bands
1.52
1.50
frequency
white light
prism
Prisms
Entering
q1
Exiting
q3
q2
For air/glass interface, we
use n(air)=1, n(glass)=n
q4
Prisms
Overall Deflection
f
q1
q3
q4
q2
• At both deflections the amount of downward deflection
depends on n (and the prism apex angle, f).
• The overall downward deflection goes like,
g ~ A(f) + B n
• Different colours will bend different amounts !
Lecture 28, ACT 3
White light is passed through a
prism as shown. Since n(blue)
> n(red) , which colour will end
up higher on the screen ?
A) BLUE
B) RED
?
?
LIKE SO!
In second rainbow
pattern is reversed
Total Internal Reflection
– Consider light moving from glass (n1=1.5) to air (n2=1.0)
n1
incident
ray
q1 qr
reflected
ray
GLASS
q2
refracted
ray
n2
AIR
ie light is bent away from the normal.
as q1 gets bigger, q2 gets bigger, but
q2 can never get bigger than 90 !!
2
In general, if sin q1  sin qC  (n2 / n1), we have NO refracted ray;
we have TOTAL INTERNAL REFLECTION.
For example, light in water which is incident on an air surface with
angle q1 > qc = sin-1(1.0/1.5) = 41.8 will be totally reflected. This
property is the basis for the optical fibers used in communication.
ACT 4: Critical Angle...
An optical fiber is
cladded by another
dielectric. In case I this
is water, with an index
of refraction of 1.33,
while in case II this is
air with an index of
refraction of 1.00.
Compare the critical
angles for total internal
reflection in these two
cases
a) qcI>qcII
b) qcI=qcII
c) qcI<qcII
water n =1.33
Case I
qc
glass n =1.5
water n =1.33
air n =1.00
Case II
qc
glass n =1.5
air n =1.00
ACT 5: Fiber Optics
The same two fibers are
used to transmit light
from a laser in one
Case I
room to an experiment
in another. Which
makes a better fiber,
the one in water (I) or
the one in air (II) ?
a) IWater
b) IIAir
Case II
water n =1.33
qc
glass n =1.5
water n =1.33
air n =1.00
qc
glass n =1.5
air n =1.00
Problem
You have a prism that from the side forms a
triangle of sides 2cm x 2cm x 22cm, and
has an index of refraction of 1.5. It is
arranged (in air) so that one 2cm side is
parallel to the ground, and the other to the
left. You direct a laser beam into the prism
from the left. At the first interaction with the
prism surface, all of the ray is transmitted
into the prism.
a) Draw a diagram indicating what happens to
the ray at the second and third interaction
with the prism surface. Include all reflected
and transmitted rays. Indicate the relevant
angles.
b) Repeat the problem for a prism that is
arranged identically but submerged in water.
A) Prism in air
Solution
• At the first interface q=0o, no deflection of initial light direction.
• At 2nd interface q=45o, from glas to air ?
• Critical angle: sin(qc)=1.0/1.5 => qc= 41.8o < 45o
• Thus, at 2nd interface light undergoes total internal reflection
• At 3rd interface q=0o, again no deflection of the light beam
B) Prism in water (n=1.33)
• At the first interface q=0o, the same situation.
• At 2nd interface now the critical angle: sin(qc)=1.33/1.5 => qc= 62o > 45o
• Now at 2nd interface some light is refracted out the prism
•
n1 sin(q1) = n2 sin(q2)
=> at q2 = 52.9o
• Some light is still reflected, as in A) !
• At 3rd interface q=0o, the same as A)