Today’s Lecture
Refraction
Refraction of Light
• The incident ray, the
reflected ray, the refracted
ray, and the normal all lie
in the same plane
• The angle of refraction, θ2,
depends on the properties
of the two media and
speeds of light in them,
v1,2 = c/n1,2
Frequency and Wavelength at Refraction.
• As light travels from one medium to
another, its frequency does not
change
– Both the wave speed and the
wavelength do change
– The wavefronts do not pile up, nor
are created or destroyed at the
boundary, so ƒ must stay the
same
1
2
1
2
1
2
f = f
v
λ
=
v
λ
λ1 v1 c / n1 n2
= =
=
λ2 v2 c / n2 n1
So we see that
Light and Diamond
Diamond has a refractive index of n=2.42. For light with a wavelength of
λ=589nm find the (a) frequency, (b) light speed, and (c) wavelength in diamond.
The frequency is continuous across the boundary!
Refraction occurs because the
speeds of light, v1 and v2, are
different in the two media,
why?
The shaded areas share a
common hypotenuse at the
at the interface of length
As noted on the last slide, the wave frequency and period
must be the same on both sides, but the velocity is different.
Snell’s Law
•
In terms of indices of refraction the
law of refraction becomes
•
Sine of angle of incidence (refraction)
is inversely proportional to the index
of refraction of the medium
Snell’s law of refraction is written in a form
symmetric to the incident and refracted
beams:
http://www.physics.uoguelph.ca/applets/Intro_physics/refraction/LightRefract.html
• For a vacuum, n = 1
• For other media, n > 1
• n is a unitless ratio, n = c/v
Media with high n, like
diamond, are sometimes
called “optically dense”.
Example of Snell’s Law
What is the maximum θ1 for which the beam will
emerge through the bottom of the polystyrene
cylinder?
At the maximum θ1 the beam will emerge at the
edge of the polystyrene cylinder. From the previous
table, n1 = nair =1 and n2 = npolys = 1.49.
The sine of the exit angle is found from:
Using Snell’s law:
Refraction of Light.
When light refracts into a material,
where the index of refraction is
higher, the angle of refraction is
less than the angle of incidence
The ray bends toward the
normal
When light refracts into a material,
where the index of refraction is
lower, the angle of refraction is
greater than the angle of incidence
The ray bends away from the
normal
Light Passing Through a Slab of Glass
n1
sin θ 2 = sin θ1
n2
n2
sin θ 4 = sin θ 3
n1
θ3 = θ 2
n2 n1
sin θ 4 =
sin θ1
n1 n2
Hence:
A light beam incident on a transparent slam emerges
traveling in the same direction but displaced from its
original path.
θ 4 = θ1
Light Passing Through a Slab of Glass
We now know that θ1 = θ4, but what
about the amount of displacement?
Define the length of the path of the light
ray through the slab as l. Then the
thickness of the slab is d=l cosθ2.
In terms of l the displacement is:
For a 2cm thick glass slab (n=1.52) with an incident angle of 42o we find:
and
Color of the Sun From Under Water.
• The Sun looks yellow, since its radiation intensity has a
maximum at λ = 550nm, which is yellow light.
•
Wavelength of this yellow light in water will be λ’ = λ /nwater =
λ/1.33 = 413nm, which corresponds to violet light.
• Is the Sun going to look violet from under water?
• Of course not! The only thing that
matters is the frequency inside
your eye, which is independent
of the n of your vitreous humor.
How do the Refractive Indices Compare?
A material has an index of refraction that increases continuously
from top to bottom. Of the three paths shown in the figure below, which
path will be the path of a light ray as it goes through the material?
(b). When light goes from one material into one having a higher index of
refraction, it refracts toward the normal line of the boundary between the
two materials. If, as the light travels through the new material, the index
of refraction continues to increase, the light ray will refract more and
more toward the normal line.
Reflection and Refraction Indices
Sparkling Diamonds
• For a light beam incident upon a boundary between two
transparent media at 90°:
Ir
• Glass – n1 = 1.5,
= 0.04
Ii
I
• Diamond – n1 = 2.4, r = 0.17
Ii
n1
n2
Δn Δn
3
3
Δn
3
Δn = n2 − n1
Total intensity of light reflected from all 3 interfaces
(Δn/3 ) 2
1
(Δn ) 2
I r = 3 ⋅ Ii
= ⋅ Ii
2
(n 1 + n 2 )
3 (n 1 + n 2 ) 2
By using multiple interfaces we can significantly reduce
reflections – antireflection coatings.
When light refracts into a material,
where the index of refraction is
lower, the angle of refraction is
greater than the angle of incidence
The ray bends away from the
normal
Snell’s law of refraction :
n1 sin θ1 = n2 sin θ2
We have
n1 > n2
What if θ1 is so large that
n1 sin θ1 > n2
Then we always have
n1 sin θ1 > n2 sin θ2
?
Critical Angle
• A particular angle of
incidence will result in an
angle of refraction of 90°
– This angle of incidence is
called the critical angle
• For angles of incidence greater than the critical angle, the
beam is entirely reflected at the boundary
– This ray obeys the Law of Reflection at the boundary
• Total internal reflection occurs only when light attempts to move
from a medium of higher index of refraction to a medium of
lower index of refraction
Demo of critical angle
Whale Watch
??
And what does the whale see beyond the cone??
Reflections back into the water, other creatures of the deep.
• Critical angle - an angle of
incidence which result in an
angle of refraction of 90°
n2
sin θ c =
for n1 > n2
n1
Glass and air
n2
1
=
n1 1.5
θ c = sin −1 (2 / 3) = 42°
Water and air
1
n2
=
n1 1.33
θ c = sin (0.75) = 49°
1
n2
Air and vacuum
=
n1 1.00022
−1
θ c = 88.8°
• Critical angle - an angle of incidence which result in an
angle of refraction of 90°
Air and vacuum
1
n2
=
n1 1.00022
θ c = 88.8°
Cold air and hot air with 10% lower index of refraction
n2 1.0002
=
n1 1.00022
θ c = 89.6°
It is still 0.4° from the surface!
Cold air and hot air with 10% lower index of refraction
n2 1.0002
=
n1 1.00022
θ c = 89.6°
It is still 0.4° from the surface!
How do we get total internal
reflection in the lab?
Would a flat-parallel slab work?
No, we need a prism with
different angles at entrance
and exit surfaces!
Those prisms are excellent reflectors. No silver coating needed!
Optical fibers:
Total internal reflection at the boundaries
between the core and cladding.
sin θ c >
ncladding
ncore
θ c = 90° − φ2
sin φ2 = sin φ1 / ncore
At high angles on incidence (grazing angles) only small differences
of indices of refraction between the core and cladding are needed.
They are made of two different kinds of glass.
Dispersion
• The index of refraction in anything
except a vacuum depends on the
wavelength of the light
• This dependence of n on λ is called
dispersion
• The index of refraction for a material
usually decreases with increasing
wavelength
• The angle of refraction when light
enters a material depends on the
wavelength of the light
• Violet light refracts more than red
light when passing from air into a
material
Dispersion
• The angle of refraction when light
enters a material depends on the
wavelength of the light
• Violet light refracts more than red
light when passing from air into a
material
Prism Spectrometer
• A prism spectrometer uses a
prism to cause the
wavelengths to separate
• The instrument is commonly
used to study wavelengths
emitted by a light source
Examples of Spectra.
Spectra of emission and absorption;
line spectra and continuous spectra…
Using Spectra to Identify Gases
• All hot, low pressure gases
emit their own characteristic
spectra
• The particular wavelengths
emitted by a gas serve as
“fingerprints” of that gas
• Some uses of spectral
analysis
– Identification of molecules
– Identification of elements in
distant stars
– Identification of minerals
Refraction in a Prism
θ
What is the difference in the
directions of the two beams, red
and violet?
There is obviously no refraction
at the left (entrance) interface,
because the angle on incidence
is 0.
n1 sin θ1 = n2 sin θ 2
Always satisfied if θ1 = θ 2 = 0
Right interface – angle of incidence is α = 40°. When n400 = 1.538
and n700 = 1.516 the angles of refraction are:
θ700 = sin-1(n700 sin 40o) = sin-1(1.516 x .642) = 77.02o
θ400 = sin-1(n400 sin 40o) = sin-1(1.538 x .642) = 81.34o
The difference in refracted angles is:
θ 400 − θ 700 = 4.32°
Refraction With a Negative Index
Metamaterials!
From Snell’s law a negative index
implies a negative refracted angle!
2
1
− 2
The index can only be negative if
both ε and μ are both negative! This
does not occur in nature.
What about the velocity of the
refracted wave?
http://www.ee.duke.edu/~drsmith/negative_index_about.htm
Refraction With a Negative Index
Metamaterials!
When there is dispersion, ω = ω(k),
and a wave is expressed as:
When A(k) is peaked at k=k0 then:
Substituting this into the wave form:
The pulse (which carries the energy) propagates at vg = dω/dk with a velocity
called the group velocity.
http://www.ee.duke.edu/~drsmith/negative_index_about.htm