Interaction light and
substance. Thermal
radiation bioobjects.
Wavefronts
At a given time, a wave's "wavefronts" are the planes where the
wave has its maxima.
A plane wave's wavefronts are equally spaced, one wavelength apart.
And they're perpendicular to the propagation direction.
A spherical wave is also a solution to
Maxwell's equations.
E(r,t) = (E0/r) Re exp i(kr – wt)
where k is a scalar, and
r is the radial co-ordinate.
Unlike a plane wave, whose amplitude remains constant as it
propagates, a spherical wave weakens. Its irradiance goes as 1/r2.
A plane wave impinging on a molecule
scatters into a spherical wave.
Scattering from an individual molecule is weak, but many such
scatterings can add up--especially if interference is constructive.
Spherical waves often
combine to form plane
waves.
A plane wave impinging on a
surface will produce a reflected
plane wave because all the
spherical wavelets interfere
constructively along a flat
surface.
What happens to light when it
encounters a surface?
It is scattered by the surface molecules. But a beam can remain a beam
if there is a direction for which constructive interference occurs.
Constructive interference occurs
for a reflected beam if the angle
of incidence = the angle of reflection.
Constructive interference occurs
for a transmitted beam if the sine of
the angle of incidence = sine of the
angle of "refraction." (Snell's Law)
Refraction and Snell's Law
AD = BD/sin(qi)
AD = AE/sin(qt)
qi
So:
BD/sin(qi) = AE/sin(qt)
But:
BD = vi Dt = (c0/ni) Dt
& AE = vt Dt = (c0/nt ) Dt
qt
So: (c0/ni) Dt/sin(qi) = (c0/nt) Dt/sin(qt)
Or:
ni sin(qi) = nt sin(qt)
Snell's Law for many layers
n1 sin q1   n2 sin q 2   n3 sin q3   ...  nm sin q m 
So we can ignore the intermediate layers!
Snell's Law explains many everyday effects
The refractive index varies with density (and hence temperature)
Prisms disperse light
Because the refractive index depends on wavelength, the refraction
angle also depends on wavelength.
Because n generally
decreases with wavelength (dn/dl < 0), the
shorter the wavelength,
the greater the refraction angle.
Differentiating implicitly w.r.t. l:
Or:
dq t dn sin(q i )

d l d l cos(q t )
cos(q t )
dq t dn

sin(q i )
dl dl
Prism dispersion
Rainbows result from refraction and
reflection of sunlight in water droplets
Note that there can be two rainbows, and the top one is inverted.
And the sky is much brighter below the bottom one.
Rainbow explanation: Light in a
spherical droplet
Light paths
Light can
enter a
droplet at
different
distances
from its
edge.
Water
droplet
Path leading
to minimum deflection
~180° deflection
Minimum deflection angle (~138°);
rainbow radius = 42°
We must compute the angle of the emerging light as a function of the
incident position.
Plotting deflection angle vs.
wavelength is the key.
Because n varies with wavelength, the
minimum deflection angle varies with color.
Lots of violet deflected at this angle
Lots of red deflected at this angle
Lots of light of all colors is deflected by >138°,
so the region below rainbow is bright and white.
Explanation of 2nd rainbow
The 2nd (upper) rainbow results from light entering the droplet
in its lower half and making 2 internal reflections in the droplet.
Water droplet
Minimum deflection angle (~232.5°)
yielding a rainbow radius of 52.5°.
Because energy is lost at each reflection, the 2nd rainbow is weaker.
3rd and 4th rainbows are even weaker, are more spread out, and are
toward the sun.
5th rainbow overlaps the 2nd, and 6th is below the 1st, but are too weak
to see.
Coherent vs. Incoherent light scattering
Coherent light scattering: scattered wavelets have nonrandom
relative phases in the direction of interest.
Incoherent light scattering: scattered wavelets have random
relative phases in the direction of interest.
Example:
Forward scattering is coherent—
even if the scatterers are randomly
arranged in space.
Path lengths are equal.
Off-axis scattering is incoherent
when the scatterers are randomly
arranged in space.
Path lengths are random.
Coherent vs. Incoherent Scattering
Coherent scattering:
N
Total complex amplitude, Acoh  1  N . Irradiance, I  A2. So: Icoh  N2
m 1
N
Incoherent scattering: Total complex amplitude, Aincoh   exp(iq m )
m 1
The irradiance
I
2
N
 exp(iq
m 1
m
)

N
N
 exp(iq ) exp(iq )
m 1
m
n 1
n
 N N

 N N

   exp[i (q m  q n )]     exp[i (q m  q n )] 
N
 m 1 n 1
 m  n  m 1 n 1
mn
qm=qn
qm  qn
So incoherent scattering is weaker than coherent scattering, but not zero.
Why the sky is blue
Air molecules scatter light, and the scattering is proportional to w4
Blue light is scattered out of the beam, leaving yellow light behind,
so the sun appears yellow.