Transcript destructive interference

Ch 16
Interference
Diffraction is the bending of waves
around obstacles or the edges of
an opening. Huygen’s Principle Every point on a wave front acts as
a source of tiny wavelets that move
forward with the same speed as the
wave; the wave front at a later
instant is the surface that is tangent
to the wavelets.
Interference alters the
intensity (brightness) of
light, just as it effects the
loudness of sound. The
waves combine following
the principle of linear
superposition.
When identical waves
cross in phase the waves
reinforce each other:
constructive interference.
When the waves are out of
phase they cancel each
other: destructive
interference.
If the wave sources
are coherent
sources, the
interference will
continue without
change.
In 1801 Thomas Young
(English) performed a
historic experiment
showing that two
overlapping light
waves interfere
with each other.
Monochromatic light passes
through a single narrow slit
and falls on two closely
spaced narrow slits S1 and
S2. These two slits act as
coherent sources of light that
interfere with each other.
When both light rays travel
the same distance, they
constructively interfere and
produce a bright spot.
When one light ray travels 1l
further than the other, they
still constructively interfere
and produce a bright spot.
But if one light ray travels
1/2 l further than the other,
they destructively interfere
and produce a dark spot.
Thus a series of bright and
dark areas is produced.
If illuminated with
monochromatic light,
successive pairs of slit images
will appear on either side of the
principle image. The first pair
are called the first-order images,
the second pair are the secondorder images.
d is the distance
between the slits,
and is called
the grating constant
l is the wavelength
θ is the diffraction angle
Monochromatic light is shined
on two slits. The distance
between the two slits is 0.030
mm. The second order bright
fringe is measured on a viewing
screen at an angle of 2.15° from
the central maximum.
Determine the wavelength of
the light.
Ex. 2 - Red light (l = 664 nm
in vacuum) is used in Young’s
experiment with the slits separated
by a distance d = 1.20 x 10-4 m.
The screen is located at a distance
from the slits of L = 2.75 m. Find
the distance y on the screen
between the central bright fringe
and the third-order bright fringe.
Diffraction gratings have as many
as 12,000 equally spaced parallel
grooves. When illuminated with
white light, each slit produces a
new wave front. These wave
fronts interfere and produce pairs
of continuous spectra equally
spaced on opposite sides of the
principle image.
If illuminated with
monochromatic light,
successive pairs of slit images
will appear on either side of the
principle image. The first pair
are called the first-order images,
the second pair are the secondorder images.
d is the distance
between the slits,
and is called
the grating constant
l is the wavelength
θ is the diffraction angle
To find the location of
the destructive
interference, replace m
with m + 1/2.
l = d sin q/(m + ½)
If white light is used rather than
monochromatic light, a central
white band results. Outside the
central point, fringes of all colors
result. At the central point all
colors are incident producing
white light. Outside the central
white fringe there is a bright
fringe for each value of l.
In these light colored bands,
red is farther out than the other
colors. Violet is closest to the
white central spot. Red light’s l is
larger than violet’s; therefore by
sin q = ml/d (l = d sin q/m) , the
red band is farther from the center.
There is one group of colored
fringes for each value of m.
Thin transparent soap films,
oil slicks, and wedge-shaped
films of air show varying
patterns of colors when
viewed by reflected white
light. Monochromatic light
produces light and dark
bands.
Some light is reflected at the top surface,
some is refracted and then partially
reflected at the bottom surface, eventually
refracted again at the top surface and
exiting parallel to the original reflected ray.
If the film is 1/4 wavelength
in thickness, the second ray
exits 1/2 wavelength behind
the originally reflected ray.
This should destructively
interfere and the film should
appear dark.
If the film is 1/2
wavelength in thickness,
the second ray should be
one whole wavelength
behind and constructive
interference should occur.
Observation shows the
opposite effect of that
expected. This occurs when
air is on both sides of the
film. Very thin layers
of soap film are very dark,
although we would expect
them to be bright.
Thomas Young explained
this by suggesting that
one of the reflected waves
undergoes a 180° phase
change during reflection.
The reflection where the
medium beyond has a
greater index of refraction
(top) undergoes the phase
shift. When the medium
beyond has a lower index
of refraction there is no
phase shift.
Thus the rule for
constructive and
destructive interference
for thin films bounded on
both sides by a medium
of lower index of
refraction is this:
Maximum constructive
interference occurs if the
optical path difference is an
odd number of half
wavelengths
(film thickness is an odd
number of quarter
wavelengths) and:
Maximum destructive
interference occurs when
the optical path difference
is a whole number of
wavelengths (film thickness
is an even number of
quarter wavelengths).
A vertical cross section of
soap film meets the odd
and even quarter
wavelength thicknesses
at successive intervals
down the film for
different colors.
Optically flat glass plates
separated by a thin film of
air produce regular
patterns of interference
fringes. irregular surfaces
produce irregular patterns.
An interferometer
can be used to
measure the
smoothness of
glass surfaces.
Ex. 4 - (a) A thin film of gasoline floats on a
puddle of water. Sunlight falls almost
perpendicularly on the film and reflects into
your eyes. Destructive interference
eliminates the blue color (l = 469 nm) , so
the film has a yellow hue. If the refractive
indices of blue light in gasoline and water
are 1.40 and 1.33 respectively, determine
the minimum nonzero thickness t of the film.
(b) Repeat part (a) assuming that the
gasoline is on glass (nglass = 1.52)
instead of water.
Ex. 5 - Under natural
conditions, thin films have a
multicolored appearance that
often changes while you are
watching them. Why are such
films multicolored, and what
can be inferred from the fact
that the colors change in time?
Ex. 6 - (a) Assume that green light
(l = 552 nm) strikes two glass
plates (n = 1.52) nearly
perpendicularly. Determine the
number of bright fringes that occur
between the place where the plates
touch and the edge of the sheet of
paper (thickness = 4.10 x 10-5 m).
(b) Explain why there is a dark
fringe where the plates touch.
Single-slit diffraction - A slit that
is only a few wavelengths wide
will produce alternating light and
dark bands. The very center of
the image is almost equidistant
from all parts of the slit, so it is
very bright because all
interference is constructive.
At points below and above
this bright band, where the
distance to each edge of the
slit differs by whole
wavelength, the difference in
the edge and the center
differs by 1/2 a wavelength.
For every point from the
bottom of the slit to the center,
there is a point from the center
to the top that is 1/2
wavelength different in
distance from the screen.
Therefore, all light is canceled
and the band is dark.
At two points further from the
center, the distance to each
edge of the slit differs by 1.5
wavelengths. In this case three
wavelets arrive, all offset by 1/2
a wavelength. Two of these
interfere constructively, the third
provides the light band (dimmer
than the center light band).
If the difference in distance
to the two edges is two
wavelengths, a dark band
results. If the difference is
2.5 wavelengths, a light band
again results, etc. The
intensity of the light bands
decreases with distance.
The angle to the normal
at which the bright fringes
(constructive interference)
are located can be found
using this equation:
sin q = (m + ½)l/d. d is the
distance between the slits;
m is the order of the image.
The formula for the
dark fringes is:
sin q = ml/d.
The dark fringes must
be halfway between the
bright fringes.
The intensity of
the light bands
decreases with
distance.
A Michelson interferometer splits
light rays into two parts then
reunites them after making one
part travel a longer distance.
At integer + 1/2 l differentials,
the reunited rays destructively
interfere. At integer l differentials,
the reunited rays interfere
constructively. These distances can
be measured to find the l of light.