Microsoft Word - Phys 11_diffaraction_316_2021.docx
Phys 316: Diffraction grating (lab 11)
Purpose:
To measure the wavelength of light with a diffraction grating and use the grating to identify
some elements.
Apparatus:
Mercury vapor lamp, a coarse diffraction grating (~1000 lines/cm), a fine diffraction grating
(~5000 lines/cm), a holder for mounting the grating on some form of support, a meter stick.
Theory:
Before we get started with the diffraction grating, it would be helpful to start with a simpler case,
that of a double slit. Thomas Young did the following experiment in XXXXXXXXXXTwo na
ow slits A and B are
cut in an opaque screen and illuminated by a light source S. When the light from the two slits is received
on a white screen OP the illumination is found to vary in a peculiar manner, consisting of a
ight band at
O bordered on both sides by alternate dark and
ight bands. The
ight bands are due to constructive
interference, the dark bands are due to destructive interference. The central band is
ight because it is
the same distance from A and B, and the light is in phase at point O. However, at some position P, the
distance BP is greater than that of AP by one half wavelength, and the light a
iving from the two slits is
opposite in phase, resulting in darkness. As point P moves farther from O, it will a
ive at a place where
BP is one wavelength longer than AP, and a
ight band will appear again. If we keep this up some more,
we will find more dark and
ight bands. Or to say it another way,
Δ = nλ (constructive interference)
Δ = (n + 2)λ (destructive interference)
where λ is the wavelength of the light, and n is any
integer.
A diffraction grating consists of an opaque
material in which a large number of equally spaced slits or lines have been scratched, thus allowing light
to pass through (lots and lots of double slits). When these were first made, lines were actually scratched
y a machine. These days, gratings are produced by exposing a photographic film using laser light and
these exposed films became the slits through which light can pass.
The discussion of what happens to a wave as it passes through a diffraction grating can be
described by making use of the principles of wave motion as suggested by Huygens. In 1678, Huygens
suggested that a wave maybe thought of as propagating itself by means of an infinite number of secondary
wavelets. Think of a wave moving through space. Every point in the wave front becomes a source of
wavelets. The sum of all the wave wavelets at any time constitutes an advancing wave front. Diffraction
phenomena are due
the interference between wavelets from different portions of the same wave front.
In the above figure, let MN represent a plane wave front (the waves are in phase along this line)
in a monochromatic beam of light incident upon a grating. The slits are represented by A, B, C, and D, the
distance between the slits is d, and Δ for the adjacent slits is the same. The inte
uption of the wave by
the grating gives rise to a set of wavelets with their origins at the slits. These secondary wavelets are in
phase at their origins. Any envelope of the secondary wavelets makes up a new wave front indicating the
propagation of light along a line normal to the wave front. The first envelope to consider is the one
epresented by the line M'N' parallel to the original wave front. The only effect of the grating upon this
part of the light is to reduce its intensity by cutting out parts of the beam. So a lens placed after the
grating as shown below would produce a direct image at O precisely (except for the reduction in intensity)
as if the grating were not there. This is known as a Fraunhofer pattern.
Another envelope, represented by the AJ line can be
constructed so that it includes a given wave from A, the first
preceding wave from B, the second preceding wave from C, and
so on. Thus, AJ is also a plane wave front traveling in the
direction indicated by the normal from the slits to AJ.
Therefore, after passing through the grating, light is
propagated not only in the original direction but at an angle
determined by the condition for constructive
interference.
Since Δ = d sin θ, where θ is the angle the diffracted wave front makes with the plane of the grating
(or the amount the diffracted ray is bent), we can write
d sin θ = nλ (a.k.a. the grating formula)
If n is one, the wave front gives rise to an image at I1 and I1', called a first order image. If n is 2,
the wave front AK is diffracted through a greater angle and a second order image is produced at I2 and I2'.
Thus, immediately on either side of the central image O are the first order images flanked by the images
of successively increasing orders.
If the light source is monochromatic (one value of λ) the angle θ is definite and monochromatic
images of the slit are formed at I1, I2, and so on. If the source is heterochromatic (like a mercury lamp)
the various wavelengths are diffracted at different angles and each image becomes a spectrum, composed
of a series of monochromatic images, with the blue end nearest to the central image and the red end
farthest away. The central image is not a spectrum but retains the composite color of the source, since
at O the retardation is zero for all wavelengths.
If we think about the grating formula, we can see that the number of orders that can be
produced is limited by the grating space d, since θ cannot exceed 90. The highest order is then given by:
So a coarse grating (large d) gives rise to a large number of orders, while a finely ruled grating maybe
limited to one or two orders. At the same time, it can be shown that the spectra produced by a fine grating
is spread out more than one formed by a coarse grating. So, what type of grating you use depends on
what you want to do. There are trade off's
Procedure:
For your lab today we will be using just green, blue, orange light.
Open:
https:
ophysics.com/l5b.html
Its your set up in diffraction grating.
You could change:
Lens to grating distance,
Grating lines per mm
Wavelength
This is just an example.
Also, this example show how to calculate I’1.
Find distance between I’1 and I1 and divide in half
Choose the green light first. λ = 5.461 x 10-7 meters for green
Choose lens to “Grating distance” as 7 m and
“Grating lines per mm” as 400 mm
Check the button “Grating in place”.
Record I1 and I1’ to the table and calculate an angle as average if you know both I1 and I1’
(hint from your trig how to get your angle: tan-1 (Opposite / Adjacent) = θ(angle)
Color
Distance
to I1
Distance
to I1'
Angle average
(θ)
Grating
Constant d
Green
Calculate grating constant distance d from our grating lines per mm _________________
__________________________________
(show calculations)
Calculated value for wavelength λ (Recall: nλ = d sin θ, n=1) _________________________
Calculate your percent e
or between your calculated value and known value.
λ = 5.461 x 10-7 meters for green
Δ% = _ ________________________
Show all calculations in detail.
Change your “lines per mm” to 300 lines/mm and submit all numbers again:
Color
Distance
to I1
Distance
to I1'
Angle average
(θ)
Grating
Constant d
Green
Calculate constant grating distance d from our grating lines per mm.
__________________________________
Calculate λ (Recall: nλ = d sin θ )_____________________________
Calculate your e
or between calculated and know value for green wavelength:
XXXXXXXXXXΔ% = _ ________________________
Show all calculations in detail.
Now change your wavelength to
λ = 4.458 x 10-7 meters for blue
and submit all I1 and I1’ to the table and use last number grating constant d
Now change your wavelength to
and λ = 6.50 x 10-7 meters for orange
Color
Distance
to I1
Distance
to I1'
average (θ)
Wavelength
(λ)
Blue
Orange
Calculate λ for blue (Recall: nλ = d sin θ)_________________________
Calculate λ for orange (Recall: nλ = d sin θ)_________________________
find the percent difference between your values and the accepted values
Blue___________________________________
Orange__________________________________
Part II
Source of white light containing all visible wavelengths is shining behind the gas. When photons
(bursts of light) from the light source make their way through this gas, some of them can interact
with the atoms provided that they have just the right frequency to bump an electron of that
element up to a higher energy level.
Photons at those particular frequencies are thus abso
ed by the gas and all those other
frequencies would come through okay. Then the spectrum of