Lesson Plans Resolution Interferenc
Resolution and Interference Patterns
A. Objective:
To observe how changing aperture size and wavelength affects resolution
B. Introduction
What is
resolution?
Resolution is the ability to
distinctly make out two point objects. Resolution is the smallest angle (measured in degrees) two objects can
be apart and still be able to be seen as two distinct objects. The sun has an angular resolution of 0.5
degrees. As objects get smaller and
smaller, degrees are divided into arc minutes and further divided into arc
seconds. There are 60 arc minutes in
one degree and 60 arc seconds in one arc minute.
How do we calculate resolution?
To calculate resolution, we use the equation:
(q) = 1.22 (l/d)
q = resolution in radians
d = diameter
of telescope
When we observe an object, we
receive its electromagnetic radiation whether we use radio waves, visible
light, or X-rays. The light is
collected and forms a beam (figure 1). The
width of this beam is determined by wavelength of radiation being observed and
diameter of the aperture. The wider the
beam is, the less chance that you will be able to resolve another object that
is close by, since its beam will be superimposed on the main beam of the other
object. Only when each object’s beam is
distinct will you be able to pick out both objects. The beam is caused by constructive interference of the light
waves and has its strongest intensity in the main beam. Constructive interference occurs when two
wavelengths of light overlap in phase (figure 2). Other mini-beams (side lobes) of less intensity are made to either
side of the main beam. The main beam is flanked on either side by minimums
where destructive interference occurs (figure 2).
How does wavelength effect on resolution?
Electromagnetic radiation spans a long range
of wavelengths. The most familiar class
of electromagnetic radiation is visible light, which ranges from 400-700
nanometers (10-9 m). Let us
compare this form of radiation to radio waves, which range from kilometer to
millimeter wavelengths. According to
the above equation, visible light gives the better resolution, but why? Imagine looking at a brick wall, and you
send out a radio wave. Radio waves are
as big as individual bricks, and when they return, you will only be able to
make out an area as big as the wave, so you would be able to see the individual
bricks but nothing more in detail. However, if you use visible light, which has wavelength in the
nanometers, those tiny waves can fit into all the nooks and crannies, and when
the waves bounce off the brick back to you, you’ll be able to see all the
details of the brick. The difference
between these to forms of light helps determine its beamwidth.
How does diameter of aperture effect resolution?
You would probably guess that the diameter
effects resolution based on the ability to catch more radiation. While this gives a brighter image, it would
not help the resolution. The larger
dish helps limit the diffraction of light as it enters the dish. Diffraction is the apparent bending of light
at the edge of obstacles and is a fundamental property of light. The diffraction of light gives rise to an
Airy disc, which is similar to looking at the beamwidth pattern from the top
down. As the diffraction effect gets
bigger, the beamwidth gets bigger, thus larger dishes are needed to combat this
effect.
C.
Materials
1. String – 2 pieces per group. 1 Meter each
2. pins – 4
3. sheet of cardboard
4. markers
1.Place two pins 30 centimeters
apart about 20 centimeters in from one end of cardboard sheet. This will represent your aperture
2. Place 2 pins just across the
aperture boundaries 20 cm from aperture at one end (the near end). This will represent the light source
3. Loop and knot a piece of string around the pins in step 2. The string will represent the light waves.
4. With the red marker, mark off
wavelengths at 5 cm each along entire length of string
5. At 1 meter from aperture, draw a
line parallel to the aperture on the cardboard. This is your receiving area.
6. From the receiving area, pull the
strings taut, and guide them up until they overlap by one-half wavelength (use
one strings marks as the reference and guide the other string until its mark is
halfway between the other strings marks). This is one of your minimums and corresponds to destructive
interference.
7. Repeat 6 going down and mark the
other minimum. The boundary of the two
minimums measures your beamwidth. Record the beamwidth.
8. Using the blue marker, mark off
wavelengths at 10 cm along length of string and. repeat steps 6 &7 and
record beamwidth
9. Now decrease the aperture to 15
cm and perform steps 6 & 7 using either the 5 cm or 10 cm wavelength. Record beamwidth
10. Increase aperture to 60 cm and
perform steps 6& 7. Use the same
wavelength you used in step 9. Record beamwidth.
1. Explain how and why your
beamwidth changed when you varied wavelength?
2. Explain how and why your
beamwidth changed when you varied the aperture?
3. Using
your results, what advantage (disadvantage) does a radio astronomer encounter
with respect to an optical astronomer?
4. Suppose
your object is much bigger than your beamwidth; how would you observe the
object?

Figure 2
