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Resolution and Interference Patterns

Objective Introduction

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.

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).

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. 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
D. Procedure – use figure to construct interference apparatus (figure forthcoming)
  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.
E. Discussion
  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 1 Figure 2
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