1.  The Very Large Array is a radio interferometer.
         a. What is the maximum diameter of the array?

                   D = 22 miles = 35 km

         b. Using this diameter, and a wavelength of lambda = 6 cm, 
                calculate the angular resolution of the VLA.

             
                    theta  =  (70 degrees)  *  (0.06 m / 35000 m)

                           =  0.00012 degrees

         c. How does it compare to the resolution of the HST? 

                 HST has resolution in the visible of about

                    theta  =  0.000015 degrees

                 which is about 10 times better than the VLA
                 at this wavelength; but the VLA can observe
                 at somewhat shorter wavelengths.

                                
 2. The Spitzer Telescope looks at infrared light.

         a. What is the diameter of its mirror?

                 D = 0.85 m

         b. What is the longest wavelength of light detected by Spitzer?

                 lambda = 180 micrometers = 180 x 10^(-6) m  = 0.00018 m

         c. Calculate the angular resolution of Spitzer at this wavelength.

                    theta  =  (70 degrees)  *  (0.00018 m / 0.85 m)

                           =  0.015 degrees

         d. How does it compare to the resolution of the HST? 

                 HST has resolution in the visible of about

                    theta  =  0.000015 degrees

                 which is about 1000 times better than Spitzer
                 at this wavelength; but recall that this is the
                 longest wavelength Spitzer can detect, so the
                 resolution is lowest here.



   3. The Compton Gamma Ray Observatory looked at very high energy 
      radiation from space. You can find the URL of its web site 
      in the text above. There is a table of instrument properties 
      available from that URL.

         a. What was the effective collecting area of BATSE in "Large Area"
            mode, for light of energy 0.1 MeV?

                  1800 square centimeters

         b. Convert the collecting area into a rough diameter of the
            telescope, assuming a circular shape.

                                             2
                  For a circle, area = pi * R

                  Solving for radius of BATSE, we find radius

                        R  = approx 24 cm

                  so diameter D must be 

                        D  = approx 48 cm  = 0.48 m
      

         c. Gamma rays of energy 0.1 MeV have a very short wavelength, about
            1.2 x 10^(-11) meters. If BATSE were limited by diffraction, 
            what would its angular resolution be?

                    theta  =  (70 degrees)  *  (1.2 x 10^(-11) m / 0.48 m)

                           =  1.8 x 10^(-9) degrees

                    This would be about 10,000 times better than
                    the resolution of HST!


         d. Look in the instrument properties table for the "Position
            Localization" of strong sources with BATSE. 
            What was the actual angular resolution of BATSE? 

                    theta  =  3 degrees   in reality