A radio telescope consists of two antennas separated by a distance of 218.0 m. Both antennas are tuned to a particular frequency, such as 65.0 MHz. The signals from each antenna are fed into a common amplifier, but one signal first passes through a phase adjuster that delays its phase by a chosen amount so that the telescope can look in different directions. When the phase delay is zero, plane radio waves that are incident vertically on the antennas produce signals that add constructively at the amplifier. What should the phase delay be so that signals coming from an angle 16.0 ° with the vertical (in the plane formed by the vertical and the line joining the antennas) will add constructively at the amplifier? The speed of light is 3.000×108 m/s.
Explanation / Answer
Given that the seperation is d=218.0m the angle =16 deg then the path difference for constructive interfrence is S= dsin =218 x sin 16 = 60.08m the frequency is f=65 x 106 Hz the speed of light is C=3 x 108 m/s hence the wave length =C/f = 4.615 m hence the phase difference is =2/ x path difference =2/4.615 *60.08 = 81.819 degree the angle =16 deg then the path difference for constructive interfrence is S= dsin =218 x sin 16 = 60.08m the frequency is f=65 x 106 Hz the speed of light is C=3 x 108 m/s hence the wave length =C/f = 4.615 m hence the phase difference is =2/ x path difference =2/4.615 *60.08 = 81.819 degree the angle =16 deg then the path difference for constructive interfrence is S= dsin =218 x sin 16 = 60.08m the frequency is f=65 x 106 Hz the speed of light is C=3 x 108 m/s hence the wave length =C/f = 4.615 m hence the phase difference is =2/ x path difference =2/4.615 *60.08 = 81.819 degree