Question

An astronomical radio telescope consists of two antennas separated by a distance of 206 m. Both antennas are tuned to the frequency of 24 MHz. The signals from each antenna are fed into a common amplifier, but one signal first passes through a phase selector 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 θ = 9° with the vertical (in the plane formed by the vertical and the line joining the antennas) will add constructively at the amplifier? Hint: Radio waves travel at 3.00  10⁸ m/s.

An astronomical radio telescope consists of two antennas separated by a distance of 206 m. Both antennas are tuned to the frequency of 24 MHz. The signals from each antenna are fed into a common amplifier, but one signal first passes through a phase selector 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 0 = 90 with the vertical (in the plane formed by the vertical and the line joining the antennas) will add constructively at the amplifier? Hint: Radio waves travel at 3.00 x 108 m/s. 16.19 X rad Wavefronts Phase selector Amplifier Calculate the required phase shift from the path difference and the wavelength of the radio waves. eBook

Please help, I only have two attempts left on this question. Thank you!

Answer 1

Remember: Phase shift is given by:

S S = 27-

Where

$\lambda$ = wavelength of radio wave = c/f

f = frequency of waves = 24 MHZ = 24*10^6 Hz

c = speed of radio waves = 3.00*10^8 m/s

$\lambda$ = 3.00*10^8/(24*10^6) = 12.5 m

= Path difference = ?

S

from given figure we can see that:

sin $\theta$ = /d

S

S
= d*sin $\theta$ = 206*sin 9.0 deg = 32.2255 m

S
= 32.2255 m*(1 $\lambda$ /12.5 m) = 2.578*$\lambda$

S
= 2$\lambda$ + 0.578$\lambda$

So, here actual path difference is only 0.578$\lambda$

So, phase shift will be:

S S = 27-

$\delta$ = 2*pi*0.578*$\lambda$/$\lambda$

$\delta$ = 2*pi*0.578

$\delta$ = 3.63 rad

Let me know if you've any query.