Question

Two radio antennas separated by d = 294 m as shown in the figure below simultaneously broadcast identical signals at the same wavelength. A car travels due north along a straight line at position x = 1090 m from the center point between the antennas, and its radio receives the signals. Note: Do not use the small-angle approximation in this problem.

(a) If the car is at the position of the second maximum after that at point O when it has traveled a distance y = 400 m northward, what is the wavelength of the signals? __m

(b) How much farther must the car travel from this position to encounter the next minimum in reception? __m

Answer 1

Two radio antennas separated by d = 294 m as shown in the figure below simultaneously broadcast identical signals at the same wavelength.

400 m 294 m 1090 m

The path difference of the optical paths from two radio antennas will be given as :

using a formula, we have

$\Delta$$\delta$ = d sin $\theta$

we know that, $\theta$ = tan^-1 [(400 m) / (1090 m)] = 20.1 degree

then, we get

$\Delta$$\delta$ = (294 m) sin 20.1⁰

$\Delta$$\delta$ = 101 m

(a) If the car is at position of second maximum after that at point O, then wavelength of signals will be given as :

$\Delta$$\delta$ = m$\lambda$ = 2 $\lambda$

$\lambda$ = (101 m) / 2

$\lambda$ = 50.5 m

(b) Using a condition, we have

d sin $\theta$ = (m + 1/2) $\lambda$             

where, m = fringe order = 2

sin $\theta$ = (2 + 0.5) (50.5 m) / (294 m)

$\theta$ = sin^-1 (0.4294)

$\theta$ = 25.4 degree

Therefore, y = L tan $\theta$ = (1090 m) tan 25.4⁰

y = 517.5 m

The car travel from this position to encounter the next minimum in reception at a distance which is given as :

$\Delta$y = (517.5 m) - (400 m)

$\Delta$y = 117.5 m