Two radio antennas separated by d = 282 m as shown in the figure below simultane

Question

Two radio antennas separated by d = 282 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 = 1480 m from the center point between the antennas, and its radio receives the signals. Do not use the small-angle approximation in this problem. 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? How much farther must the car travel from this position to encounter the next minimum in reception?

Explanation / Answer

distance travelled by the wave from upper antenna

r1 = sqrt(1480^2+(400-141)^2) = 1502.5 m

distance travelled by the wave from lower antenna

r2 = sqrt(1480^2+(400+141)^2) = 1575.8 m

path difference = r2-r1

path difference = m*lambda

for secon order maximum m = 2

1575.8-1502.5 = 2*lambda

lambda = 36.65 m <<<<<-------answer

++++++++++++++

at some point

r1 = sqrt(1480^2+(y-141)^2)

r2 = sqrt(1480^2+(y+141)^2)

path difference = r2 - r1

path difference = (m + 1/2 ) *lambda

for m = 0

sqrt(1480^2+(y+141)^2) - sqrt(1480^2+(y-141)^2) = 36.65/2

y = 96.81 m

for m = 1

sqrt(1480^2+(y+141)^2) - sqrt(1480^2+(y-141)^2) = 3*36.65/2

y = 295.45 m

for m = 2

sqrt(1480^2+(y+141)^2) - sqrt(1480^2+(y-141)^2) = 5*36.65/2

y = 510.5 m

the car has to travel 510.5 - 400 = 110.5 farther         <<<<<<<<<------answer

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.