Question

An AM radio station has a frequency of 1000 kHz; it uses two identical antennas at the

same elevation, 150 m apart.

a) In what direction is the intensity maximum, considering points in a horizontal plane?

b) Calling the intensity in (a) I₀, determine in terms of I₀ the intensity in directions making

angles of 30^o, 45^o, 60^o and 90^o to the direction of maximum intensity.

Answer 1

given that :

frequency of AM radio station, f = 1000 kHz

distance between two identical antennas at same elevation, d = 150 m

(a) the maximum intensity that has considering points in a horizontal plane at different angles which are given as :

using a formula,

= 45⁰ +                   { eq. 1 }

where, = 30⁰ , 45⁰ , 60⁰ & 90⁰

inserting the value of '' in above eq. & get a different phase angle,

₁ = 75⁰  

₂ = 90⁰

₃ = 105⁰

₄ = 135⁰

(b) Intensity is given as :

I₁ = I₀ / 2                         { eq.2 }

I₂ = I₁ Cos² (30⁰)

I₂ = 0.75 I₀ / 2                                  (from eq.2)

I₂ = 0.375 I₀

I₃ = I₂ Cos² (45⁰)

I₃ = 0.5 I₂

I₃ = (0.5) (0.375 I₀) = 0.187 I₀

I₄ = I₃ Cos² (60⁰)

I₄ = 0.25 I₃

I₄ = (0.25) (0.187 I₀) = 0.046 I₀

I₅ = I₄ Cos² (90⁰)

I₅ = 0