Question

Two radio antennas A and B radiate in phase. Antenna B is a distance of 110 m to the right of antenna A. Consider point Q along the extension of the line connecting the antennas, a horizontal distance of 30.0 m to the right of antenna B. The frequency, and hence the wavelength, of the emitted waves can be varied.

a)What is the longest wavelength for which there will be destructive interference at point ?
b)What is the longest wavelength for which there will be constructive interference at point ?

Answer 1

Answer:

a) maximum wavelength for destructive interference = 220 m

b) maximum wavelength for constructive interference = 110 m

Explanation:

The antennas are placed such that

A ------110 m------ B --30 m-- Q

For destructive interference, the difference in path length of the waves from each antenna is related to the wavelength through.

(m + ½)λ = |d₁ - d₂|

For maximum wavelength, m = 0 and

(λ/2) = |d₁ - d₂|

And for Constructive interference, the difference in path length of the waves from the two antennas is related to the wavelength through

mλ = |d₁ - d₂|

For maximum wavelength, m = 0

λ = |d₁ - d₂|

d₁ = path length from Q to A = 110+30 = 140 m

d₂ = path length from Q to B = 30 m

|d₁ - d₂| = 140 - 30 = 110 m

So, maximum wavelength for destructive interference is given as

(λ/2) = |d₁ - d₂| = 110

λ = 110 × 2 = 220 m

maximum wavelength for Constructive interference is given as

λ = |d₁ - d₂| = 110

λ = 110 m

Hope this Helps!!!

Answer 2

The wavelengths will be:

(A) 220 m

(B) 110 m

According to the question,

Horizontal distance,

• 30 m

AQ,

• 110+30 = 140

(A)

For destructive:

→   $\frac{\lambda}{2} = 110$

→ $\frac{3 \lambda}{2} =110$

or,

→ $\frac{S \lambda}{2} =110$

For $\lambda_{max}$,

→ $\frac{\lambda}{2} = 110$

  $\lambda = 220 \ m$

(B)

For constructive:

We know

• $\lambda = 110$
• $2 \lambda = 110$
• $3 \lambda = 110$

then,

$\lambda_{max} = 110 \ m$

Thus the above answer is correct.

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