Answer:
a. 200 m
b. 100 m
Step-by-step explanation:
Solution:-
- We will first draw three points marked A,B and Q from left most to right most.
- We are told that the antennas at A and B radiate in phase. This means the radio-waves emitted by each antenna are synchronous in terms of ( frequency and wavelength ).
- We will denote the common wavelength of coherent sources of radio-waves ( A and B ) with λ.
- The relation between the wavelength ( λ ) and the path difference between the source and observation point ( Q ) for the case of destructive interference is:
AQ - BQ = n*λ/2
Where,
n: The order of wavelength
AQ: The distance between antenna A and point Q
BQ: The distance between antenna B and point Q
- The point Q is positioned ( 100 + 50 ) m away from antenna A and 50 m from antenna B. Hence,
150 - 50 = n*λ/2
- To determine the longest wavelength ( λ ) to meet destructively at point Q with the given path difference. The order of wavelength ( n ) must be minimum ( 1 ). Therefore,
100 = λ/2
λ = 200 m .... Answer
- The relation between the wavelength ( λ ) and the path difference between the source and observation point ( Q ) for the case of constructive interference is:
AQ - BQ = n*λ
Where,
n: The order of wavelength
AQ: The distance between antenna A and point Q
BQ: The distance between antenna B and point Q
- The point Q is positioned ( 100 + 50 ) m away from antenna A and 50 m from antenna B. Hence,
150 - 50 = n*λ
- To determine the longest wavelength ( λ ) to meet constructively at point Q with the given path difference. The order of wavelength ( n ) must be minimum ( 1 ). Therefore,
100 = λ
λ = 100 m .... Answer