Asked by Ramesh

A receiving antenna becomes misaligned by 30° to its original position. The power of the received signal in this new position is 12µW.
Wavelength = 0.016 m
I calculated the power received in the original position as 1.04 x 10^-5 W

What I need to find out next is the minimum time between the wave leaving the transmitting antenna and its reception.

Thanks
Answered by Ramesh
Never mind, I figured out how to do it and got 1.2 x 10^-18 s
Answered by Anonymous
can you post the process?
Answered by Ramesh
I had this question bookmarked and happened to come back.

c = fλ
3 x 10^8 = f(0.016)
f = 18,750,000,000 Hz

E = hf
E = (6.63 x 10^-34)(18,750,000,000)
E = 1.24 x 10^-23 J

P = E/t
1.04 x 10^-5 = (1.24 x 10^-23)/t
t = 1.2 x 10^-18 s
Answered by Ramesh
Apparently my teacher was wrong?? I finally looked at the markscheme while studying for the mocks and it's TOTALLY different. And a lot simpler.

Use:
I = Initial I(cosθ)^2
Arrange that into:
Initial I = I/(cosθ)^2

(12 µ)/(cos30)^2
1.6 x 10 ^-5 W
Answered by Ramesh
The time should be 1.9 x 10^-4 s using that new power.
Answered by Bob
How did you get the time using that power though?
Answered by Anonymous
where is that equation coming from I've never seen it: I = Initial I(cosθ)^2
also assuming its true putting it into a calculator with those values does not give you 1.6 x 10 ^-5 W i am so confused
Answered by Someone
The time taken is speed of light/ distance. In the figure the distance is given to be 56 km (56 x 10^3 m) and the speed of light is 3 x 10^8 ms^-1, so the time taken is ≈ 1.9 x 10^-4 seconds.
Answered by Someone
you can see that it is a one mark question do don't over think it.
Answered by Jason
It's actually a lot simpler, the answer is very simple, you are given the distance between antennas to be 56km ie 56000m, since v is that of light and equal to 3 times 10^8 you simply do 56000/3*10^8= 1.9 * 10^-4
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