00:01
Okay, so the signal will be out of face when the path difference equals an odd number of half wavelengths, so destructive interference, right? and so we can represent the distance from between s1 and s2 is d.
00:17
So d is 175 meters.
00:19
Y is what we don't know.
00:23
It's the length on the y -axis.
00:26
And then lambda, which is wavelength, is just speed of light over frequency.
00:33
So that's 3 times 10 to the 8 meters per second over 6 times 10 to the 6 hertz, 6 megahertz.
00:40
So that's 50 meters for the wavelength.
00:45
And so basically what's left now is to find the left -hand side.
00:51
So m plus 1 half lambda, so that's for destructive interference.
00:56
On the right -hand side is equal to you use the pythagorean theorem and you have square root of y squared plus d squared right because because this is y and this is d and so this is why and this is d and so with the by taking the squares of the of the two and and putting it in a square you have the hypotenuse minus y which is the vertical length and so you can pull y to the other side and square both sides you get y squared plus d squared equal equal equals m uh squared plus two y times m plus half lambda plus m plus m plus and plus half quantity squared, lambda squared.
02:02
So these ys cancel.
02:04
And so you have an expression for y in terms of everything else, which is y equals d squared minus m plus half squared, lambda squared over two lambda times m plus half.
02:22
And so you have, we can plug in values now.
02:27
D is 175 meters...