You seem to be using formulas for destructive interference from two slits.
m = 0 is the first node, not m = 1. That is the first time when the difference in the paths from each slit is 1/2 lambda.
This problem is not the same however, although the requirement that the difference in path length between the two paths must be half a wavelength for m=0 destructive interference is the idea.
That difference in length between the path straight from the sun and the path reflected off the water must be half a wavelength or 2.5 meters
The distance from the reflection point on the ground to the receiver is 25/sin theta
That is the hypotenuse of a right triangle with angle at the receiver between straight and reflected incoming rays of 2 theta
so
I draw my reflected ray as if it went straight and hit the receiver at 25 m below ground.
Then I draw a perpendicular from the receiver to that ray below ground.
It hits a distance d from the below ground receiver image. That distance must be the one we are looking for, half a wavelength or 2.5 meters
The hypotenuse is from the above ground receiver to the below ground one, 50 m
then sin theta = 2.5/50 = .05
theta = 2.87 degrees
sin theta = .1
theta =
Hey Damon, I posted this question a while back and you helped me figure out the first part of it.
Assume that radio waves have a frequency of 6.0 × 10^7 Hz, and that the radio receiver is
25 m above the surface of the sea. What is the smallest angle theta above the horizon that
will give destructive interference of the waves at the receiver?
You helped me figure out that the distance was 50.
So now to find the angle I used this equation
y' = (m+1/2) ((lambda) L/d)
and since m = 1 will give destructive intereference..am i right?
I did
25 = (1 + 1/2) (5L/50)
L = 166.67
Now to find the angle
I did arctan(25/166.67) = 8.5degrees. Is that right?
5 answers
You do not really need the above ground triangle with the 2 theta.
You seem to be using formulas for destructive interference from two slits.
m = 0 is the first node, not m = 1. That is the first time when the difference in the paths from each slit is 1/2 lambda.
This problem is not the same however, although the requirement that the difference in path length between the two paths must be half a wavelength for m=0 destructive interference is the idea.
That difference in length between the path straight from the sun and the path reflected off the water must be half a wavelength or 2.5 meters.
I draw my reflected ray as if it went straight and hit the receiver at 25 m below ground.
Then I draw a perpendicular from the receiver to that ray below ground.
It hits a distance d from the below ground receiver image. That distance must be the one we are looking for, half a wavelength or 2.5 meters
The hypotenuse is from the above ground receiver to the below ground one, 50 m
then sin theta = 2.5/50 = .05
theta = 2.87 degrees
sin theta = .1
theta =
You seem to be using formulas for destructive interference from two slits.
m = 0 is the first node, not m = 1. That is the first time when the difference in the paths from each slit is 1/2 lambda.
This problem is not the same however, although the requirement that the difference in path length between the two paths must be half a wavelength for m=0 destructive interference is the idea.
That difference in length between the path straight from the sun and the path reflected off the water must be half a wavelength or 2.5 meters.
I draw my reflected ray as if it went straight and hit the receiver at 25 m below ground.
Then I draw a perpendicular from the receiver to that ray below ground.
It hits a distance d from the below ground receiver image. That distance must be the one we are looking for, half a wavelength or 2.5 meters
The hypotenuse is from the above ground receiver to the below ground one, 50 m
then sin theta = 2.5/50 = .05
theta = 2.87 degrees
sin theta = .1
theta =
I don't know where you got sin theta = .1 from??
Look if I use m = 0
and the formula I used then I get L = 500m
when I do arctan(25/500) I get 2.86 degrees.
Look if I use m = 0
and the formula I used then I get L = 500m
when I do arctan(25/500) I get 2.86 degrees.
that sin theta = .1 is a typo
Yes, I got 2.87 degrees in the line above
Yes, I got 2.87 degrees in the line above
Where did you get L = 500 m from?
2 answers