How High is the Cliff if Radio Waves Interfere at 25 Degrees?

[a.a]

Homework Statement

Radio waves of wavelength 250 meters from a star reach a radio telescope by two separate paths. One is a direct path to the receiver which is situated on the edge of a cliff by the ocean. The second is by reflection off the water. The first minimum of destructive interference occurs when the star is 25 degrees above the horizon. Find the height of the cliff.

We know the path difference is 125 meters since first minimum occurs at 25 degrees, so the difference is half a wavelength.

I just found phase difference (=half wavelength) and found path difference (from: phase difference = 2pi*phase difference/wavelength)
Then I used height = phase difference*sin theta
but that didnt work
any advice

 

[Doc Al]

Draw yourself a diagram showing the two rays (one direct; one reflected) and use it to find the difference in path length between them.

 

[a.a]

I did.. I got 5504.29 as my path diff,
PD= (lambda)^2/4pi

thats from phase difference = half lambda = 2pi PD/lambda
Where am i going wrong?
After I found PD.. from my diagram i found that H= PD sin 22/sin46

 

[Doc Al]

I did.. I got 5504.29 as my path diff,
PD= (lambda)^2/4pi

thats from phase difference = half lambda = 2pi PD/lambda
Where am i going wrong?

I don't quite understand what you're doing here. As you said in your first post, the path difference must be λ/2.

After I found PD.. from my diagram i found that H= PD sin 22/sin46

Some problems with your diagram:
(1) The light rays from the star are parallel.
(2) What you've labeled as PD Δx is not the path difference.
(3) Your diagram shows the angle as 22 degrees, not 25.

 

[a.a]

Sorry, my numbers are different, the angle from horizontal is 22 and my wavelength is 263 m.. so the question would read:
Radio waves of wavelength 263 meters from a star reach a radio telescope by two separate paths. One is a direct path to the receiver which is situated on the edge of a cliff by the ocean. The second is by reflection off the water. The first minimum of destructive interference occurs when the star is 23 degrees above the horizon. Find the height of the cliff.

Okay.. does this diagram look better?
Phase difference is half lambda.. path difference is something else, isn't it?

 

[a.a]

phase difference=(2pi* path difference)/labmda

 

[Doc Al]

Sorry, my numbers are different, the angle from horizontal is 22 and my wavelength is 263 m.. so the question would read:
Radio waves of wavelength 263 meters from a star reach a radio telescope by two separate paths. One is a direct path to the receiver which is situated on the edge of a cliff by the ocean. The second is by reflection off the water. The first minimum of destructive interference occurs when the star is 23 degrees above the horizon. Find the height of the cliff.

OK.

Okay.. does this diagram look better?

Not really. I want to see two parallel rays coming from that star. (The star is very far away.) Those rays make an angle of 22 degrees with the horizontal.

Phase difference is half lambda.. path difference is something else, isn't it?

Imagine that the two rays start out with the same phase as they leave the star. The difference in their path lengths must equal λ/2.

 

[a.a]

is this better?

 

[a.a]

where would i go from here?

 

[Doc Al]

is this better?

Much better! But what you've labeled as λ/2 is not right. You need to compare that distance with the distance that the other ray travels (the hypotenuse of the triangle). The difference will be the path difference.

 

[a.a]

So, in this diagram half lamda = H-X

so how do we go about finding H or X?
nothing is given...:S
Probably a lot of trigs, do the angles in my second diagram make sence?
If so then I get:
sin 22= hyp (h)/(half lambda +X)
and I don't really know where to go from there..

 

[Doc Al]

So, in this diagram half lamda = H-X

OK.

so how do we go about finding H or X?
nothing is given...:S

H and X are part of a right triangle, so they are related. (Find the angle of that triangle.)

 

[a.a]

related by pythagorean? How does that help us?

 

[Doc Al]

related by pythagorean? How does that help us?

Even easier is to use some trig to relate H and X. Add to that the equation H - X = λ/2. Then you can solve for H. Once you have that, use some more trig to find the height of the cliff.

 

[a.a]

sorry about the late reply,
I only got as far as :
sin 46 = X/H = (H-half lambda)/H
Solved for H and got 468.54
Then: sin 22 = h/H
and got h= 175.5
Is that right?

 

[a.a]

nope...I got it wrong...?

 

[Doc Al]

sorry about the late reply,
I only got as far as :
sin 46 = X/H = (H-half lambda)/H
Solved for H and got 468.54
Then: sin 22 = h/H
and got h= 175.5
Is that right?

Looks right to me.