# Tough reflection problem

1. Apr 6, 2005

### Jacob87411

Radio waves of wavelength 250 meters from a star reach a radio telescope by two seperate 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 attached a picture for the rest as a rough diagram (Use the second one)

The distance that ray 2 travels is going to be d - x, which equals 125 meters. The angle there is the 25 the ray is reflecting at, the other two angles are 90 and 65. Im having a real hard time finding H now though, ive figured this all out but cant finish it

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2. Apr 7, 2005

### SpaceTiger

Staff Emeritus
Well, I can't see your diagram for some reason, but if I'm picturing the same situation, then you have two right triangles to consider. The first is made up of the height of the cliff, the horizontal distance from the reflection point, and the hypotenuse is the length of the line connecting the reflection point and the observatory. This is a 25-65-90 triangle.

For the other triangle, I assumed two things. First, the light rays hitting the water and the observatory are parallel. If the star is at a great distance, this is a natural assumption. Second, I assumed that points of constant phase in the wave are denoted by perpendiculars to the light rays. The combination of these two assumptions gives another triangle by drawing a line from the reflection point that's perpendicular to the incoming light rays. The hypotenuse is the same as for the other triangle, but the other two sides are given by the perpendicular line connecting the two light rays and the extra distance travelled by the light ray hitting the observatory. This is a 40-50-90 triangle.

With these two triangles, H ought to come from simple trigonometry, given that the path length difference is the hypotenuse minus the shortest side of the second triangle.

3. Apr 7, 2005

### GCT

-Well you'll need to find the equation for the magnitute of the difference in pathlengths

-also considering the fact that their will be a 180 degree phase change when the ray reflects off the water.

4. Apr 8, 2005

### Jacob87411

May I just ask how you got the triangle to be a 40-50-90?

5. Apr 8, 2005

### SpaceTiger

Staff Emeritus
Alright, I can see your figure now, so I can say what I did differently. In the diagram you drew, it looks as if the light rays are coming from a point very nearby (a few times the distance from the reflection point to the observatory). The most important thing to recognize is that the rays coming in from the left should be parallel. This is a natural consequence of the fact that the star is so far away. The 40-50-90 triangle comes when you draw a line perpendicular to both of these parallel rays (instead of your leftmost vertical line). The rays will have constant phase along this line and you'll need it to find the path length difference.

6. Apr 8, 2005

### Jacob87411

Alright thanks..i got the right answer i was just curious where the 40-50-90 came from

7. Nov 29, 2009

### a.a

I'm having trouble with this one, I don't understand how to find the value for the hyp.

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