# Radio Antennas and Young's Double slit

• STJ
In summary, the car's radio receives the signals from two antennas separated by d = 294 m. If the car is at the position of the second maximum after that at point O when it has traveled a distance y = 400 m northward, what is the wavelength of the signals? The car must travel 399.8 m further from this position to encounter the next minimum in reception.
STJ

## Homework Statement

Two radio antennas separated by d = 294 m as shown in the figure below simultaneously broadcast identical signals at the same wavelength. A car travels due north along a straight line at position x = 1 290 m from the center point between the antennas, and its radio receives the signals. Note: Do not use the small-angle approximation in this problem.

(a) If the car is at the position of the second maximum after that at point O when it has traveled a distance y = 400 m northward, what is the wavelength of the signals? ANS: 43m (Correct)

(b) How much farther must the car travel from this position to encounter the next minimum in reception? ANS: 399.8m (Incorrect)

## Homework Equations

I just used trig, this does resemble double split though but I didn't see how any double split formulas would help when trig should be enough.

## The Attempt at a Solution

I determined the wavelength(w) to be 43m, which is correct. I started by knowing that for the car to reach the next minimum it must travel some distance x past the 400m mark. S1 = ((400 + x + 147)^2 + 1290^2)^.5 m and S2 = ((400 + x - 147)^2 + 1290^2)^.5 m. The difference between the top radio and bottom radio (DS) = 235200+588x. This must equal 2.5w as the next minimum is 1 half wavelength away from the 2nd maximum. Plugging in w and solving for x I get x = -399.8. Through common sense the next minimum can't be backwards, it has to be forwards so I make it positive instead. I'm being told I am wrong though. Some hints perhaps?

Seems like you forgot about the ^.5 outside the parenthesis.

Say double slit, not double split three times in a row.

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It seems I did forget about that. After looking at this for a minute, how is it possible to solve

(1963309 + 1094x + x^2)^.5 - (1728109 + 506 + x^2)^.5 = 2.5w

for x? I don't see any way of solving that for x, so I guess I have to use some formulas or something?

EDIT: would Ydark = Ltan(theta dark) be the proper formula for this? If so is theta equal to the angular difference between the "lines" to the car from the antennas? I'm not sure where to go from here.

Square the whole equation, isolate the one remaining term that has a square root in it, square the equation again, you should find a quadratic for x. Solve it.

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There has got to be a faster way to do that problem. Doing this during a 1 hour exam would take most of my time. There is no simpler way to solve this problem?

EDIT: remember the admonition not to use small-angle approximations.

Thus, the rays from both transmitters at the 1st minimum are not parallel and you have to deal with a scalene triangle. So the difference between the two beams must be 1/2 wavelength. The rest is trig. Draw the picture and find the required distance form the starting point (midpoint between the stations projected perpendicularly to the road.)

I'm a bit surprised you got the right answer for part (a) and not for part (b).

Would help if you posted the diagram.

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STJ said:
There has got to be a faster way to do that problem. Doing this during a 1 hour exam would take most of my time. There is no simpler way to solve this problem?

It's a bit of work, but not way too bad. Solve the problem algebraically and only plug in the data at the very last step. Using vector notation simplifies the algebra somewhat

rude man said:
EDIT: remember the admonition not to use small-angle approximations.

Thus, the rays from both transmitters at the 1st minimum are not parallel and you have to deal with a scalene triangle. So the difference between the two beams must be 1/2 wavelength. The rest is trig. Draw the picture and find the required distance form the starting point (midpoint between the stations projected perpendicularly to the road.)

I'm a bit surprised you got the right answer for part (a) and not for part (b).

Would help if you posted the diagram.

You mean the difference between the two beams must be 3/2 wavelength (3rd minimum). Part (b) is much more challenging than part (a).

dauto said:
You mean the difference between the two beams must be 3/2 wavelength (3rd minimum). Part (b) is much more challenging than part (a).

Right. Sorry. But it's the second minimum, not the 3rd: 1st max (center), 1st min, 2nd max, 2nd min. At least that's my reading.

No matter: you have a scalene triangle with r2 - r1 = 3(lambda)/2 and you need to calculate the distance from the center road position to that point, then subtract 400m.

PS still no diagram ...

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## 1. What is the purpose of a radio antenna?

A radio antenna is a device used to transmit or receive radio waves. It converts electrical signals into radio waves for transmission and vice versa for receiving. It is an essential component of any radio communication system.

## 2. How do radio antennas work?

Radio antennas work by converting electrical signals into electromagnetic waves. When an electrical current flows through the antenna, it creates an electric field that extends into the surrounding space. This electric field then creates a magnetic field, and the combination of the two creates an electromagnetic wave that travels through the air.

## 3. What is the difference between a radio antenna and a TV antenna?

Although both are designed to receive radio waves, the main difference between a radio antenna and a TV antenna is the frequency range they are designed to receive. Radio antennas are designed to receive a wider range of frequencies, while TV antennas are designed to receive a specific range of frequencies used for television broadcasting.

## 4. What is Young's double slit experiment?

Young's double slit experiment is a classic experiment used to demonstrate the wave nature of light. It involves a light source, a barrier with two narrow slits, and a screen to observe the interference pattern. When light passes through the two slits, it diffracts and creates an interference pattern on the screen, indicating that light behaves like a wave.

## 5. How are radio antennas used in the Young's double slit experiment?

In Young's double slit experiment, the light source is usually a laser, and the two slits act as sources of electromagnetic waves. The interference pattern observed on the screen is a result of the superposition of the waves from the two slits. This experiment has been used to study the properties of light and has also been extended to study other types of waves, such as sound waves.

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