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Detecting a stars microwave radiation intensity

  1. Dec 12, 2008 #1
    1. The problem statement, all variables and given/known data
    A microwave detector is located 0.5m above the surface of a large lake far from the shore. As a star, emitting monochromatic microwave radiation of 21cm wavelength, rises slowly above the horizon, the detector indicates successive maxima and minima in the signal intensity. At what angle above the horizon is the star when the first maximum is received?


    2. Relevant equations

    Wave intensity, for a monochromatic plane wave;

    [tex]\frac{1}{2}c\epsilon_{0}E_{0}^{2}[/tex]

    3. The attempt at a solution

    I am completely lost on this, I don't even know where to begin. If anyone could give me some clues or pointers I would be very grateful.

    Thanks.
     
  2. jcsd
  3. Dec 12, 2008 #2

    Dick

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    The maximum must come from interference, the microwave signal direct from the star must interfere with a signal bounced off the lake. Find the path length difference between the direct path and the bounce path as a function of angle above the horizon.
     
  4. Dec 13, 2008 #3
    Hi there,

    Thankyou for your reply. Do I want the path difference between the source wave and the reflected wave to be an integer multiple of the wavelength? I suppose I want the extra distance travelled by the reflected wave to be some multiple number of wavelengths?

    I've attatched a diagram of my interpretation. The angkes in the diagram are the angle of incidence and the angle of reflection, which are always the same, could I also say the angle of reflection/incidence is also the angle above the horizon (of the star).

    Am I on the right track? Also are there any equations that would be useful (if you give me their names I can do the looking for them)? The only equations I can find relate to twin slit interference where the source separation is small and the point is far from the source (our point is only far away from one source, if I'm treating the reflected wave like a second source).
     

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  5. Dec 13, 2008 #4

    Dick

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    How much of a wavelength do you need to shift a wave to the negative of the original? That the condition for destructive interference. No fancy formulas needed here. You just need to work out the geometry to get the difference in path lengths as a function of angle.
     
  6. Dec 13, 2008 #5

    Dick

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    Your picture is ok, but misleading. A 'star' is almost infinitely far away. Draw both incoming paths to be parallel.
     
  7. Dec 14, 2008 #6
    You need half a wavelength in path difference to get to the negative of the original. I thought they were looking for the maximum intensity and hence you would want constructive interference?

    Okay. So the waves coming in are parallel (I can't make the waves parallel in the diagram, without them heading to the same place). What I need to find is the minimum angle that makes the extra distance travelled by the reflected wave a half integer multiple of the wavelength?

    To find the angle of reflection (theta):

    [tex]\sin{\theta}=\frac{0.5}{h}[/tex]
    h is the extra distance the wave has to travel,

    [tex]h=\frac{m*\lambda}{2}[/tex]
    m is some integer

    So,
    [tex]\sin{\theta}=\frac{2*0.5}{m*\lambda}[/tex]

    Is this right? I have two unkowns in the above equation, how would I eleminate one? Or can I just vary m and see when theta goes to a minimum?

    Then once I find theta I can say that the angle of reflection=angle of incidence= angle above the horizon?
     
  8. Dec 14, 2008 #7

    Dick

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    Ok if they want a maximum you need a full wavelength. But I don't agree that sin(theta)=0.5/h is the path difference. It's just the distance from the reflection point to the receiver. When the wave in the reflected path meets the water the wave in the direct path still hasn't reached the receiver. You want the difference of those two distances.
     
  9. Dec 14, 2008 #8
    Right,

    If I let r1 be the distance left for the unreflected wave to travel and r2 be the distance left for the reflected to travel then we have,

    path difference=r2-r1

    I want r2-r1= [tex]m\lambda[/tex] where m is some integer. Is this correct?

    How would go about finding r1 and r2?

    Sorry if I'm being dumb, and thanks again for all the help your giving me.
     
  10. Dec 14, 2008 #9

    Dick

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    You have to use trig to find the difference in those distances. Start by drawing a line from the reflection point perpendicular to the incoming path. Another picture might help.
     
  11. Dec 14, 2008 #10
    I can find R2,

    [tex]R_{2}=\frac{0.5}{\sin{\theta}}[/tex]

    But how do I find R1? The way I found left me with something not very useful. I first drew a horizontal line from the detector to the line joining r1 and r2. I then called the sides of the triangle this formed with r1 x and y.

    [tex]x=R_{2}\cos{\theta}[/tex]
    [tex]y=d-0.5[/tex]

    Then,
    [tex]R_{1}^{2}=x^{2}+y^{2}[/tex]
    [tex]R_{1}=\sqrt{(R_{2}^{2}\cos^{2}{\theta}+(d-0.5)^{2}}[/tex]
    [tex]R_{1}=\sqrt{0.25\cot^{2}{\theta}+(d-0.5)^{2}}[/tex]

    Which leaves me with an awkward expresion for R2-R1, which I have no idea how to solve,

    [tex]R_{2}-R_{1}=\frac{0.5}{\sin{\theta}}-\sqrt{0.25\cot^{2}{\theta}+(d-0.5)^{2}}[/tex]

    I'm geussing I have made a fatal error somewhere?
     

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  12. Dec 14, 2008 #11

    Dick

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    I've been waiting to see your jpg for a long time. It's still not passed screening. So I don't know what R1, R2, etc are. I hope they are in the picture. It's really not that hard to solve.
     
  13. Dec 15, 2008 #12

    Dick

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    You didn't follow my suggestion of showing the incoming rays as parallel, did you? It makes life easier, believe me. And what makes you think the incoming rays are in phase at the vertical line? I said to pick a line perpendicular to the incoming ray, not perpendicular to the water.
     
  14. Dec 16, 2008 #13
    I completely misinterpreted what you said, I'm sorry. And i kept thinking two lines can't be parallel and separated in space when they come from the same point... As for them being in phase, I don't know what I was thinking. I'm halfway through moving at the moment and the mayhem is affecting me I think.

    Drawing the lines parallell and taking a line orthogonal to the wave hitting the water means that when that line crosses the other wave the two waves will be in phase. Correct? This simplifies the analysis somewhat.

    Can I also treat the detector as though it were a point? Or at least that the two waves hit the same point.

    Now the distance travelled by the wave from the water to the detector is R1, the distance travelled by the unreflected wave from the point where our orthogonal line cuts it is R2, theta is the angle of reflection/incidence and beta is the angle the reflected and unreflected waves make with each other when they meet.

    [tex]R_{2}=\frac{0.5}{\sin{\theta}}[/tex]

    [tex]R_{1}=R_{2}\cos{\beta}[/tex]

    [tex]\beta=2\theta[/tex]

    [tex]R_{2}-R_{1}=\frac{0.5}{\sin{\theta}}-\frac{0.5\cos{2\theta}}{\sin{\theta}}[/tex]

    [tex]R_{2}-R_{1}=\frac{0.5}{\sin{\theta}}(1-\cos{2\theta)[/tex]

    [tex]\frac{0.5}{\sin{\theta}}(1-\cos{2\theta)=\frac{0.5}{\sin{\theta}}(1-(1-2\sin^{2}\theta)[/tex]

    [tex]\frac{0.5}{\sin{\theta}}(1-(1-2\sin^{2}\theta)=\sin\theta[/tex]

    [tex]R_{2}-R_{1}=m\lambda[/tex]

    So,

    [tex]m\lambda=\sin\theta[/tex]

    Am I on the right track now?
     

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  15. Dec 16, 2008 #14

    Dick

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    Yes, you are on the right track now. You might want to leave a hint of the units in. m*lambda=(1*meter)*sin(theta). Trying to solve sin(theta)=21cm can cause confusion.
     
  16. Dec 16, 2008 #15
    Yeah, I was thinking that, you should never have units inside a trig function.

    Wouldn't the smallest angle that satisfies that equation be 0, corresponding to m=0? Which means the first maximum occurs when the star has just risen? Or can m not be zero?
     
  17. Dec 16, 2008 #16

    Dick

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    You can have other values of m. As long as you can solve the equation. I'm just going to take a wild guess that they mean m=1. First max AFTER the star has risen.
     
  18. Dec 16, 2008 #17
    So theta, in radians, is approximately 0.21? I see what you mean-this isn't a difficult problem to solve, so long as you draw the right diagram. That seems crucial in physics problems, if you have a bad diagram you'll have a bad answer (always seems to be the case for me).

    Thankyou for all your help, and

    Merry Christmas!
     
  19. Dec 16, 2008 #18

    Dick

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    Merry XMas to you. Yeah, I get sin(theta)=0.21. So theta~0.21. That's not a guarantee it's right, of course.
     
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