Intensity of RF wave at a distance from source can be calculated by using inverse square law. But imagine an RF energy travels a distance $d1$ before being perfectly reflected from a metal surface. How much is the energy intensity at a point B, which is at a distance $d2$ from the point of reflection?

1) B lies in the line of reflection.

2) B is not located at the line of reflection. It forms an angle $\theta$ with the line of reflection.

My though process is following. Intensity must be much more along line of reflection. Probably , it should follow inverse square law along line of reflection, but i am not sure.

  • $\begingroup$ I'm curios to see your calculations, could please provide them ? $\endgroup$ Aug 31 '18 at 16:06
  • $\begingroup$ If the Point B is between the source and the mirror, are you assuming that the detector casts a shadow on the mirror? $\endgroup$ Aug 31 '18 at 16:58
  • $\begingroup$ Do you have any background in basic optics? because this is just putting a mirror in the system, thus folding the propogation path. Nothing else changes. $\endgroup$ Aug 31 '18 at 18:32
  • $\begingroup$ ^In some cases, adding a mirror at the correct location and angle can redirect 'overspill' back to the detector, and increase the intensity of radiation at that point. $\endgroup$ Aug 31 '18 at 19:46

Take a look at the image below: The Yellow side shows a point source, bouncing off a mirror. You can clearly see that the radiation intensity starts bright at the source, and then falls off according to the inverse square law. It's also visibly brighter in the area where the reflected radiation overlaps the incident radiation. At that point, there's more overall light, because it's coming from two directions. Of course, putting a detector in there can create shadows that messes with this - I'm assuming no shadows for now.

The orange half of the image shows what's know as 'extended rays', or, the point from which the reflected light rays appear to originate from, to an observer on the yellow side of the mirror. You can see here how the rays again start bright, and then fall off according to inverse square. To calculate the intensity at any point, then, you simply calculate the distance to the detector from the yellow and orange sources, work out intensity using inverse square, and subtract any shadows as necessary.

Point Source at Mirror

  • $\begingroup$ That's kind of a long-winded, and confusing, way of saying "The plate just folds the path." The "brighter area" is not brighter, because if you put a detector there, then the putative reflected rays won't exist anywhere the source rays were intercepted in the first place. $\endgroup$ Aug 31 '18 at 18:33
  • $\begingroup$ Wait a moment, I may be mistaken, but hear me out... If you place an object in front of a point source, it will have a fixed amount of intensity on it, a shadow ahead of it, and light spilling out on both sides of it. If you add a mirror on one of those sides, and redirect the 'lost' light back at the detector, then there will be more incident radiation at the detector, even accounting for shadows. $\endgroup$ Aug 31 '18 at 19:13
  • $\begingroup$ thanks for great answer. I am a bit confused about calculation part. if detector is at distance d2 from apparent source and d1 from real source, then will the intensity at detector proportional to 1/(d1+d2)^2 ? Also, why distance from real source ie d1 should come into picture? $\endgroup$
    – edsxcd
    Sep 1 '18 at 6:49
  • $\begingroup$ More information regarding the shape and size of your detector (and the shadows it casts) is required to be able to calculate anything meaningful. Does it only detect in one direction? In the case of the mirror reflecting overspill, the intensity would be proportional to 1/d1 + 1/d2. If the source was outside the cone from the yellow source to the mirror, and only receiving reflected light, then it would just be 1/d2, and if the cone was inside that cone and shading the mirror, so only direct light, it’d be 1/d1... $\endgroup$ Sep 1 '18 at 9:34
  • $\begingroup$ Thanks, I am assuming reeceiver getting only reflected energy, which means intensity it receives is W/4*pi*d2^2, where W is the wattage of actual source. $\endgroup$
    – edsxcd
    Sep 1 '18 at 14:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.