Oxford PAT 2013, Questions 14 and 15

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4 thoughts on “Oxford PAT 2013, Questions 14 and 15

  1. Hi, I thought power was directly proportional to the radius squared.
    I.e. from the Intensity equation, I = Power/Area, where the area is 4 * pi * r^2.

    So Power = Intensity multiplied by Area, and if the intensity and ‘4 pi’ stay constant, power would be directly proportional to the radius squared.

    1. The equation you write is the correct one, but you are not using it right. In the example, the radio receiver has some kind of antenna that has a fixed physical size and so covers a constant area. As you move the antenna away from the radio source, the energy from the transmitter is spread out over a larger and larger area, so the proportion of the energy that reaches the antenna is reduced accordingly. The area over which the power is spread out increases proportionally to r^2 and so the proportion of the energy that reaches the antenna varies as 1/r^2.

      Have a look at this link http://hyperphysics.phy-astr.gsu.edu/hbase/forces/isq.html, which explains the inverse square law with some nice diagrams.

    1. Hi Dan. I have applied the inverse square law, which states that the received power is inversely proportional to the square of the distance from the source.

      This question is a little bit unsatisfactory because, although the inverse square law is a genuine physical law, there are a lot of reasons why, in real life, the facts might work out differently (terrain, reflections leading to interference, atmospheric effects …). In most real life cases, attenuation and reflection by trees is a major factor, which I presume is why the question is about a desert. I think the examiners would have done better to set it in space, where environmental effects wouldn’t be a problem (there being no environment).

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