Please answer #23 in test 2 section 4.

For this one, let’s plug in. Let’s say Observer A observes an intensity of 16, so I = 16. Observer B would then observe an intensity of 1, so his I = 1.

Both observers are observing the same ratio signal, so the value of P will be the same for both. Because it won’t change, we can plug in for that, too. Let’s say P = 100. (You can really pick anything for this, but 100 is a good choice because you’ll end up taking its square root.)

Now solve both equations for r. I’m going to use r_A and r_B to keep track of which distance is which.

    \begin{align*}16&=\dfrac{100}{4\pi {(r_A)}^2}\\64\pi{(r_A)}^2&=100\\{(r_A)}^2&=\dfrac{100}{64\pi}\\r_A&=\sqrt{\dfrac{100}{64\pi}}\\r_A&=\dfrac{10}{8\sqrt{\pi}}\\r_A&=\dfrac{5}{4\sqrt{\pi}}\end{align*}

    \begin{align*}1&=\dfrac{100}{4\pi {(r_B)}^2}\\4\pi {(r_B)}^2&=100\\{(r_B)}^2&=\dfrac{100}{4\pi}\\{r_B}&=\sqrt{\dfrac{100}{4\pi}}\\r_B&=\dfrac{10}{2\sqrt{\pi}}\\r_B&=\dfrac{5}{\sqrt{\pi}}\end{align*}

The question asks for the ratio \dfrac{r_A}{r_B}. We can calculate that!


That frankenfraction is not as gnarly as it looks. Just remember that dividing by a fraction is the same as multiplying by its reciprocal, and simplify:


Comments (2)

Because P could be anything, I plugged in 4π for P, which made my math a lot easier (no gnarly frankenfraction)… I don’t always notice ‘nice’ numbers to plug in, but it really helped in this case!

Is there a way, logically, to do this problem by seeing that if A’s I value (intensity) is 16 times that of B, that A’s r value (distance) has to be the inverse square root since (looking at the equation) the r value is in the denominator & is squared? In the past on these problems, I always think afterwards, “Oh, yah, that makes sense.” I just would like to be able to do that from the beginning without getting a brain cramp.

Also, alternatively, could using the answer to #22 help in solving this question since it involves solving for r^2?

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