Are the authors saying that the observer effect plays no role in Bohr's thought experiment of the Heisenberg uncertainty principle?

Question

Here is an excerpt from Eisberg & Resnick's Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles. Here is introducing Bohr's though experiment to establish a physical origin for the Hesisenberg Uncertainty Principle.

"Let us say that we wish to measure as accurately as possible the position of a particle, like an electron. For greatest precision we use a microscope to view the electron, as in Figure 3-6. To see the electron we must illuminate it, for it is actually the light photon scattered by the electron that the observer sees. At this stage, even before any calculations are made, we can see the uncertainty principle emerge. The very act of observing the electron disturbs it. The moment we illuminate the electron, it recoils because of the Compton effect, in a way that we shall soon find cannot be completely determined. If we don't illuminate the electron, however, we don't see (detect) it. Hence the uncertainty principle refers to the measuring process itself, and it expresses the fact that there is always an undetermined interaction between observer and observed; there is nothing we can do to avoid the interaction or to allow for it ahead of time. In the case at hand we can try to reduce the disturbance to the electron as much as possible by using a very weak source of light. The very weakest we can get is to assume that we can see the electron if only one scattered photon enters the objective lens of the microscope. The magnitude of the momentum of the photon is $p = \dfrac{h}{\lambda}$. But the photon may have been scattered anywhere within the angular range $2\theta'$ subtended by the objective lens at the electron. This is why the interaction cannot be allowed for. Hence we find the x component of the momentum can vary from $p\sin\theta'$ to $-p\sin\theta'$ and is uncertain after the scattering by an amount $\Delta p=2p\sin\theta'$."

Figure 3-6 is what's shown below.

In the last few sentences, what are the authors referring to when they state "interaction cannot be allowed for"? Are they saying that the observer effect plays no role in Bohr's thought experiment? Also, how do they find that $\Delta p=2p\sin\theta$?

Answer 1

As for the second question, it's fairly simple. For a vertically emitted electron, the horizontal (x-axis) momentum is zero. But, as stated, the horizontal momentum of an electron detected at the edge of the cone is $$p_x = \frac{H}{\lambda}\sin{\theta} = p\sin{\theta} $$ but keep in mind that is simply the maximum possible, since the electron can be anywhere within the cone. Since the total angle of the cone is $2\theta$ the total uncertainty of the momentum $\Delta p$ is the momentum at one edge minus the momentum at the other, or $$\Delta p = p\sin{\theta} - (p\sin{(-\theta})) = 2p\sin{\theta} $$ since $$\sin{(-\theta )} = -\sin{\theta}$$

Answer 2

I can only answer your first question due to time constraints:

To me the phrase "interaction cannot be allowed for" is a roundabout way of saying they cannot determine the photon's exact momentum for use in the related equations.

Answer 3

I think you are probably misinterpreting the context here.

If you read the previous line carefully it says "there is always an undetermined interaction between observer and observed; there is nothing we can do to avoid the interaction or to allow for it ahead of time.

And later he just says due to the fact that photon can be scattered within the 2θ' angle you cannot allow for or compensate in advance for the interaction of photon with electron and measure electrons position exactly.