This is based on question 5d from this A-Level Paper (OCR A 2023 Paper 3):
Plane-polarised microwaves are sent out from a transmitter aerial to a parallel receiver aerial. Behind the receiver, along the axis joining the two aerials, is a metal sheet capable of reflecting the microwaves. Describe what happens to the intensity of the signal recieved as the reflective sheet is moved along the axes.
The markscheme said something to this effect:
The reflected wave will have travelled $2d$ further than the wave that reaches the receiver having travelled straight from the transmitter. So there will be a phase difference of $\Delta \phi = \frac{2d}{\lambda} 2\pi + \pi$ between the two waves - the extra $\pi$ coming from the fact that, after reflection, phase is changed by $\pi\ \mathrm{rad}$. Thus, when $d = \frac{2n+1}{4}\lambda$, $n \in \Bbb{Z^+}$ constructive intefrence occurs and the signal recieved is of maximum intensity, and when $d = \frac{n}{2}\lambda$, a minima in intensity is detected.
This answer seemed reasonable, until I thought about it a bit more.
This is what I think actually happens
- When the waves arrive at the receiver aerial, they are in phase. Some waves are absorbed by the receiver immediately, and some travel past, reach the metal sheet and are reflected back to the receiver and only then are absorbed.
- During the time it takes for the 'reflected waves' to return to the receiver, it does travel a distance $2d$ and its phase does differ - by $\Delta \phi_r = \frac{2d}{\lambda} 2\pi + \pi$ - relative to its phase at the point it passed the receiver a time $\frac{2d}{c}$ ago.
- Over this $\frac{2d}{c}$ time, the waves that get absorbed straight away have moved forward by $\frac{2d}{c} \cdot c = 2d$, so their phase differs by $\Delta \phi_a = \frac{2d}{\lambda} 2\pi$ relative to the phase of those waves that got immediately absorbed by the receiver $\frac{2d}{c}$ time ago.
- So, overall, the 2 waves which reach the receiver $\frac{2d}{c}$ time after the first waves reach it differ in phase by $\Delta \phi = \Delta \phi_r - \Delta \phi_a = \pi$ - i.e. the only reason there is phase difference when they meet is because some of the waves were reflected and their phase was changed by $\pi\ \mathrm{rad}$ - the different distances they travel have nothing to do with it! The 2 waves should always arrive out of phase and minimum intensity should always be detected, regardless of the length of $d$.
I'd like to check that my answer is more reasonable than the mark scheme's - i.e. changing $d$ shouldn't affect the 'interference pattern' detected by the receiver, right?
(Edit.) As Farcher explains below, it would make sense that a standing wave is produced in this case and in that case it's obvious my above explanation is wrong: it predicts there'd always be destructive interference at the receiver, independent of $d$. But, if a standing wave is formed, we'd obviously get nodes or antinodes at the receiver, depending on $d$. So now I'm curious as to why my attempt at the explanation is wrong.
My thoughts are that the problem is in the fourth bullet point, where I say $\Delta \phi = \Delta \phi_r - \Delta \phi_a$. (Maybe because the two waves travel in opposite directions) it might not be correct to combine the phase differences in this way? Everything else seems ok to me in my explanation, so my only other thought is that, instead we hsould find phase difference between the two wave 'trains' at the receiver by finding the sum of $\Delta \phi_r$ and $\Delta \phi_a$, instead of their difference. But, this would give $\Delta \phi = \frac{4d}{\lambda} 2\pi + \pi$ which would suggest constructive interference occurs for $d = \frac{2n+1}{8}\lambda, n\in \Bbb{Z^+}$ and destructive interference for $\frac{n}{4}\lambda$, which still disagrees with the markscheme!
Any thoughts would be appreciated.
