I recently came across a question where a wave huts a wall reflects and it's intensity becomes 0.64 times the original. Intensity depends on frequency and amplitude. Now after reflection could the frequency change? (The solution to the question changed amplitude which) If not why can't frequency change of rather what does frequency depends on
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$\begingroup$ When a wave interacts with a wall, there are three possible effects: reflection, absorption and transmission. A perfect rigid wall will perfectly reflect the wave, since transmission and absorption depends on the wall movement. If you lose intensity in reflection, by energy conservation it means some of the other two effects are happening. The reflected wave away has the same frequency. $\endgroup$– RuffoloCommented Oct 15 at 14:14
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Intensity depends on frequency and amplitude.
This is not a great way to think about it.
Intensity is directly related to amplitude.
But the relation between intensity and frequency is not direct.
The photon energy depends on the frequency. But the intensity depends on both the photon energy and the photon flux (number of photons passing per unit time and over a unit of surface area). So regardless of the frequency we can increase or decrease the intensity arbitrarily by increasing or decreasing the photon flux.
If you have an imperfect reflection then not all of the incoming photons are reflected. Some may be absorbed or transmitted. Therefore the reflected wave has lower photon flux and thus lower intensity.
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$\begingroup$ Thank you, but I still haven't understood why frequency can't change? $\endgroup$ Commented Oct 18 at 14:22
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$\begingroup$ @Mainframe, frequency can't change because of the boundary condition that the E fields of the incident and reflected waves must have a fixed relationship at the plane of reflection. Or because in a linear system all responses will have the same frequency as the source. $\endgroup$ Commented Oct 18 at 14:50